Programming Chaska

347. Top K Frequent Elements - LeetCode - Medium

Problem Description:

Given an integer array nums and an integer k, return the k most frequent elements. You may return the answer in any order.

Example 1:

Input: nums = [1,1,1,2,2,3], k = 2

Output: [1,2]

Example 2:

Input: nums = [1], k = 1

Output: [1]

Constraints:

1 <= nums.length <= 10^5

-10^4 <= nums[i] <= 10^4

k is in the range [1, the number of unique elements in the array].

It is guaranteed that the answer is unique.

Approach:

1. Frequency Counting using HashMap

We traverse the array and maintain a frequency map using HashMap<Integer, Integer>.
This map stores each unique element as a key and its occurrence as the value.
Time Complexity: O(n) (since we iterate over nums once).

2. Using a Max-Heap (PriorityQueue) for Sorting

We use a PriorityQueue (max-heap) to sort elements based on their frequency.
The comparator ensures that the elements with higher frequency are at the top.
Time Complexity: O(n log n) (due to sorting operations in the heap).

3. Extracting the Top K Frequent Elements

We poll the top k elements from the heap and store them in the result array.
Time Complexity: O(k log n) (polling k elements from the heap).

JAVA:

Time Complexity : 0(n log n) due to heap operations.

Space Complexity: 0(n) for storing the frequency map and heap.

class Solution {
    public int[] topKFrequent(int[] nums, int k) {
        Map < Integer, Integer > map = new HashMap < > ();
        for (int n: nums) map.put(n, map.getOrDefault(n, 0) + 1);
        Queue < Map.Entry < Integer, Integer >> pq = new PriorityQueue < > ((a, b) - > b.getValue() - a.getValue());
        pq.addAll(map.entrySet());
        int[] res = new int[k];
        while (k-- > 0) res[k] = pq.poll().getKey();
        return res;
    }
}

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Perfect number Program in C | C++ | JAVA

noreply@blogger.com (Taranjeet Singh Kalsi) — Thu, 30 May 2024 14:33:00 +0000

Start Practicing LeetCode with Programming Chaska. I have explained my approach to the problems in simple way. I have also shared the optimized code so that you can learn some new.

Perfect number Program in C | C++ | JAVA

Perfect Number: A natural number that is equal to the sum of its all factors (excluding number as its factor itself), is called a Perfect Number.

For Example: Number --> 6
factors of 6 are -> 1,2 and 3
->1 + 2 + 3 = 6.
So, 6 is a Perfect Number.
Some of the Perfect numbers are 28, 496, and 8128.

Algorithm to check whether a given number is a Perfect number or not:

Step 1: Accept a number.
Step 2: Find factors of number starting from 1 to number (Do not include the number itself as a factor). If given number is divisible by number smaller than itself, then its a factor.
Step 3: Add all the factors of a number and store it in a variable.
Step 4: If the sum of all the factors of a number is equal to the number itself then Display Perfect number else Display Not a Perfect number.
Step 5: Stop
Implementation in C:

Code 1: This is the general code in C with time complexity of O(n).


#include <stdio.h>
int main()
{
    int number, sum = 0;
    printf("Enter number here: ");
    scanf("%d", &number);
    for (int i = 1; i < number; i++)
    {
        if (number % i == 0)
            sum = sum + i;
    }
    if (sum == number)
        printf("The given number %d is a Perfect Number", number);
    else
        printf("The given number %d is not a Perfect Number", number);
    return 0;
}

Implementation in C++:

Code 2: This is the general code in C++ with time complexity of O(n).

#include <iostream>

using namespace std;

int main()

{

  int number, sum = 0;

  cout << "Enter number here: ";

  cin >> number;

  for (int i = 1; i < number; i++)

  {

    if (number % i == 0)

      sum = sum + i;
  }

  if (sum == number)

    cout << "The given number" << number << " is a Perfect Number";

  else

    cout << "The given number" << number << " is not a Perfect Number";

  return 0;
}

Implementation in JAVA:

Code 3: This is the general code in JAVA with time complexity of O(n).


import java.util.Scanner;
 
public class PerfectNumber { // class scope starts
 
    public static void main(String[] args) { //main method starts
 
        Scanner in = new Scanner(System.in);
 
        System.out.println("Enter your number");
 
        int number = in.nextInt();
 
        in.close();
 
        int sum = 0;
 
        for (int i = 1; i < number; i++) { //starting for loop from 1 to number
 
            if (number % i == 0)
 
                sum = sum + i;
 
        }
 
        if (sum == number)
 
            System.out.println("The given number " + number + " is a Perfect number"); //printing statement
 
        else
 
            System.out.println("The given number " + number + " is not a Perfect number"); //printing statement
 
    }
 
}

Start Practicing LeetCode with Programming Chaska. I have explained my approach to the problems in simple way. I have also shared the optimized code so that you can learn some new.

Optimized Code in C | C++ | JAVA:

Code 4: This is optimized code to check perfect numbers. Here we are using mathematical property for factor checking. The factors of a number starting repeating when we go beyond its square root.

Example:

Factors of 6 :

6 = 1 * 6

6 = 2 * 3

------------ (crossed square root of 6)

6 = 3 * 2

6 = 6 * 1

We can see that 23 is same as 32, so we will not go beyond square root.

Time complexity - O(sqrt(n))


    // Logic in C | C++ | JAVA

    public bool checkPerfectNumber(int num) { // change return type bool to boolean for JAVA
        if(num == 1) return false;

        int sum = 1;
        for(int i = 2; i * i <= num; i++) {
            if(num % i == 0) {
                sum += i + (num / i); // num/i to find other factor
            }
        }

        return sum == num; // if sum equals num return true else false
    }

Start Practicing LeetCode with Programming Chaska. I have explained my approach to the problems in simple way. I have also shared the optimized code so that you can learn some new.

If you have any doubts/questions related to the Program, if you want to give any suggestions, or want an explanation of the code, feel free to comment below. I hope that my code is understandable, if not then you can tell me in the comments so that I will try to make it easier to understand.

Group Anagrams - LeetCode Solution and Approach

noreply@blogger.com (Taranjeet Singh Kalsi) — Mon, 22 Apr 2024 19:12:00 +0000

49. Group Anagrams - LeetCode - Medium

Solve this problem in LeetCode

Problem Description:

Given an array of strings strs, group the anagrams together. You can return the answer in any order.

An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

Example 1:

Input: strs = ["eat","tea","tan","ate","nat","bat"]

Output: [["bat"],["nat","tan"],["ate","eat","tea"]]

Example 2:

Input: strs = [""]

Output: [[""]]

Example 3:

Input: strs = ["a"]

Output: [["a"]]

Constraints:

1 <= strs.length <= 104

0 <= strs[i].length <= 100

strs[i] consists of lowercase English letters.

Approach:

We need to place same anagrams in groups.

We can sort the elements in array and easily check for similar anagrams. But we will use another approach that is slightly more efficient.

Rather than sorting each string, we can calculate frequency of each character and store in variable say freq. We can now easily form a string of freq and hence it will be same for same frequency of characters. We can now make groups based on string frequency.

We will iterate through given string array.
We will count the character frequency of current string and store in variable say freq.
We will convert value of freq variable to String to act as key. This key will be same when a string has same number of characters.
We will add the current string in the group based on key. And keep on iterating.
After iterating, we will get the groups of anagrams in values of Map.

JAVA:

Time Complexity : 0(nk), where n is number of strings, and k is length of each string

Space Complexity: 0(n)

class Solution {

    public List <List<String >> groupAnagrams(String[] strs) {
        if (strs.length == 0 || strs == null) return new ArrayList <>();

        Map <String, List<String>> map = new HashMap <>();

        for (String str: strs) {
            char[] freq = new char[26];
            for (char ch: str.toCharArray()) freq[ch - 'a']++; // calculating frequency

            String key = String.valueOf(freq); // converting freq to string to act as key
            map.computeIfAbsent(key, k->new ArrayList<>()).add(str); // adding string to its group based on key
        }
        return new ArrayList<>(map.values());
    }
}

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Remove Duplicates from Sorted Array - LeetCode Solution and Approach

noreply@blogger.com (Taranjeet Singh Kalsi) — Mon, 15 Apr 2024 20:27:00 +0000

26. Remove Duplicates from Sorted Array - LeetCode - Easy

Solve this problem in LeetCode

Problem Description:

Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place such that each unique element appears only once. The relative order of the elements should be kept the same. Then return the number of unique elements in nums.

