Software Development

Save Flowers

2023-03-05T20:43:00.003+05:30

Save Flowers

// { Driver Code Starts

#include<bits/stdc++.h>

using namespace std;

class Array

{

public:

template <class T>

static void input(vector<T> &A,int n)

{

for (int i = 0; i < n; i++)

{

scanf("%d ",&A[i]);

}

template <class T>

static void print(vector<T> &A)

{

for (int i = 0; i < A.size(); i++)

{

cout << A[i] << " ";

}

cout << endl;

}

};

// } Driver Code Ends

class Solution {

public:

int saveFlowers(int n, vector<int> &arr, string s) {

// vector<int> v(n+1, 0);

int count =0;

for(int i=n-1; i>=0; i--){

if(s[i]=='1'){

count =i;

break;

}

int curr_sum =0, flag =1, ans =0, last_element=1000;

for(int i=count; i>=0; i--){

if(s[i]=='1'){

last_element = min(arr[i], last_element);

curr_sum = curr_sum + arr[i];

flag =0;

}else {

if(!flag && last_element < arr[i]){

curr_sum = curr_sum + arr[i] - last_element;

flag = 1;

}

last_element =1000;

}

ans = max(curr_sum, ans);

}

return ans;

}

};

// { Driver Code Starts.

int main(){

int t;

scanf("%d ",&t);

while(t--){

int n;

scanf("%d",&n);

vector<int> arr(n);

Array::input(arr,n);

string s;

getline(cin,s);

Solution obj;

int res = obj.saveFlowers(n, arr, s);

cout<<res<<endl;

}

// } Driver Code Ends

Nice Array

2023-03-05T20:41:00.001+05:30

Nice Array

Solution:

class Solution {

public:

int niceSubarray(int n, vector<int> &arr) {

// kadane algorithm

// sort(arr.begin(), arr.end());

int len=0, mx=0, j=1, count =0;

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

if(arr[i]==i+1){

count++;

}

if(count == n) return n;

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

if(arr[i]==j){

j++;

len++;

}else {

mx = max(len, mx);

j=1;

len =0;

}

return mx;

}

Maximum Weight Node

2023-01-20T23:23:00.001+05:30

Maximum Weight Node

Given a maze with N cells. Each cell may have multiple entry points but not more than one exit (i.e entry/exit points are unidirectional doors like valves).
You are given an array Edge[] of N integers, where Edge[i] contains the cell index that can be reached from cell i in one step. Edge[i] is -1 if the ith cell doesn't have an exit.
The task is to find the cell with maximum weight (The weight of a cell is the sum of cell indexes of all cells pointing to that cell). If there are multiple cells with the maximum weight return the cell with highest index.

Note: The cells are indexed with an integer value from 0 to N-1. If there is no cell pointing to the ith cell then the weight of the i'th cell is zero.

Example 1:

Input:
N = 4
Edge[] = {2, 0, -1, 2}
Output: 2
Explanation: 
1 -> 0 -> 2 <- 3
weight of 0th cell = 1
weight of 1st cell = 0 
(because there is no cell pointing to the 1st cell)
weight of 2nd cell = 0+3 = 3
weight of 3rd cell = 0
There is only one cell which has maximum weight
(i.e 2) So, cell 2 is the output.

Solution:

//{ Driver Code Starts

//Initial Template for C++

#include<bits/stdc++.h>

using namespace std;

// } Driver Code Ends

//User function Template for C++

class Solution

{

public:

int maxWeightCell(int N, vector<int> Edge)

{

vector<int> ind(N+1, 0);

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

if(Edge[i]!=-1)

ind[Edge[i]]= ind[Edge[i]]+ i;

}

int a = *max_element(ind.begin(), ind.end());

int ans =0;

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

if(a==ind[i]){

ans = i;

}

return ans;

}

};

//{ Driver Code Starts.

int main(){

int tc;

cin >> tc;

while(tc--){

int N;

cin >> N;

vector<int> Edge(N);

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

cin>>Edge[i];

}

Solution obj;

int ans=obj.maxWeightCell(N, Edge);

cout<<ans<<endl;

}

return 0;

}

// } Driver Code Ends

Find the first node of loop in linked list

2023-01-18T21:37:00.002+05:30

Find the first node of loop in linked list

Given a singly linked list of N nodes. Find the first node of the loop if the linked list has a loop. If a loop is present return the node data of the first node of the loop else return -1.