Consider the number of unique elements of nums to be k, to get accepted, you need to do the following things:

Change the array nums such that the first k elements of nums contain the unique elements in the order they were present in nums initially. The remaining elements of nums are not important as well as the size of nums.
Return k.

Custom Judge:

The judge will test your solution with the following code:

int[] nums = [...]; // Input array

int[] expectedNums = [...]; // The expected answer with correct length

int k = removeDuplicates(nums); // Calls your implementation

assert k == expectedNums.length;

for (int i = 0; i < k; i++) {

assert nums[i] == expectedNums[i];

}

If all assertions pass, then your solution will be accepted.

Example 1:

Input: nums = [1,1,2]

Output: 2, nums = [1,2,_]

Explanation: Your function should return k = 2, with the first two elements of nums being 1 and 2 respectively.

It does not matter what you leave beyond the returned k (hence they are underscores).

Example 2:

Input: nums = [0,0,1,1,1,2,2,3,3,4]

Output: 5, nums = [0,1,2,3,4,_,_,_,_,_]

Explanation: Your function should return k = 5, with the first five elements of nums being 0, 1, 2, 3, and 4 respectively.

It does not matter what you leave beyond the returned k (hence they are underscores).

Constraints:

1 <= nums.length <= 3 * 10^4

-100 <= nums[i] <= 100

nums is sorted in non-decreasing order.

Approach:

In this problem we have to remove the duplicates in-place. The elements are in sorted order, so the duplicates will be in sequence. Now the problem is reduced to finding next unique element, which is quiet simpler.

We will store the first unique element in a variable say prevUnique.
We will iterate through given array and check if number is not same as previous unique element stored in prevUnique. Now we have found new unique element. We will store the element at index say p which is equal to number of unique elements.
At last we will update the previous unique element with new unique and will continue iterating.
After iterating, we will get the number of unique elements stored in index p.

JAVA:

Time Complexity : 0(n)

Space Complexity: 0(1)

class Solution {
    public int removeDuplicates(int[] nums) {
        int prevUnique = nums[0], p = 1;
        for (int n : nums) {
            if (n != prevUnique) {
                nums[p++] = n;
                prevUnique = n;
            }
        }
        return p;
    }
}

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Longest Consecutive Sequence - LeetCode Solution and Approach

noreply@blogger.com (Taranjeet Singh Kalsi) — Mon, 15 Apr 2024 20:06:00 +0000

128. Longest Consecutive Sequence - LeetCode - Medium

Solve this problem in LeetCode

Problem Description:

Given an unsorted array of integers nums, return the length of the longest consecutive elements sequence.

You must write an algorithm that runs in O(n) time.

Example 1:

Input: nums = [100,4,200,1,3,2]

Output: 4

Explanation: The longest consecutive elements sequence is [1, 2, 3, 4]. Therefore its length is 4.

Example 2:

Input: nums = [0,3,7,2,5,8,4,6,0,1]

Output: 9

Constraints:

0 <= nums.length <= 10^5

-10^9 <= nums[i] <= 10^9

Approach:

In this problem we have to find length of the longest consecutive elements sequence. In simple words, we need to find the largest range of numbers. We can use Set to store numbers, as we can use contains method in approx O(1) time. Set will also remove duplicates numbers.

We will find the starting point of the range and will iterate through the end of range to find its length. Once we get the length of range, we will find the maximum length among other lengths.

If given array contains no element or only 1 element, the longest consecutive sequence will be 0 or 1 respectively. So need to check further.
We will store the array numbers in Set and will iterate through set.
During iterating, we will check if set doesn't contain previous number (n-1). It means its starting of a range.
Now we found the starting of range. We will iterate through the range and count number of elements. After iterating we will take the highest count and store in variable say maxK.
We will continue iterating set elements. After iterating we will get the result in maxK.

JAVA:

Time Complexity : 0(n)

Space Complexity: 0(n)

class Solution {
    public int longestConsecutive(int[] nums) {
        if (nums.length <= 1) return nums.length;

        int maxK = 0;
        Set<Integer> set = new HashSet<>();
        
        for (int n : nums) set.add(n);

        for (int n : set) {
            if (!set.contains(n - 1)) { // if number is starting of value of range
                int nextValue = n + 1; 
                int k = 1;
                while (set.contains(nextValue++)) k++; // counting next values
                maxK = Math.max(maxK, k); // storing maximum count
                
            }
        }

        return maxK;
    }
}

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Valid Anagram - LeetCode Solution and Approach

noreply@blogger.com (Taranjeet Singh Kalsi) — Sat, 13 Apr 2024 20:25:00 +0000

242. Valid Anagram - LeetCode - Easy

Solve this problem in LeetCode

Problem Description:

Given two strings s and t, return true if t is an anagram of s, and false otherwise.

An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

Example 1:

Input: s = "anagram", t = "nagaram"

Output: true

Example 2:

Input: s = "rat", t = "car"

Output: false

Constraints:

1 <= s.length, t.length <= 5 * 10^4

s and t consist of lowercase English letters.

Approach:

Two words are Anagram if frequency of each character is same in both words. So we can check if both words have the same character's frequency. We can use Map but for more optimization we will use array as input will only consist of lowercase English Letters (26).

We will initialize Array of 26 int. The default value of each int element is 0.
We will iterate through first word and increase value of array by 1 on character's index. And keep on iterating.
After iterating first word, we will iterate through second word and instead of increasing, we will decrease value of array by 1 on character's index because we want to balance out positive with negative(we will get 0 if frequency of characters in both words is same). And keep on iterating.
After iterating both words, we will iterate through Array of 26 int and return false if any element doesn't equals 0. After iterating if all elements are 0, then return true.

JAVA:

Time Complexity : 0(n)

Space Complexity: 0(1)

class Solution {
    public boolean isAnagram(String s, String t) {
        int[] freq=new int[26];
        for(char c:s.toCharArray())
            freq[c-'a']++;
        for(char c:t.toCharArray())
            freq[c-'a']--;
        for(int n:freq)
        if(n!=0) return false;
        return true;
    }
}

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Contains Duplicate - LeetCode Solution and Approach

noreply@blogger.com (Taranjeet Singh Kalsi) — Sat, 13 Apr 2024 19:34:00 +0000

217. Contains Duplicate - LeetCode - Easy

Solve this problem in LeetCode

Problem Description:

Given an integer array nums, return true if any value appears at least twice in the array, and return false if every element is distinct.

Example 1:

Input: nums = [1,2,3,1]

Output: true

Example 2:

Input: nums = [1,2,3,4]

Output: false

Example 3:

Input: nums = [1,1,1,3,3,4,3,2,4,2]

Output: true

Constraints:

1 <= nums.length <= 10^5

-10^9 <= nums[i] <= 10^9

Approach:

The problem becomes easy if you use Set Data Structure. Set has inbuilt methods for checking if set already contains given value or not. In JAVA we have contains method which returns true if given number is present in Set. We also have the add method to add given number to Set. The add method returns true if given number is added, and false if given number is already present in Set.

We will initialize a Set. We will iterate through given array and add current number to the Set. If Set already contains given number, the add method returns false, so we will return true. And we will keep iterating.
If we completed iterating all the numbers, means the set doesn't contains duplicates, so we will return false.

JAVA:

Time Complexity : 0(n)

Space Complexity: 0(n)

class Solution {
    public boolean containsDuplicate(int[] nums) {
        Set<Integer> set=new HashSet<>();
        for(int n:nums)
        if(!set.add(n)) return true;
        return false;
    }
}

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Longest Substring Without Repeating Characters - LeetCode Solution and Approach

noreply@blogger.com (Taranjeet Singh Kalsi) — Sat, 13 Apr 2024 15:05:00 +0000

3. Longest Substring Without Repeating Characters - LeetCode - Medium

Solve this problem in LeetCode

Problem Description:

Given a string s, find the length of the longest substring without repeating characters.

Example 1:

Input: s = "abcabcbb"

Output: 3

Explanation: The answer is "abc", with the length of 3.

Example 2:

Input: s = "bbbbb"

Output: 1

Explanation: The answer is "b", with the length of 1.

Example 3:

Input: s = "pwwkew"

Output: 3

Explanation: The answer is "wke", with the length of 3.

Notice that the answer must be a substring, "pwke" is a subsequence and not a substring.

Constraints:

0 <= s.length <= 5 * 10^4

s consists of English letters, digits, symbols and spaces.

Approach:

The problem is to find maximum length of substring without duplicate characters. So, in other words, we need to find the maximum length of sliding window without duplicate characters.

We can use Map to keep track of duplicates Characters but I will take int array to store indexes of characters.
We will iterate through the characters of string and if we found a duplicate character, we will set the starting pointer (i) of sliding window to index of duplicate character (that is stored in array) or keep i if i is at greater index.
We will update the index of the current character and store it in array.
We will calculate the length of the sliding window and store the maximum length in a variable say maxL.
We will increase the right pointer by 1 and keep iterating.
We will get the answer in maxL.

JAVA:

Time Complexity : 0(n)

Space Complexity: 0(1)

class Solution {
    public int lengthOfLongestSubstring(String s) {
        int i = 0, j = 0, l = s.length(), maxL = 0;
        int[] chars = new int[128];
        while (j < l) {
            if (chars[s.charAt(j)] != -0) // if we find duplicate character 
                i = Math.max(i, chars[s.charAt(j)]); // then set starting pointer (i) as index of duplicate character, or keep i if i is greater 
            chars[s.charAt(j)] = j + 1; // update index of current character
            maxL = Math.max(maxL, j - i + 1); // set maxL to size of sliding-window or keep maxL if maxL is greater
            j++;
        }
        return maxL;
    }
}

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Add Two Numbers - LeetCode Solution and Approach

noreply@blogger.com (Taranjeet Singh Kalsi) — Sat, 13 Apr 2024 12:23:00 +0000

2. Add Two Numbers - LeetCode - Medium

Solve this problem in LeetCode

Problem Description:

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example 1:

Input: l1 = [2,4,3], l2 = [5,6,4]

Output: [7,0,8]

Explanation: 342 + 465 = 807.

Example 2:

Input: l1 = [0], l2 = [0]

Output: [0]

Example 3:

Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]

Output: [8,9,9,9,0,0,0,1]

Constraints:

The number of nodes in each linked list is in the range [1, 100].

0 <= Node.val <= 9

It is guaranteed that the list represents a number that does not have leading zeros.

Approach:

As digits are stored in reverse order in Linked List. We will iterate through the both at same time and get the current Digits. We can now easily add the digits and store the carry if any. We will add the sum of current Digits to answer and keep iterating.

JAVA:

Time Complexity : 0(n)

Space Complexity: 0(n)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode sum = new ListNode();
        ListNode header = sum; //storing Header
        int carry = 0, t = 0;
        while (l1 != null || l2 != null || carry != 0) {
            // calculating sum of current Digits and storing in t
            if (l1 != null && l2 != null) t = l1.val + l2.val + carry;
            else if (l1 == null && l2 != null) t = l2.val + carry; // not taking l1 if null
            else if (l2 == null && l1 != null) t = l1.val + carry; // not taking l2 if null
            else if (l1 == null && l2 == null) t = carry; // not taking l1 and l2 if null

            carry = t / 10; // calculating carry from t

            if (l1 != null) l1 = l1.next; // iterating through linkedlist 1
            if (l2 != null) l2 = l2.next; // iterating through linkedlist 2

            sum.next = new ListNode(t % 10); // storing last digit of sum of current digits to LinkedList
            sum = sum.next;
        }

        return header.next;
    }
}

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Two Sum - LeetCode Solution and Approach

noreply@blogger.com (Taranjeet Singh Kalsi) — Sat, 13 Apr 2024 11:31:00 +0000

1. Two Sum - LeetCode - Easy

Solve this problem in LeetCode

Problem Description:

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

You can return the answer in any order.

Example 1:

Input: nums = [2,7,11,15], target = 9

Output: [0,1]

Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].

Example 2:

Input: nums = [3,2,4], target = 6

Output: [1,2]

Example 3:

Input: nums = [3,3], target = 6

Output: [0,1]

Constraints:

2 <= nums.length <= 10^4

-10^9 <= nums[i] <= 10^9

-10^9 <= target <= 10^9

Only one valid answer exists.

Approach:

If we try to find sum of each 2 integers in array, we will have to iterate through array in nested way and ends up getting time complexity of 0(n^2).

Instead we can store the value of target-currentValue in Map.

As, anyUpcomingValue + currentValue = Target, so anyUpcomingValue = Target - currentValue

So if we found anyUpcomingValue equals to previously stored target-currentValue, we will get the pair. But we want indexes of those pairs, so we will store the index as value in Map.

JAVA:

Time Complexity : 0(n)

Space Complexity: 0(n)

class Solution {
    public int[] twoSum(int[] nums, int target) {
        Map < Integer, Integer > map = new HashMap();
        for (int i = 0; i < nums.length; i++) {
            if (map.containsKey(nums[i])) // if such pair found
                return new int[] {map.get(nums[i]), i}; // then return pair's indexes
            map.put(target - nums[i], i); // storing key as target-currentValue and value as currentIndex 
        }
        return new int[] {-1, -1}; // return [-1,-1] if no such pair found
    }
}

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Accenture Interview Experience | Coding Questions

noreply@blogger.com (Taranjeet Singh Kalsi) — Mon, 04 Sep 2023 20:42:00 +0000

Accenture Interview Experience:

In this post, I am sharing my personal Accenture interview experience which I got through an off-campus Accenture drive for ASE in July 2023. ASE stands for Associate Software Engineer, which offers a 4.5 LPA salary in 2023. Let's take a look at the eligibility criteria, Selection Process, and my experience.

Eligibility Criteria :

All streams/branches of B.E/B.Tech/M.E/M.Tech, MCA, and M.Sc. (CSE, IT only) from 2022 & 2023 years of pass outs with full-time education only (Part-time and distance educations are not considered)

Note: MBA /PGDBM candidates are not eligible to apply for this role

No active backlogs during application and/or during onboarding.
You should have completed your relevant degree (eligible for this job role), in the stipulated duration of the degree. Hence, there should be no gaps during the course of your degree.

Eg: You must complete your B.Tech in 4 years or M.Tech/MSc in 2 years

Your highest educational qualification only will be considered
Candidate should not have appeared for the Accenture Recruitment assessment/interview process in the last three months.
Candidate should not have more than 11 months of experience.
Candidates must be eligible to work in India either by citizenship/ have relevant work permit documents.
Please note that Bhutan and Nepal nationals can work in India without the need to obtain a work visa. All other foreign nationals require a work visa or an Overseas Citizenship of India (OCI) or Person of Indian Origin (PIO) card to work in India
You are willing to work in any Business Unit/Service line within the company
You are willing to join /relocate to any Accenture office across India

Job Type: Full Time

Location: Bangalore, Hyderabad, Pune, Mumbai, Chennai, Gurgaon, Kolkata, Indore, Jaipur, Coimbatore

Selection Process :

You will get an email invitation to take part in online tests if you are qualified.
The assessment has two phases that must be completed within the allotted time.
Mock Assessment

You have to give a 20-minute mock test. It is required that you take a mock exam, which will check that your computer, software, and internet are up to par.

Assessment #1: Cognitive and Technical Assessment

You have 90 minutes to finish this assessment. There will be 90 questions. You can switch between questions easily. The level of questions was easy.

50 questions are included in the cognitive and technical assessment on:

English ability
Critical thinking and Problem-solving
Abstract Reasoning

40 questions are included in the technical assessment on the following topics:

Common Application and MS Office

Pseudocode

Fundamentals of Networking, Security, and Cloud

Results of the cognitive and technical assessments will be communicated to you within 5 to 10 minutes of completion. You are urged to keep sitting and wait for the outcome. Coding evaluation will start right away after the result notification. For consideration in subsequent stages, you must meet the section and subsection cutoff in our assessment.

After passing the cognitive and technical assessments, candidates are required to take the mandatory coding assessment. The process will come to an end for candidates who do not pass the cognitive and technical examinations or do not take part in them.

Assessment #2: Coding assessment

This test will take 45 minutes to complete 2 Coding Questions. One question was easy and another was of medium difficulty level.

You will be given two questions to answer in your choice of the following languages.

C++, Dot Net, Java, and Python.

Assessment #3: Communication assessment

The assessment takes about 30 minutes to complete.

To take part in this evaluation, you will receive a separate email containing deadlines and login information.

You will be evaluated on the following criteria as part of the communication assessment.

Sentence Mastery - You will be given jumbled sentences and you have to place them in the correct order.
Vocabulary - We have to read the sentence which is currently showing on the screen. This is to test your non-verbal skills like pronunciation.
Fluency - You will be given 2 questions like "Are Arts and Music important in education? Why?", "Should it be necessary to play sports in school? Why?", etc. You will have 30 seconds to frame your story in mind and after that, the recording starts.
Pronunciation - You will be given only 1 chance to listen to a sentence and have to repeat it exactly. Be sure to pronounce the words exactly as you have listened.