Example 1:

Input:

Output: 3
Explanation:
We can see that there exists a loop 
in the given linked list and the first
node of the loop is 3.

Solution:

//{ Driver Code Starts

//Initial Template for C++

#include <bits/stdc++.h>

using namespace std;

struct Node

{

int data;

Node* next;

Node(int val)

{

data = val;

next = NULL;

}

};

void loopHere(Node* head, Node* tail, int position)

{

if(position==0) return;

Node* walk = head;

for(int i=1; i<position; i++)

walk = walk->next;

tail->next = walk;

}

// } Driver Code Ends

//User function Template for C++

/*struct Node

{

int data;

struct Node *next;

Node(int x) {

data = x;

next = NULL;

}

class Solution

{

public:

int findFirstNode(Node* head)

{

Node *slow =head;

Node *fast = head;

int flag=0;

int ans=-1;

while(fast && slow){

slow = slow->next;

fast = fast->next->next;

if(fast==slow){

flag =1;

break;

}

if(flag==0) return -1;

Node *temp = head;

int n=1;

while(n>0){

if(temp==slow){

n=0;

return temp->data;

}

slow = slow->next;

temp = temp->next;

}

return ans;

}

};

//{ Driver Code Starts.

int main()

{

int t;

cin>>t;

while(t--)

{

int n, num;

cin>>n;

Node *head, *tail;

cin>> num;

head = tail = new Node(num);

for(int i=0 ; i<n-1 ; i++)

{

cin>> num;

tail->next = new Node(num);

tail = tail->next;

}

int pos;

cin>> pos;

loopHere(head,tail,pos);

Solution ob;

int ans = ob.findFirstNode(head);

cout<<ans<<"\n";

}

return 0;

}

// } Driver Code Ends

Absolute difference divisible by K

2023-01-08T18:31:00.004+05:30

Absolute difference divisible by K

Given an array of integers of size n and an integer k, find all the pairs in the array whose absolute difference is divisible by k.

Example 1:

Input:
n = 3
arr[] = {3, 7, 11}
k = 4
Output:
3
Explanation:
(11-3) = 8 is divisible by 4
(11-7) = 4 is divisible by 4
(7-3) = 4 is divisible by 4

Example 2:

Input:
n = 4
arr[] = {1, 2, 3, 4}
k = 2
Output :
2
Explanation:
Valid pairs are (1,3), and (2,4).

Solution:

long long countPairs(int n, int arr[], int k) {

int a[n];

map<int, int> m;

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

a[i] = (arr[i]+k)%k;

m[a[i]]++;

}

long long int ans =0;

for(auto i=m.begin(); i!=m.end(); i++){

int no = i->second;

ans = ans + (no*(no-1)/2);

}

return ans;

}

fhdfn

Maximum Value

2023-01-02T14:23:00.001+05:30

Maximum Value

//{ Driver Code Starts
//Initial Template for C++
#include <bits/stdc++.h>
using namespace std;

/* A binary tree node has data, pointer to left child
   and a pointer to right child */
struct Node {
    int data;
    struct Node* left;
    struct Node* right;
};


// } Driver Code Ends
//User function Template for C++
/* Tree node structure  used in the program

struct Node
{
    int data;
    struct Node* left;
    struct Node* right;
}; */

class Solution {
  public:
    vector<int> maximumValue(Node* root) {
   
    queue<Node *> q;
    vector<int> ans;
    if (!root)
        return ans;
    q.push(root);

    while (!q.empty())
    {
        vector<int> v;
        int size = q.size();
     //   ans.push_back(q.front()->data);

        while (size--)
        {

            Node *t = q.front();
            v.push_back(t->data);
            q.pop();
            if (t->left)
                q.push(t->left);
            if (t->right)
                q.push(t->right);
        }
        int a = *max_element(v.begin(), v.end());
        ans.push_back(a);
    }
    return ans;
    }
};

//{ Driver Code Starts.