Interview

There was only 1 interview in the ASE role which lasted 20-25 minutes. It was not a technical interview (Not even a single question asked from OOPs, DBMS, DSA, etc.). It was a typical managerial round. You should prepare the managerial questions like what would you do in a particular situation. The panelist was very friendly and nice and made me comfortable like I was talking to my friend. This was my easiest interview so far.

The panelist started with my introduction followed by the project I worked on in my college. Also asked if I have worked or led a team, how do you handle disputes in a team?, Who is your mentor, and tell how he/she helped you in any challenge? , Describe a project you are proud of. There were several questions and ultimately the panelist asked if I had any questions.

My interview was on 1 August and I got the interview result after 1 month.

So that's my Accenture Interview Experience. I will be sharing more interview experiences in the future. So keep an eye out and don't forget to share this and help your friends. I love to hear from your side if this post helps you in any way.

Hexaware Interview Experience | Coding Questions

noreply@blogger.com (Taranjeet Singh Kalsi) — Mon, 17 Jul 2023 16:14:00 +0000

Hexaware Interview Experience:

In this post, I am sharing my personal Hexaware interview experience which I got through PGET Program in 2022. PGET stands for Premier Graduate Engineer Trainee, which offers 6 LPA salary in 2022. There is another GET program which stands for Graduate Engineer Trainee that offers 4 LPA. My college was eligible for PGET Program. Let's take a look at the eligibility criteria, Selection Process, and my experience.

Eligibility Criteria :

1. 60 % throughout (10th to highest Degree).

2. Branches eligible: CS, IT, EC, EE & EX

Selection Process :

Round 1- Aptitude Test:

This is the first elimination round which contains the Aptitude, Reasoning, and Verbal section. Each section contains 20 questions. So a total of 60 questions are to be solved in 60 minutes. The level of questions is more difficult than I expected, and so, I got eliminated from this round. Almost half of my friends are able to get through this round. The second round of Coding unlocks only for those who passed the first round.

This round was conducted on 19th July 2023. Later in September, I got a notice from my college that Hexaware cut down the cutoff of the Aptitude round and I am now qualified for the next round.

Round 2- Coding Test:

This is the second elimination round which contains 2 Coding questions that are to be solved within 40 minutes. The level of questions is medium, (not too tough like graph, tree). I got a problem based on set union and intersection property and another was based on framing formula to solve the question which requires an easy level of understanding of basic Mathematics.

Both of my codes passed all the test cases. I was the only one from my college to successfully clear the coding round and got forwarded to the next round of Communication. I personally experienced that the 40 minutes time limit is less, so you should keep an eye on the timer.

Round 3- Communication Round:

This is the third elimination round which contains 4 sections, to be answered in 45 minutes:

Reading- This section is to test your pronunciation. A sentence is displayed on the screen and we got a few seconds to prepare it in mind and start recording. We have to read the sentence which is currently showing on the screen. This is to test your non-verbal skills like pronunciation, pitch, tone, and volume.
Repeating- This section is to test your listening skills and short-term memory. You will be given only 1 chance to listen to a sentence and you have to start recording and repeat it exactly. Be sure to pronounce the words exactly as you have listened. The next sentences will keep getting longer, so make sure to listen properly.
Understanding- This section is to test your understanding and listening skills. You will be given only 1 chance to listen to the story question. You have to understand and remember the story and the questions will be based on the story.
Speaking- This section is to test your fluency and storytelling. You will be given questions like "Describe your best friend", "What have you learned from college?", etc. You will have 30 seconds to frame your story in mind and after that, the recording starts.

I suggest you be silent for 2 seconds in starting and ending of the recording.

Technical Interview:

This round lasted 50 minutes. The panelist was an experienced guy. The interview started with my introduction and the interviewer asked me to explain my project. He got deep into the project and I had to explain in detail and answer some questions related to the project like "Why do you use Relational DBMS instead of a traditional Database such as file?", "Where do you store your data?", "How's your project useful?", "How it is different from the already existing system?" and more.

After answering project-related questions, he moved on to the DSA, OOPs, DBMS, and Networking and asked many basic questions. The questions were so basic that it becomes difficult to answer without using technical terms. For example, He asked me "What is a Data-Structure?" to which I replied that it is the type of structure in the memory to store data and I gave many examples. He was not convinced after several tries. So, I confidently pitched him to give me any DSA-related coding question to which he gave me an easy-level question. I easily solved and explained it and he was convinced.

I successfully cleared the round and got forwarded to the second interview.

Executive Committee Interview:

This round lasted 30 minutes. The panelist was a member of the Executive committee in Hexaware. He was very friendly. We started with the introduction and my project. W had a normal conversation. At last, he asked a few questions related to the project and was impressed. It was my best interview experience so far.

HR Round:

This round lasted 10 minutes. The HR asked me to introduce myself. He asked a few typical HR questions like "Why do you want to join Hexaware?" "What do you know about Hexaware?". After, He explained the job role, training details, and salary details.

I got selected for the PGET program. The interviews level were easy compared to the Aptitude, Coding, and Communication round.

So that's my Hexaware Interview Experience. I will be sharing more interview experiences in the future. So keep an eye out and don't forget to share this and help your friends. I love to hear from your side if this post helps you in any way.

Cisco Ideathon Interview Experience | Coding Questions

noreply@blogger.com (Taranjeet Singh Kalsi) — Mon, 26 Jun 2023 18:29:00 +0000

Cisco Interview Experience:

In this post, I am sharing my personal Cisco interview experience which I got through Cisco Ideathon 2022. The program is a great opportunity for students with potential and creativity to be a part of Cisco’s technological innovation. The qualifiers of the Cisco Ideathon are awarded an internship opportunity followed by full-time placement within Cisco. Let's take a look at the eligibility criteria, Selection Process, and my experience.

Eligibility Criteria :

1. 7.0 CGPA in B.Tech. till the current semester. No criteria for 10th and 12th

2. Students must have completed any One Certification from CISCO as mentioned below :

(a) CCNA Module 1 or Networking Essentials

(b) DevNet or Python

Selection Process :

Registration:

There are 2 options to select for Online Assessment -

Advanced Networking
Advanced Software

You can select any in which you feel you are good enough. I had chosen advanced software. I got 90 minutes for the online assessment.

Round 1 (Online Assessment):

This is the first screening round. It contains 4 sections:

Aptitude section - This section consists of aptitude, reasoning, and verbal questions. It contains 10 questions with 15 minutes time limit.
Networking section - This section consists of 10 questions based on basic networking. You should have a strong understanding of networking concepts to answer the questions.
Coding section - It contains only 1 programming question which is of medium difficulty.
Advanced Networking or Advanced Software - As I selected advanced software, I got 1 extra coding question of medium difficulty level but more difficult than the question in the Coding section. If you select Advance Networking, you will get Networking MCQs.

I successfully cleared the online assessment round and got forwarded to the next round of idea submission.

Idea Submission:

I got a mail which contains 2 problem statements. We have to choose only one problem statement and make a PPT of our idea/solution to solve the problem. One question is based on a Networking problem that any user might face. And another problem is based on a real-life problem. I chose a real-life problem and proposed a solution in PPT. You have to present the PPT in the Technical rounds, so make sure to prepare your presentation and prepare answers to questions that can be asked by the interviewer.

TR-1 Round (Technical Interview-1):

This round lasted 1 hour. The panelist was a friendly guy and made me comfortable. In TR-1, the panelist asked me to introduce myself. After the introduction, he asked me to present and explain the PPT. He asked several questions and pointed out the problems with my idea. I was prepared for that and was able to satisfy him.

After the presentation, he asked many basic questions about Networking and also asked me how all layers of TCP/IP protocol work, in detail and step by step. After the networking question, he asked basic questions on DSA, OOPS, DBMS, and Operating Systems. The panelist tries to get deep into all questions means the next questions are related to the previous questions that I answered. In DSA, there were also questions from Graph and Tree but were basic like ways to traverse a Tree.

I successfully cleared the round and got forwarded to the second interview.

TR+MR -2 Round (Technical and Manager Interview -2):

This round lasted 45 minutes. In the second round, the panelist asked Technical and Managerial questions. Also, he asked me to present the PPT. He asked several networking questions similar to the previous round.

After Technical questions, he asked me several situation-based questions (Managerial questions). You should prepare for such Managerial questions in advance. And finally, he asked about my hobbies and discussed them. I was able to satisfy him and got forwarded to the HR round.

HR Round:

This round lasted 15 minutes. The HR asked me to quickly introduce myself. He asked several typical HR questions like where do you see yourself in 5 years? , What are your plans for future study? , What do you know about Cisco and its competitors? , Why do you want to join Cisco? Explain your job role in Cisco? and Will you work overtime?

Lastly, HR asked told me about the internship and stipend and asked me if I have any questions. I asked what type of projects I will work on in the internship which he didn't disclose.

Note: All interviews are on the same day and I got the final results on 3rd August. I was not selected for the internship.

So that's my Cisco Interview Experience. I will be sharing more interview experiences in the future. So keep an eye out and don't forget to share this and help your friends. I love to hear from your side if this post helps you in any way.

Persistent Interview and Internship Experience | Coding Questions

noreply@blogger.com (Taranjeet Singh Kalsi) — Thu, 08 Dec 2022 12:46:00 +0000

Persistent Interview Experience:

In this post, I am sharing my personal persistent interview and 6 Months of internship experience. There is a Martian program by Persistent that provide 5 weeks of training followed by an opportunity for an internship. In my college, Persistent hired me through the Martian program. Let's take a look at the eligibility criteria, Selection Process, and my experience.

Eligibility Criteria :

Minimum 60% in 10th, 12th/Diploma, B.E/BTech (current aggregate)
Branches - CS, IT
No active backlogs

Selection Process :

Martian Entry Round (Assessment Test-1):

This is the first screening round. It contains aptitude, reasoning, verbal, and technical MCQs, and 2 programming questions of easy and medium levels. The aptitude and reasoning questions are of moderate level. After clearing this round I got selected for 5 weeks Martian Internship in which I learned Computer Networks, DBMS, Data Structure and Algorithms, Operating Systems, and Programming Languages (JAVA/C++/Python) through Persistent courses. After completion of the 5 weeks of Martian Training, we have to clear the final Martian assessment to get the internship certificate and further qualify for interviews.

Martian Assessment (Assessment Test-2):

The final Martian assessment test consists of MCQs based on Computer Networks, DBMS, Data Structure and Algorithms, Operating Systems, and Programming Languages (JAVA or C++, or Python).

Again there are 2 Coding Questions of Easy and Medium Difficulty. I cleared this test and got forwarded for the next rounds. We get the internship certificate after passing the final Martian assessment.

TR-1 Round (Technical Interview):

In TR-1, the panelist asked basic questions on OOPs, Computer Networks, Operating Systems, DBMS, DSA, and JAVA (the language I choose among C++, Python, and JAVA). He tries to get deep into questions means the next question will be related to the previously answered questions. He asked questions on multi-threading, inheritance, polymorphism, and more basic concepts. I had explained overloading vs overriding in detail with code and he was convinced. I cleared this interview and got forwarded to the next interview.

TR-2 Round (Coding Interview):

In TR-2, the interviewer just gave me a coding question and asked me to solve it within 30 minutes. The question complexity is medium. The question is based on Hashmaps and Priority queues or Sorting the Hashmaps. I successfully passed all the test cases and explained my code to the interviewer and cleared this round.

HR Round:

In this round, I gave my introduction to HR. Then there are common HR questions like "Where do you see yourself in 5 years?", "What do you know about our company?". Lastly, HR gave me feedback. I cleared this round and got selected for 6 months internship at Persistent Systems.

My Tips for the Interview:

Study the basics of all core subjects such as OS, DBMS, Computer Networks, DSA, and Software Engineering.
Prepare for the aptitude and reasoning test as it is the first elimination test and also the opportunity for higher packages was given to those who topped the test.
Prepare and practice the medium-level coding questions based on Arrays, Strings, Hash-Maps, Sorting, Priority Queues, and more.
Be confident in the interview.
Don't give wrong answers if you don't know about it. Be honest in a smart way.
Don't try to be quick, take your time and answer the questions with patience with the use of hand gestures.
Write your introduction and answers to the basic HR Questions and prepare them well. And Also Study the background of the company.
Give a clear, concise, and complete introduction of yourself, and try to include the main skills you have or learned during college and a little bit about the projects and internships that you have done.
The most important thing is don't be stressed if you're not selected. There might be many more way better opportunities waiting for you. So be confident, give your best, and grab the right opportunity.

Persistent 6 Months Internship Experience:

After clearing so many rounds I finally got selected for the 6 Months Internship (called GEMS training) from January to June. The Training has majorly 3 sections:

Foundational Concepts Training: This training is common for all. The training duration is 1 Month. It contains courses such as OOPs, SQL, SDLC, and GIT. After Foundational training, we have to pass its Assessment with a minimum of 70%.
Communication and Power Skills Training: This training is common for all. The course is based on enhancing communication skills and social skills. The Training duration is 2 Months( February and March). After Communication training, we have to pass its assessment named Power Skills Objective with a minimum of 70%.
Core Track Training: After foundational training, we got allocated to different training tracks such as JAVA, Salesforce, Dotnet, UI, and more. I got the JAVA track which contains courses such as Core JAVA, Maven, Spring Boot, HTML & CSS, and JavaScript. After Core Track training, you have to pass a final Core Track assessment which contains objective and subjective questions. My exam was in June and I successfully cleared it and got an internship certificate.

Note- We got 3 attempts in all exams in the GEMS training and the minimum passing percentage is 70%.

So that's my Persistent Interview and Internship Experience. I will be sharing more interview experiences in the future. So keep an eye out and don't forget to share this and help your friends. I love to hear from your side if this post helps you in any way.

Expected move with Solution

noreply@blogger.com (Taranjeet Singh Kalsi) — Mon, 13 Jun 2022 08:23:00 +0000

Expected move:

Problem Description:

You are given a fair coin and two integers X and N.

You repeatedly flip the coin and perform the following operation based on the result:

· If it lands on heads, then X will increase by 1. However, if X=N then X will not change.

· If it lands on tails, X will decrease by 1.

You stop as soon as the value of X becomes 1.

Find the expected number of coin flips.

Note that it is guaranteed that under the given constraints, the answer is an integer and does not exceed 2⋅10^18

Input:

· The first line contains a single integer T — the number of test cases. Then the test cases follow.

· The first and only line of each test case contains two space-separated integers X and N.

Output:

For each testcase, output the expected number of coin flips.

Constraints

· 1≤T≤10^5

· 1≤X≤N≤10^9

Sample Input 1:

1 3

2 2

12 12

568 57800

Sample Output 1:

132

65223144

EXPLANATION:

In the first test case, X is already 1. So, the expected number of the coin flips, in this case, is 0.

In the second test case, if the coin lands on the head, X will not change because X is equal to N. Whenever the coin lands on the tail, X will decrease by 1 and the game ends as X becomes 1. The expected number of coin flips to get the first tail is 2. So, the answer for the case is 2.

Solution:

C++:

#include <iostream>
using namespace std;

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        unsigned long long int x, n, moves = 0;
        cin >> x >> n;
        moves = (x - 1) * (2 * n - x);
        cout << moves << endl;
    }
    return 0;
}

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Reduce to zero with Solution

noreply@blogger.com (Taranjeet Singh Kalsi) — Mon, 13 Jun 2022 07:56:00 +0000

Reduce to zero:

Problem Description:

Chef has two integers X and Y. Chef wants to perform some operations to make both X and Y zero simultaneously. In one operation, Chef can either:

· set X:=2⋅X

· or set Y:=2⋅Y

· or set X:=X−1 and Y:=Y−1

Chef is a little busy preparing for the contest. Help him find the minimum number of operations required to make both X and Y equal to 0 simultaneously. If it is impossible to do so after any number of operations, print −1.

Input:

· The first line contains a single integer T — the number of test cases. Then the test cases follow.

· The first and only line of each test case contains two space-separated integers X and Y.

Output:

For each test case, print the minimum number of operations required to make X and Y zero simultaneously. If it is impossible to do so after any number of operations, print −1 instead.

Constraints

· 1≤T≤3⋅10^5

· 0≤X, Y≤10^18

Sample Input 1:

1 2

99999999999999999 99999999999999999

Sample Output 1:

99999999999999999

EXPLANATION:

Let (x, y) denote the values of X and Y respectively.

Test case 1: Three operations are required: (1,2)→ X:=X⋅2→ (2,2) → Y:=Y−1, X:=X−1→ (1,1)→ Y:=Y−1, X:=X−1→ (0,0)

It can be proven that it is impossible in fewer operations.