Node* newNode(int val) {
    Node* temp = new Node;
    temp->data = val;
    temp->left = NULL;
    temp->right = NULL;
    return temp;
}
Node* buildTree(string str) {
    // Corner Case
    if (str.length() == 0 || str[0] == 'N') return NULL;

    // Creating vector of strings from input
    // string after spliting by space
    vector<string> ip;

    istringstream iss(str);
    for (string str; iss >> str;) ip.push_back(str);

    // Create the root of the tree
    Node* root = newNode(stoi(ip[0]));

    // Push the root to the queue
    queue<Node*> queue;
    queue.push(root);

    // Starting from the second element
    int i = 1;
    while (!queue.empty() && i < ip.size()) {

        // Get and remove the front of the queue
        Node* currNode = queue.front();
        queue.pop();

        // Get the current node's value from the string
        string currVal = ip[i];

        // If the left child is not null
        if (currVal != "N") {

            // Create the left child for the current node
            currNode->left = newNode(stoi(currVal));

            // Push it to the queue
            queue.push(currNode->left);
        }

        // For the right child
        i++;
        if (i >= ip.size()) break;
        currVal = ip[i];

        // If the right child is not null
        if (currVal != "N") {

            // Create the right child for the current node
            currNode->right = newNode(stoi(currVal));

            // Push it to the queue
            queue.push(currNode->right);
        }
        i++;
    }

    return root;
}
/* Function to get Largest value at each level of Binary Tree */
vector<int> maximumValue(struct Node* tree);


int main() {
    int t;
    scanf("%d\n", &t);
    while (t--) {
        string s;
        getline(cin, s);
        Node* root = buildTree(s);
        Solution ob;
        vector<int> answer = ob.maximumValue(root);
        for(auto i:answer)
            cout<<i<<" ";
        cout<<endl;
    }
    return 0;
}
// } Driver Code Ends

520. Detect Capital

2023-01-02T14:05:00.003+05:30

520. Detect Capital

We define the usage of capitals in a word to be right when one of the following cases holds:

All letters in this word are capitals, like "USA".
All letters in this word are not capitals, like "leetcode".
Only the first letter in this word is capital, like "Google".

Given a string word, return true if the usage of capitals in it is right.

Example 1:

Input: word = "USA"
Output: true

Example 2:

Input: word = "FlaG"
Output: false

Solution:

class Solution {
public:
    bool detectCapitalUse(string word) {

     int sam=0, cap=0;
     int n = word.length();
        for(int i=0; i<n; i++){
            if(word[i]>= 'a' && word[i] <='z'){
                sam++;
            }
            if(word[i]>='A'&& word[i]<='Z'){
                cap++;
            }
        }
       if(sam ==n|| cap==n)  return true;
       if(word[0]>='A' && word[0]<='Z' && sam == n-1) return true;
       return false;

       
    }
};

Find the Town Judge leetcode (997)

2022-10-31T20:51:00.003+05:30

997. Find the Town Judge

In a town, there are n people labeled from 1 to n. There is a rumor that one of these people is secretly the town judge.

If the town judge exists, then:

The town judge trusts nobody.
Everybody (except for the town judge) trusts the town judge.
There is exactly one person that satisfies properties 1 and 2.

You are given an array trust where trust[i] = [ai, bi] representing that the person labeled ai trusts the person labeled bi.

Return the label of the town judge if the town judge exists and can be identified, or return -1 otherwise.

Example 1:

Input: n = 2, trust = [[1,2]]
Output: 2

Example 2:

Input: n = 3, trust = [[1,3],[2,3]]
Output: 3

Example 3:

Input: n = 3, trust = [[1,3],[2,3],[3,1]]
Output: -1

Constraints:

1 <= n <= 1000
0 <= trust.length <= 104
trust[i].length == 2
All the pairs of trust are unique.
ai != bi
1 <= ai, bi <= n

Solution:

class Solution {

public:

int findJudge(int n, vector<vector<int>>& edges) {

int ans =-1;

vector <int> ind(n+1);

vector <int> out(n+1);

for(int i=0; i<edges.size(); i++){

out[edges[i][0]]++;

ind[edges[i][1]]++;

}

for(int i=0; i<ind.size(); i++){

if((ind[i]==n-1)&&(out[i]==0)){

ans = i;

// break;

}

return ans;

}

};

Happy Coding ;

Move Last Element to Front of a Linked List

2022-10-16T18:56:00.004+05:30

Move Last Element to Front of a Linked List

You are given the head of a Linked List. You have to move the last element to the front of the Linked List and return the list.