Solution:

C++:

#include <bits/stdc++.h>
using namespace std;

int main()
{
    long t;
    cin >> t;
    while (t--)
    {
        unsigned long long int x, y;
        cin >> x >> y;
        if (x > y)
        {
            unsigned long long int t = x;
            x = y;
            y = t;
        }

        if (x == y)
            cout << x << endl;
        else if (x == 0 || y == 0)
            cout << "-1" << endl;
        else
        {
            unsigned long long int operations = 0, nOf2mul = 0;
            if (x < y)
            {
                nOf2mul = abs(log2(y) - log2(x));

                operations += nOf2mul;
                x = x * pow(2, nOf2mul);
            }

            if (x >= y)
            {
                operations--;
                x /= 2;
            }
            unsigned long long int prev = x - (y - x);
            x -= prev;
            y -= prev;
            operations += prev;
            operations++; // operation for 2x
            operations += y;
            cout << operations << endl;
        }
    }
    return 0;
}

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Possible GCD with Solution

noreply@blogger.com (Taranjeet Singh Kalsi) — Mon, 13 Jun 2022 07:26:00 +0000

Possible GCD:

Problem Description:

Chef has two distinct positive integers A and B.

Chef wonders how many distinct values are possible for the expression gcd(A+X, B+X), where X can take any non-negative integer value.

Help Chef find this value.

Here, gcd stands for Greatest Common Divisor.

Input:

· The first line contains a single integer T — the number of test cases. Then the test cases follow.

· The first and only line of each test case contains two distinct space-separated integers A and B.

Output:

For each testcase, output the number of distinct values of the expression gcd(A+X, B+X), where X can take any non-negative integer value.

Constraints

· 1≤T≤1000

· 1≤A, B≤10^9

· A≠B

Sample Input 1:

1 2

12 8

Sample Output 1:

EXPLANATION:

Test case 1: Here gcd(1+X,2+X)=1 no matter what value of X you choose.

Test case 2:

· If we take X=0, gcd(12,8)=4.

· If we take X=1, gcd(13,9)=1.

· If we take X=2, gcd(14,10)=2.

It can be shown that no other value of gcd is possible.

Solution:

C++:

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        long long int a, b, gcdCount = 0, d, i;
        cin >> a >> b;
        d = abs(a - b);
        for (i = 1; i <= sqrt(d); i++)
        {
            if (d % i == 0)
            {
                if (d / i == i)
                    gcdCount++;
                else
                    gcdCount += 2;
            }
        }

        cout << gcdCount << endl;
    }
    return 0;
}

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Divisible by i with Solution

noreply@blogger.com (Taranjeet Singh Kalsi) — Mon, 13 Jun 2022 07:07:00 +0000

Divisible by i:

Problem Description:

You are given an integer N.

Construct a permutation P of length N such that

· For all i (1≤i≤N−1), i divides abs(Pi+1−Pi)abs(Pi+1−Pi).

Recall that a permutation of length NN is an array where every integer from 1 to N occurs exactly once.

It can be proven that for the given constraints at least one such P always exists.

Input:

· The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.

· The only line of each test case contains an integer N - the length of the array to be constructed.

Output:

For each test case, output a single line containing N space-separated integers P1, P2,…, PN denoting the elements of the array P.

If there exist multiple such arrays, print any.

Constraints

· 1≤T≤5⋅10^4

· 2≤N≤10^5

· The sum of N over all test cases does not exceed 10^5.

Sample Input 1:

Sample Output 1:

1 2

2 1 3

EXPLANATION:

Test case 1: A possible array satisfying all the conditions is [1,2]:

· For i=1: abs(A2−A1)=abs(2−1)=1 is divisible by 1.

Test case 2: A possible array satisfying all the conditions is [2,1,3]:

· For i=1: abs(A2−A1)=abs(1−2)=1 is divisible by 1.

· For i=2: abs(A3−A2)=abs(3−1)=2 is divisible by 2.

Solution:

C++:

#include <iostream>
using namespace std;

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int n, t;
        cin >> n;
        t = n;
        int a[n];
        for (int i = n - 1; i >= 0; i -= 2)
        {
            a[i] = n;
            n--;
        }
        int j;
        if (t % 2 == 0)
            j = 0;
        else
            j = 1;
        for (j; j < t; j += 2)
        {
            a[j] = n;
            n--;
        }
        for (int i = 0; i < t; i++)
            cout << a[i] << " ";
        cout << endl;
    }
    return 0;
}

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Equal Strings with Solution

noreply@blogger.com (Taranjeet Singh Kalsi) — Mon, 13 Jun 2022 06:58:00 +0000

Equal Strings:

Problem Description:

Given a string A of length N consisting of lowercase English alphabet letters.

You are allowed to perform the following operation on the string A any number of times:

· Select a non-empty subsequence S of the array [1, 2, 3,…, N] and any lowercase English alphabet αα;

· Change Ai to αα for all i∈S.

Find the minimum number of operations required to convert A into a given string B of length N consisting of lowercase English alphabet letters.

Input:

· The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.

· The first line of each test case contains an integer N - the length of the string A and B.

· The second line of each test case contains the string A.

· The third line of each test case contains the string B.

Output:

For each test case, output the minimum number of operations required to convert string A into string B.

Constraints

· 1≤T≤10^4

· 1≤N≤10^5

· Sum of N over all test cases does not exceed 10^5.

Sample Input 1:

aaa

bab

abcd

aaab

Sample Output 1:

EXPLANATION:

Test case 1:

· In the first operation, we can choose S={1} and α= c. After this operation, A will become cb.

· In the second operation, we can choose S={2} and α= d. After this operation, A will become cd.

Test case 2:

· In the first operation, we can choose S={1,3} and α= b. After this operation, A will become bab.

Test case 3:

· In the first operation, we can choose S={2,3} and α= a. After this operation, A will become aaad.

· In the second operation, we can choose S={4} and α= b. After this operation, A will become aaab.

Solution:

C++:

#include <bits/stdc++.h>
using namespace std;
bool uniqueChars(int x, string s)
{
    for (int i = 0; i < x; i++)
    {
        if (s[i] == s[x])
            return false;
    }
    return true;
}
int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int n, k = 0;
        string a, b, u = "";
        cin >> n >> a >> b;
        // finding unique characters in b
        for (int i = 0; i < n; i++)
        {
            if (uniqueChars(i, b))
            {
                u += b[i];
            }
        }
        // checking if any unique character need to be replaced
        for (int i = 0; i < u.length(); i++)
        {
            char x = u[i];
            bool flag = true;
            for (int i = 0; i < n; i++)
            {
                if (b[i] == x && !(a[i] == b[i]))
                    flag = false;
            }
            if (flag)
                k++;
        }

        cout << u.length() - k << endl;
    }
    return 0;
}

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Alternate Additions with Solution

noreply@blogger.com (Taranjeet Singh Kalsi) — Sun, 12 Jun 2022 16:21:00 +0000

Alternate Additions:

Problem Description:

Chef has 2 numbers A and B (A<B).

Chef will perform some operations on A.

In the ith operation:

· Chef will add 1 to A if i is odd.

· Chef will add 2 to A if i is even.

Chef can stop at any instant. Can Chef make A equal to B?

Input:

· The first line contains a single integer T — the number of test cases. Then the test cases follow.

· The first and only line of each test case contains two space separated integers A and B.

Output:

For each test case, output YES if Chef can make A and B equal, NO otherwise.

Note that the checker is case-insensitive. So, YES, Yes, yEs are all considered same.

Constraints

· 1≤T≤1000

· 1≤A<B≤10⁹

Sample Input 1:

1 2

3 6

4 9

10 20

Sample Output 1:

YES

EXPLANATION:

Test case 1: Chef may perform one operation to make A equal to B: 1−+1→2

Test case 2: 3−+1→4−+2→6

Test case 3: It can be shown that it is impossible to make A and B equal.

Test case 4: 10−+1→11−+2→13−+1→14−+2→16−+1→17−+2→19−+1→20

Solution:

C++:


#include <iostream>
using namespace std;

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int a, b;
        cin >> a >> b;
        int diff = b - a;
        if (diff % 3 == 2)
            cout << "NO" << endl;
        else
            cout << "YES" << endl;
    }
    return 0;
}

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Subscriptions with Solution

noreply@blogger.com (Taranjeet Singh Kalsi) — Sun, 12 Jun 2022 16:04:00 +0000

Subscriptions:

Problem Description:

A new TV streaming service was recently started in Chefland called the Chef-TV.
A group of N friends in Chefland want to buy Chef-TV subscriptions. We know that 6 people can share one Chef-TV subscription. Also, the cost of one Chef-TV subscription is X rupees. Determine the minimum total cost that the group of NN friends will incur so that everyone in the group is able to use Chef-TV.