Example 1:

Input:
N = 5
List = {2,5,6,2,1}
Output:
{1,2,5,6,2}
Explanation:
In the given linked list, the last element is 1,
after moving the last element to the front the
linked list will be {1,2,5,6,2}.

//{ Driver Code Starts

//Initial Template for C++

#include <bits/stdc++.h>

using namespace std;

class ListNode{

public:

int val;

ListNode *next;

ListNode(int x){

val=x;

next=NULL;

}

};

// } Driver Code Ends

//User function Template for C++

class Solution{

public:

ListNode *moveToFront(ListNode *head){

if(head==NULL) return NULL;

else if(head->next ==NULL) return head;

else if(head->next->next ==NULL){

ListNode *temp =head, *t=head;

head = head ->next;

t->next= NULL;

head->next= temp;

}

else {

ListNode *temp = head, *t =head->next, *t1=head;

while(temp->next!=NULL){

temp = temp->next;

}

while(t->next->next!=NULL){

t = t->next;

}

t->next =NULL;

head = temp;

head->next = t1;

return head;

}

};

//{ Driver Code Starts.

int main(){

int t;

cin>>t;

while(t--){

int n;

cin>>n;

vector<ListNode*> a(n);

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

int x;

cin>>x;

a[i]=new ListNode(x);

if(i!=0){

a[i-1]->next=a[i];

}

ListNode *head=a[0];

Solution ob;

head=ob.moveToFront(head);

while(head!=NULL){

cout<<head->val;

if(head->next!=NULL){

cout<<" ";

}

head=head->next;

}

cout<<endl;

}

return 0;

}

// } Driver Code Ends

Happy Coding 😊.

Heap sort

2022-09-28T01:47:00.002+05:30

Heap Sort C++ Code :

#include <iostream>

using namespace std;

void heapify(int a[], int n, int i)
{
    int largest = i;

    int l = 2 * i + 1;

    int r = 2 * i + 2;

    if (l < n && a[l] > a[largest])

        largest = l;

    if (r < n && a[r] > a[largest])

        largest = r;

    if (largest != i)
    {

        swap(a[i], a[largest]);

        heapify(a, n, largest);
    }
}

void heapSort(int a[], int n)
{
    for (int i = n / 2 - 1; i >= 0; i--)

        heapify(a, n, i);

    for (int i = n - 1; i >= 0; i--)

    {
        swap(a[0], a[i]);

        heapify(a, i, 0);
    }
}

void printArray(int a[], int n)
{
    for (int i = 0; i < n; i++)

        cout << a[i] << " ";

    cout << "\n";

}

int main()
{
    cout << "\nCode With Khushpreet" << endl;

    int n;

    cout << "Enter the number of element: "<<endl;

    cin >> n;

    cout << "Enter the element: " << endl;

    int a[n];

    for (int i = 0; i < n; i++)
    {
        cin >> a[i];

    }
    heapSort(a, n);

    cout << "Sorted array is \n";
    
    printArray(a, n);
}

Time Complexity

Worst Case Time Complexity is: O(NlogN)

Average Case Time Complexity is: O(NLogN)

Best Case Time Complexity is: O(NLogN)

Happy Coding 😊.

adjacency list of a bidirectional graph

2022-08-04T18:39:00.006+05:30

Given the adjacency list of a bidirectional graph. Your task is to copy/clone the adjacency list for each vertex and return a new list.

Example 1:

Input:

Output: 
0-> 1-> 4 
1-> 0-> 2-> 3-> 4 
2-> 1-> 3 
3-> 1-> 2-> 4 
4-> 0-> 1-> 3
Explanation:
As 0,1 and 3 is connected to 4 so 4th row
of the list containing 4 and its connected
nodes 0,1 and 3 and we have to add those in
sorted order and same for every row.

Example 2:

Input:


Output: 
0-> 2-> 3 
1-> 2 
2-> 0-> 1 
3-> 0
Explanation:
As 2 and 3 is connected to 0 so 0th row
of the list containing 0 and its connected 
nodes 2 and 3 and we have to add those in
sorted order and same for every row.