Input:

· The first line contains a single integer T — the number of test cases. Then the test cases follow.

· The first and only line of each test case contains two integers N and X — the size of the group of friends and the cost of one subscription.

Output:

For each test case, output the minimum total cost that the group will incur so that everyone in the group is able to use Chef-TV.

Constraints

· 1≤T≤1000

· 1≤N≤100

· 1≤X≤1000

Sample Input 1:

1 100

12 250

16 135

Sample Output 1:

100

500

405

EXPLANATION:

· Test case 1: There is only one person in the group. Therefore he will have to buy 1 subscription. Therefore the total cost incurred is 100.

· Test case 2: There are 12 people in the group. Therefore they will have to buy 2 subscriptions. Therefore the total cost incurred is 500.

· Test case 3: There are 16 people in the group. Therefore they will have to buy 3 subscriptions. Therefore the total cost incurred is 405.

Solution:

C++:


#include <iostream>
#include <cmath>
using namespace std;

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int n, x;
        cin >> n >> x;
        cout << ceil(n / 6.0) * x << endl;
    }
    return 0;
}

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Jio Interview Experience | Coding Questions

noreply@blogger.com (Taranjeet Singh Kalsi) — Fri, 10 Jun 2022 18:10:00 +0000

JIO Interview Experience:

In this post, I am sharing my personal JIO interview experience. Jio came to my college for an on-campus hiring drive. Let's take a look at the eligibility criteria, Job Description, Selection Process, and my experience.

Eligibility Criteria :

Minimum 60% in 10th, 12th/Diploma, B.E/BTech (current aggregate)
No year drops during the engineering
No active backlogs

Job Description :

Based on the skills of candidates and the requirements of the organization, selected candidates from all three levels will be assigned to different Business Units. You will work in fields like Software development, Testing, Automation Testing, Operations, DevOps, UI/UX, Project Management, Product Management, and Data Analytics. You will get a chance to learn the latest technologies like AI/ML/Data Science etc.

Selection Process :

The selection process consists of an online exam followed by two TR+HR interviews. If a candidate fails to pass the online exam then the candidate still has a chance to qualify for the interviews through the GD round. If a candidate fails in GD or any one of the TR+HR rounds, the candidate will be out of the selection process. If the candidate passes the online test, the candidate is directly qualified for the TR+HR interviews.

Online Exam: It is the first round of the selection process. It consists of MCQ Questions based on Logical Reasoning, Verbal Ability, Domain Knowledge, and Programming. There are two Coding Questions of easy and medium complexity.

My Experience:

In Online Test there are several graph-based statistics questions, Verbal Ability Questions, and Domain Knowledge questions from SQL, OS, and OOP concepts. Both Coding questions are easy. One is based on performing Array Operations and the other is based on guessing numbers(formula-based). I cleared the online test and qualified for the first TR+HR interview.

In the First TR+HR interview, the interviewer asked a few HR questions like "What are your Hobbies?", "Where do you see yourself in 5 years?" and more. The interviewer asked various basic Technical questions about Cloud Computing and its disadvantages, IoT, C vs C++, C++ vs JAVA, Virtual Functions, Interface in JAVA, Multiple Inheritance in JAVA, Constructors, Pointers, and more. My interview lasted about 10 minutes and it was a good experience.

I was only able to answer about 70-80% of the questions but somehow I qualified for the final TR+HR interview. The Final interview is easier than the previous interview. The interviewers asked very basic questions based on my Resume on OS, DBMS, Networking, and Data Structures. And after that they asked HR questions like "Why do we hire you?", "Why do you want to join our company?" and more. I answered all the questions correctly and I got selected.

My Tips for the Interview:

Study the basics of all core subjects such as OS, DBMS, Computer Networks, DSA, and Software Engineering.
Prepare and practice the basic coding questions like Fibonacci, Palindrome, and more.
Be confident in the interview.
Don't give wrong answers if you don't know about it. Be honest in a smart way.
Don't try to be quick, take your time and answer the questions with patience with the use of hand gestures.
Write your answers to the basic HR Questions and prepare them well. And Also Study the background of the company.
Give a clear, concise, and complete introduction of yourself, and try to include the main skills you have or learned during college and a little bit about your projects.
The most important thing is don't be stressed if you're not selected. There might be many more way better opportunities waiting for you. So be confident, give your best, and grab the right opportunity.

So that's my experience in JIO Drive. I will be sharing more interview experiences in the future. So keep an eye out and don't forget to share this and help your friends. I love to hear from your side if this post helps you in any way.

Wordle with Solution

noreply@blogger.com (Taranjeet Singh Kalsi) — Tue, 15 Mar 2022 15:31:00 +0000

Wordle:

Problem Description:

Chef invented a modified wordle.

There is a hidden word S and a guess word T, both of length 5.

Chef defines a string M to determine the correctness of the guess word. For the ith index:

If the guess at the ith index is correct, the ith character of M is G.

If the guess at the ith index is wrong, the ith character of M is B.

Given the hidden word S and guess T, determine string M.

Input:

First line will contain T, number of test cases. Then the test cases follow.
Each test case contains of two lines of input.
First line contains the string S - the hidden word.
Second line contains the string T - the guess word.

Output:

For each test case, print the value of string M.

You may print each character of the string in uppercase or lowercase (for example, the strings BgBgB, BGBGB, bgbGB, and bgbgb will all be treated as identical).

Constraints

1≤T≤1000
|S|=|T|=5
S, T contain uppercase english alphabets only.

Sample Input 1:

ABCDE

EDCBA

ROUND

RINGS

START

STUNT

Sample Output 1:

BBGBB

GBBBB

GGBBG

EXPLANATION:

Test Case 1: Given string S=ABCDE and T=EDCBA. The string M is:

Comparing the first indices, A≠E, thus, M[1]=B.

Comparing the second indices, B≠D, thus, M[2]=B.

Comparing the third indices, C=C, thus, M[3]=G.

Comparing the fourth indices, D≠B, thus, M[4]=B.

Comparing the fifth indices, E≠A, thus, M[5]=B.

Thus, M=BBGBB.

Test Case 2: Given string S=ROUND and T=RINGS. The string M is:

Comparing the first indices, R=R, thus, M[1]=G.

Comparing the second indices, O≠I, thus, M[2]=B.

Comparing the third indices, U≠N, thus, M[3]=B.

Comparing the fourth indices, N≠G, thus, M[4]=B.

Comparing the fifth indices, D≠S, thus, M[5]=B.

Thus, M=GBBBB.

Solution:

C++:


#include <iostream>
using namespace std;

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        string s, g;
        cin >> s;
        cin >> g;
        for (int i = 0; i < s.length(); i++)
        {
            if (s[i] == g[i])
                cout << "G";
            else
                cout << "B";
        }
        cout << endl;
    }
    return 0;
}

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Xor Palindrome with Solution

noreply@blogger.com (Taranjeet Singh Kalsi) — Sun, 13 Feb 2022 16:55:00 +0000

Xor Palindrome:

Problem Description:

A (1-indexed) binary string S of length N is called a xor palindrome if the value of Si⊕S(N+1−i) is the same for all 1≤i≤N.

For example, 0, 1111 and 0101 are xor palindromes, while 1110 and 110101 are not.

You are given a binary string S of length N. Determine if it is possible to rearrange it to form a xor palindrome or not.

Input:

· The first line of input contains a single integer T — the number of test cases. The description of T test cases follows.

· The first line of each test case contains an integer N — the length of the binary string S.

· The second line of each test case contains the binary string S containing 0s and 1s only.

Output:

For each test case, output YES if it is possible to rearrange S to convert it into a xor palindrome. Otherwise output NO.

You may print each character of YES and NO in uppercase or lowercase (for example, yes, yEs, Yes will be considered identical).

Constraints

• 1≤T≤1000

• 1≤N≤105

• S is a binary string, i.e, contains only the characters 0 and 1

• It is guaranteed that the sum of N over all test cases does not exceed 2⋅105.

Sample Input 1:

0011

001

0001

Sample Output 1:

Yes

EXPLANATION:

Test case 1: 00 is already a xor palindrome. [The value of Si⊕S(N+1−i) is 0 for all 1≤i≤N.]

Test case 2: 0011 is already a xor palindrome. [The value of Si⊕S(N+1−i) is 1 for all 1≤i≤N.]

Test case 3: 001 can be rearranged to form 010 which is a xor palindrome. [The value of Si⊕S(N+1−i) is 0 for all 1≤i≤N.]