Solution:

//{ Driver Code Starts

#include <bits/stdc++.h>

using namespace std;

// } Driver Code Ends

class Solution

{

public:

// Function to return the adjacency list for each vertex.

vector<vector<int>> printGraph(int V, vector<int> adj[])

{

vector<vector<int>> v;

for (int i = 0; i < V; i++)

{

cout << i;

for (int j = 0; j < adj[i].size(); j++)

{

cout << "-> " << adj[i][j];

}

cout << endl;

}

return v;

}

};

//{ Driver Code Starts.

int main()

{

int tc;

cin >> tc;

while (tc--)

{

int V, E;

cin >> V >> E;

vector<int> adj[V];

for (int i = 0; i < E; i++)

{

int u, v;

cin >> u >> v;

adj[u].push_back(v);

adj[v].push_back(u);

}

Solution obj;

vector<vector<int>> ans = obj.printGraph(V, adj);

for (int i = 0; i < ans.size(); i++)

{

for (int j = 0; j < ans[i].size() - 1; j++)

{

cout << ans[i][j] << "-> ";

}

cout << ans[i][ans[i].size() - 1];

cout << endl;

}

return 0;

}

// } Driver Code Ends

Happy Coding 🤭 .

Puri Sharma Pathania book pdf

2022-07-12T13:44:00.001+05:30

Puri Sharma Pathania book pdf Download.

Author of the book :

Madan S Pathania, Br Puri, Lr Sharma

Details about book :

Quantum Mechanics
Chemical bonding molecular quantum mechanics
Group theory
molecular spectroscopyhe gaseous state
The liquid state: physical properties of liquids
The gaseous state part 2: real gases
liquid crystals; The mesomorphic States
The third law of thermodynamics
Phase equilibria
Chemical equilibria
The nernst distribution law
Electrochemistry part 3: electrified inferences
Electrochemistry part 2: electromotive force of Galvanic cells
Photochemistry
Catalysis
Surface chemistry
Mass spectrometry
Macromolecules
Irreversible thermodynamics

Happy Coding 😊.

C++ Mcq

2022-07-08T18:44:00.003+05:30

Type - MCQs : C++ Multiple Choice Questions

3. Which of the following is used for implementing the late binding

a.Operator Functions
b. Constant Functions
c. Virtual Functions
d. Both A and B.

a. Distributed
b. Multiple
c. Multl-level
d. Hierarchal

Please, write your score in the comment box.

Hope you enjoy the session, so may sign-up for free to get latest updates.

Thankyou.

constructor and destructor in cpp

2022-07-04T10:54:00.007+05:30

Assign Tasks

2022-06-11T22:13:00.002+05:30

Assign Tasks Geeks For Geeks

STUDYEVERTHING

Example 1

Input:
N = 3
A = [6, 3, 6]
B = [2, 6, 8]
Output: 2
Explanation: Assign 1st task to 2nd person,
2nd task to 1st person and 3rd task to 3rd person.
Days taken by 1st, 2nd and 3rd person will be:
2, 1, 1.

Example 2

Input:
N = 2
A = [100, 100]
B = [1, 200]
Output: 100
Explanation: Since the work required by both
tasks is same, we can assign them in any order

// { Driver Code Starts
// Initial Template for C++

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

// } Driver Code Ends
// User function Template for C++

class Solution
{
public:
    long long solve(int n, long long A[], long long B[])
    {

        int C[n];
        sort(A, A + n);
        sort(B, B + n);
        int flag = 1;
        int max;

        int count = 0;
        while (flag > 0)
        {
            max = 0;
            for (int i = 0; i < n; i++)
            {
                if (A[i] - B[i] < 0)
                {
                    A[i] = 0;
                }
                else
                {
                    A[i] = A[i] - B[i];
                    // count++;
                }
                if (A[i] > max)
                {
                    max = A[i];
                    // cout<<max<<" ";
                }
            }
            // cout<<max<<" ";
            if (max <= 0)
            {
                flag = 0;
            }
            count++;
        }
        return count;
    }
};

// { Driver Code Starts.

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        cin >> n;
        long long arr[n], brr[n];
        for (int i = 0; i < n; i++)
        {
            cin >> arr[i];
        }
        for (int i = 0; i < n; i++)
        {
            cin >> brr[i];
        }
        Solution ob;
        long long ans = ob.solve(n, arr, brr);
        cout << ans << "\n";
    }
    return 0;
}
// } Driver Code Ends

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Happy Coding 😊.

Share and Earn

2022-06-09T16:24:00.005+05:30

Share and Earn

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Kth smallest element

2022-06-06T16:32:00.003+05:30

Given an array arr[] and an integer K where K is smaller than size of array, the task is to find the Kth smallest element in the given array. It is given that all array elements are distinct.

STUDYEVERTHING

// { Driver Code Starts
//Initial function template for C++

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

 // } Driver Code Ends
//User function template for C++

class Solution{
    public:
    // arr : given array
    // l : starting index of the array i.e 0
    // r : ending index of the array i.e size-1
    // k : find kth smallest element and return using this function
    int kthSmallest(int arr[], int l, int r, int k) {
      int ans;
      sort(arr, arr+r+1);
      for(int i=0; i<=r; i++){
          if(i+1==k){
              ans = arr[i];
          }
      }
      return ans;
    }
};

// { Driver Code Starts.
 
int main()
{
    int test_case;
    cin>>test_case;
    while(test_case--)
    {
        int number_of_elements;
        cin>>number_of_elements;
        int a[number_of_elements];
        
        for(int i=0;i<number_of_elements;i++)
            cin>>a[i];
            
        int k;
        cin>>k;
        Solution ob;
        cout<<ob.kthSmallest(a, 0, number_of_elements-1, k)<<endl;
    }
    return 0;
}  // } Driver Code Ends

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Happy Coding 😊.

Equilibrium Point

2022-05-28T22:19:00.005+05:30

Given an array A of n positive numbers. The task is to find the first Equilibium Point in the array.

Equilibrium Point in an array is a position such that the sum of elements before it is equal to the sum of elements after it.

STUDYEVERTHING

// { Driver Code Starts
#include <iostream>
using namespace std;

// } Driver Code Ends
class Solution
{
public:
    // Function to find equilibrium point in the array.
    // a: input array
    // n: size of array
    int equilibriumPoint(long long a[], int n)
    {
        int start = a[0], end = a[n - 1];
        int j = n - 2, index = -1;
        int i = 1;
        // cout << a[n - 1] << a[n - 2];
        while (j > i)
        {
            if (start < end)
            {
                start = start + a[i];
                //  cout << start << " ";
                i++;
                //     cout << i << " ";
                //  if ((start == end) && ((j - i) == 2))
                //  {
                // index = i;
                //  }
            }
            if (start > end)
            {
                end = end + a[j];
                // cout << end << " ";
                //    cout << j << " ";
                j--;
                //    if ((start == end) && ((j - i) == 2))
                //    {
                //       index = i;
                //        cout << i << " ";
                //   }
            }
            if (start == end)
            {
                //  cout << i;
                if (((j - i) <= 0) && (j >= i))
                {
                    index = i + 1;
                    //         cout << i << " ";
                }
                start = start + a[i];
                i++;
            }
        }
        //  cout << " ";
        //  cout << index;
        if (n == 1)
        {
            index = 1;
        }
        return index;
    }
};

// { Driver Code Starts.

int main()
{

    long long t;

    // taking testcases
    cin >> t;

    while (t--)
    {
        long long n;

        // taking input n
        cin >> n;
        long long a[n];

        // adding elements to the array
        for (long long i = 0; i < n; i++)
        {
            cin >> a[i];
        }

        Solution ob;

        // calling equilibriumPoint() function
        cout << ob.equilibriumPoint(a, n) << endl;
    }
    return 0;
}
// } Driver Code Ends

// { Driver Code Starts

// Initial template for C++

#include <bits/stdc++.h>

using namespace std;

// } Driver Code Ends

// User function template for C++

class Solution {

public:

long long int binarysearch(int arr[], int n, int k) {

long long int l = 0, r = n - 1, mid, ans;

while (l <= r)

{

mid = (r - l) / 2 + l;

if (arr[mid] == k)

{

ans = mid;

l = n;

}

else if (arr[mid] > k)

{

r--;

}

else

{

l++;

}

int ans2=-1;

if(ans == mid){

ans2 = ans;

}

return ans2;}

};

// { Driver Code Starts.

int main() {

int t;

cin >> t;

while (t--) {

int N;

cin >> N;

int arr[N];

for (int i = 0; i < N; i++) cin >> arr[i];

int key;

cin >> key;

Solution ob;

int found = ob.binarysearch(arr, N, key);

cout << found << endl;

}

// } Driver Code Ends

Happy Coding 😊.