Test case 4: It can be proved that 0001 can not be rearranged to form a xor palindrome.

Solution:

C++:

#include <iostream>
using namespace std;

int main()
{
    int t, n;
    string s;
    cin >> t;
    while (t--)
    {
        cin >> n;
        cin >> s;
        int count0 = 0, count1 = 0;
        for (int i = 0; i < n; i++)
        {
            if (s[i] == '0')
                count0++;
            else if (s[i] == '1')
                count1++;
        }
        if (count0 % 2 != 0 && count1 % 2 != 0 && count0 != count1)
            cout << "NO" << endl;
        else
            cout << "YES" << endl;
    }
    return 0;
}

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Avoid Fixed Points with Solution

noreply@blogger.com (Taranjeet Singh Kalsi) — Sat, 12 Feb 2022 07:11:00 +0000

Avoid Fixed Points:

Problem Description:

Chef has a sequence of N integers A=[A1, A2,…, AN]. He can perform the following operation any number of times (possibly, zero):
·       Choose any positive integer K and insert it at any position of the sequence (possibly the beginning or end of the sequence, or in between any two elements).
For example, if A=[5,3,4] and Chef selects K=2, then after the operation he can obtain one of the sequences [2,5,3,4],[5,2,3,4],[5,3,2,4] , or [5,3,4,2].
Chef wants this sequence to satisfy the following condition: for each 1≤i≤∣A∣, Ai≠i. Here, ∣A∣ denotes the length of A.

Help Chef to find the minimum number of operations that he has to perform to achieve this goal. It can be proved that under the constraints of the problem, it's always possible to achieve this goal in a finite number of operations.
Input:
·       The first line of input contains an integer T, denoting the number of test cases. The description of T test cases follows.
·       The first line of each test case contains an integer N.

·       The second line contains N space-separated integers A1, A2,…, AN.

Output:
For each test case, print a single line containing one integer — the minimum number of operations that Chef has to perform to achieve the given condition.
Constraints
·       1≤T≤10⁴
·       1≤N≤10⁵
·       1≤Ai≤10⁹
·       Sum of N over all test cases does not exceed 2⋅10⁵.

Sample Input 1:
3
3
2 4 5
3
4 1 3
4
3 2 4 2

Sample Output 1:
0
1
2

EXPLANATION:
Test case 1: The given sequence does not contain any index i such that Ai=i. Hence Chef does not have to perform any operation.
Test case 2: In the given sequence, A3=3. Chef can choose K=2 and insert it before the first element, making the sequence A=[2,4,1,3], which does not contain any index i for which Ai=i.
Test case 3: In the given sequence, A2=2. Chef can perform the following sequence of operations:
·       Choose K=5 and insert it before the first element. The sequence becomes A=[5,3,2,4,2], and now A4=4.
·       Choose K=3 and insert it between the third and fourth element. The sequence becomes A=[5,3,2,3,4,2], which does not contain any index i for which Ai=i.

It can be verified that there is no way to satisfy the given condition in less than two operations.

Solution:

C++:

#include <iostream>
#include <vector>
using namespace std;

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int n, minOperations = 0, input;
        vector<int> a;
        cin >> n;

        for (int i = 0; i < n; i++)
        {
            cin >> input;
            a.push_back(input);
        }

        for (int i = 0, tempi = 0; i < n; i++, tempi++)
        {
            if (a[i] == (tempi + 1))
            {
                tempi++;
                minOperations++;
            }
        }

        cout << minOperations << endl;
    }
    return 0;
}

Programming Chaska

Top K Frequent Elements- LeetCode Solution and Approach

347. Top K Frequent Elements - LeetCode - Medium

Problem Description:

Approach:

JAVA:

Keep Learning and Practicing with Programming Chaska !!!

Perfect number Program in C | C++ | JAVA

Perfect number Program in C | C++ | JAVA

For Example: Number --> 6factors of 6 are -> 1,2 and 3->1 + 2 + 3 = 6.So, 6 is a Perfect Number.Some of the Perfect numbers are 28, 496, and 8128.

Algorithm to check whether a given number is a Perfect number or not:

Code 1: This is the general code in C with time complexity of O(n).

Code 2: This is the general code in C++ with time complexity of O(n).

Code 3: This is the general code in JAVA with time complexity of O(n).

Start Practicing LeetCode with Programming Chaska. I have explained my approach to the problems in simple way. I have also shared the optimized code so that you can learn some new.

Optimized Code in C | C++ | JAVA:

Code 4: This is optimized code to check perfect numbers. Here we are using mathematical property for factor checking. The factors of a number starting repeating when we go beyond its square root.

Example:

Factors of 6 :

6 = 1 * 6

6 = 2 * 3

6 = 3 * 2

6 = 6 * 1

We can see that 2*3 is same as 3*2, so we will not go beyond square root.

If you have any doubts/questions related to the Program, if you want to give any suggestions, or want an explanation of the code, feel free to comment below. I hope that my code is understandable, if not then you can tell me in the comments so that I will try to make it easier to understand.

Group Anagrams - LeetCode Solution and Approach

49. Group Anagrams - LeetCode - Medium

Problem Description:

Approach:

JAVA:

Keep Learning and Practicing with Programming Chaska !!!

Remove Duplicates from Sorted Array - LeetCode Solution and Approach

26. Remove Duplicates from Sorted Array - LeetCode - Easy

Problem Description:

Approach:

JAVA:

class Solution { public int removeDuplicates(int[] nums) { int prevUnique = nums[0], p = 1; for (int n : nums) { if (n != prevUnique) { nums[p++] = n; prevUnique = n; } } return p; }}

Keep Learning and Practicing with Programming Chaska !!!

Longest Consecutive Sequence - LeetCode Solution and Approach

128. Longest Consecutive Sequence - LeetCode - Medium

Problem Description:

Approach:

JAVA:

Keep Learning and Practicing with Programming Chaska !!!

Valid Anagram - LeetCode Solution and Approach

242. Valid Anagram - LeetCode - Easy

Problem Description:

Approach:

JAVA:

class Solution { public boolean isAnagram(String s, String t) { int[] freq=new int[26]; for(char c:s.toCharArray()) freq[c-'a']++; for(char c:t.toCharArray()) freq[c-'a']--; for(int n:freq) if(n!=0) return false; return true; }}

Keep Learning and Practicing with Programming Chaska !!!

Contains Duplicate - LeetCode Solution and Approach

217. Contains Duplicate - LeetCode - Easy

Problem Description:

Approach:

JAVA:

class Solution { public boolean containsDuplicate(int[] nums) { Set<Integer> set=new HashSet<>(); for(int n:nums) if(!set.add(n)) return true; return false; }}

Keep Learning and Practicing with Programming Chaska !!!

Longest Substring Without Repeating Characters - LeetCode Solution and Approach

3. Longest Substring Without Repeating Characters - LeetCode - Medium

Problem Description:

Approach:

JAVA:

Keep Learning and Practicing with Programming Chaska !!!

Add Two Numbers - LeetCode Solution and Approach

2. Add Two Numbers - LeetCode - Medium

Problem Description:

Approach:

JAVA:

Keep Learning and Practicing with Programming Chaska !!!

Two Sum - LeetCode Solution and Approach

1. Two Sum - LeetCode - Easy

Problem Description:

Approach:

JAVA:

Keep Learning and Practicing with Programming Chaska !!!

Accenture Interview Experience | Coding Questions

Accenture Interview Experience:

Eligibility Criteria :

Selection Process :

For Example: Number --> 6
factors of 6 are -> 1,2 and 3
->1 + 2 + 3 = 6.
So, 6 is a Perfect Number.
Some of the Perfect numbers are 28, 496, and 8128.

We can see that 23 is same as 32, so we will not go beyond square root.

`class Solution { public int removeDuplicates(int[] nums) { int prevUnique = nums[0], p = 1; for (int n : nums) { if (n != prevUnique) { nums[p++] = n; prevUnique = n; } } return p; } }`

`class Solution { public boolean isAnagram(String s, String t) { int[] freq=new int[26]; for(char c:s.toCharArray()) freq[c-'a']++; for(char c:t.toCharArray()) freq[c-'a']--; for(int n:freq) if(n!=0) return false; return true; } }`

`class Solution { public boolean containsDuplicate(int[] nums) { Set<Integer> set=new HashSet<>(); for(int n:nums) if(!set.add(n)) return true; return false; } }`

You will get an email invitation to take part in online tests if you are qualified.
The assessment has two phases that must be completed within the allotted time.
Mock Assessment