How to install IBM SPSS Statistics 26

2022-05-05T16:39:00.002+05:30

How to install IBM SPSS Statistics 26 (for free)

STUDYEVERTHING

Things keep in mind while downloading IBM spss statistics :

Turn off your internet otherwise software will not activated.
Turn off window defender Protection.
Turn off any other anti-virus.

Steps to download IBM statistics Spss 26:

Download Spss by click on the button given below:

Install IBM statistics Spss 26:

Click on setup and run as administrator

Copy lserver file

And paste it in IBM spss statics 26 folder.

Now explore SPSS for free and life time access.

Happy Coding 😊.

Tally App

2021-12-23T20:48:00.003+05:30

Tally App

STUDYEVERTHING

Main activity

package com.example.tallycounter;

import androidx.appcompat.app.AppCompatActivity;

import android.os.Bundle;
import android.view.View;
import android.widget.Button;
import android.widget.TextView;

public class MainActivity extends AppCompatActivity {

    private TextView textView;

    private Button increaseBIN;
    private Button decreaseBIN;
    private Button resetBIN;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

        setUI();
        increaseBIN.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {
                String newText = Integer.toString(Integer.parseInt(textView.getText().toString()) + 1);
                textView.setText(newText);
            }
        });
        decreaseBIN.setOnClickListener(new View.OnClickListener() {
            public void onClick(View v) {
                if (Integer.parseInt(textView.getText().toString()) != 0) {
                    String newText = Integer.toString(Integer.parseInt(textView.getText().toString()) - 1);
                    textView.setText(newText);
                }
            }

        });

    }

    public void resetTextView(View view) {
        textView.setText("0");
    }

    private void setUI() {
        textView = findViewById(R.id.textView);
        increaseBIN = findViewById(R.id.increase);
        decreaseBIN = findViewById(R.id.decrease);
        resetBIN = findViewById(R.id.reset);

    }

}

string

<resources>
    <string name="app_name">tally counter</string>

    <string name="increase">+</string>
    <string name="decrease">-</string>
    <string name="reset">Reset</string>

    <string name="textView">0</string>

</resources>

Hope you find the article useful.

Live working proof.

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Heap Sort C++ Code :

adjacency list of a bidirectional graph

Given the adjacency list of a bidirectional graph. Your task is to copy/clone the adjacency list for each vertex and return a new list.

Solution:

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Puri Sharma Pathania book pdf

C++ Mcq

Type - MCQs : C++ Multiple Choice Questions

1. Which of the following can be considered as the correct syntax for declaring an array of pointers of integers that has a size of 10 in C++

2. Which of the following can be considered as the members that can be inherited but not accessible in any class

3. Which of the following is used for implementing the late binding

4. Which of the following is not a kind of inheritance?

5. Which type of approach is used by the C++ language.

constructor and destructor in cpp

Assign Tasks

Assign Tasks Geeks For Geeks

Share and Earn

Share and Earn

Kth smallest element

Given an array arr[] and an integer K where K is smaller than size of array, the task is to find the Kth smallest element in the given array. It is given that all array elements are distinct.

Equilibrium Point

Given an array A of n positive numbers. The task is to find the first Equilibium Point in the array.

Equilibrium Point in an array is a position such that the sum of elements before it is equal to the sum of elements after it.

Array Subset of another array

Given two arrays: a1[0..n-1] of size n and a2[0..m-1] of size m. Task is to check whether a2[] is a subset of a1[] or not. Both the arrays can be sorted or unsorted..

Leaders in an array

Given an array A of positive integers. Your task is to find the leaders in the array. An element of the array is a leader if it is greater than or equal to all the elements to its right side. The rightmost element is always a leader.

Minimize the sum of product

You are given two arrays, A and B, of equal size N. The task is to find the minimum value of A[0] * B[0] + A[1] * B[1] +…+ A[N-1] * B[N-1], where shuffling of elements of arrays A and B is allowed.

Check if two arrays are equal or not

Given two arrays A and B of equal size N, the task is to find if given arrays are equal or not. Two arrays are said to be equal if both of them contain same set of elements, arrangements (or permutation) of elements may be different though

Binary Search (GFG)

Given a sorted array of size N and an integer K, find the position at which K is present in the array using binary search.

How to install IBM SPSS Statistics 26

How to install IBM SPSS Statistics 26 (for free)

Things keep in mind while downloading IBM spss statistics :

Steps to download IBM statistics Spss 26:

Tally App

Tally App

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