CodeToSkill

AI PDF Query Master: Simplifying PDF Document Analysis with AI-Powered Natural Language Processing

2023-07-15T19:29:00.003+05:30

Introduction:

In today's digital age, vast amounts of information are stored in PDF documents, making it challenging for researchers, students, and professionals to extract valuable insights efficiently. However, with the advent of AI and natural language processing (NLP) technologies, analyzing and querying PDFs has become more accessible and convenient. The "AI PDF Query Master" is a web application that leverages the power of OpenAI's ChatGPT language model to enable users to interactively query uploaded PDF documents using natural language. This blog post will explore the features, benefits, and usage of the AI PDF Query Master application.

Features and Benefits:

1. Upload PDF Documents:

The AI PDF Query Master allows users to easily upload their PDF files directly within the application. Users can drag and drop multiple PDFs or use the browse files feature to import them, providing a seamless experience.

2. Interactive Chat Interface:

The application offers a user-friendly chat interface where users can ask questions, seek information, and provide instructions to retrieve relevant details from the uploaded PDF documents. By conversing with the AI-powered ChatGPT model, users can obtain precise answers and insights.

3. Natural Language Processing:

Powered by the OpenAI ChatGPT model, the AI PDF Query Master intelligently understands and responds to user queries in a conversational manner. It leverages NLP techniques to analyze the uploaded PDFs and extract pertinent information efficiently.

4. Query History and Context:

To enhance the user experience, the application maintains a history of previous queries and preserves context during the conversation. Users can refer back to previous questions, continue the flow seamlessly, and explore related topics effectively.

5. Fast and Accurate Results:

The AI PDF Query Master employs advanced NLP techniques and PDF processing capabilities to quickly analyze and extract relevant information from the uploaded PDFs. Users can expect fast and accurate results, saving them valuable time and effort.

Usage:

To get started with the AI PDF Query Master, users need to follow a few simple steps:

1. Provide OpenAI API Key:

Users must provide their OpenAI API key, which can be obtained from the OpenAI platform. This key allows the application to access the necessary resources for AI-powered processing.

2. Upload PDFs:

Users can drag and drop their PDF files or use the browse files feature to import them into the application. Multiple PDFs can be uploaded simultaneously, enabling users to analyze and query extensive document collections conveniently.

3. Start the Conversation:

Once the PDFs are uploaded, users can enter their queries or questions in the chat interface. The AI PDF Query Master will process the queries using the ChatGPT model and provide relevant answers based on the content of the uploaded PDFs.

Conclusion:

The AI PDF Query Master is a powerful web application that harnesses the capabilities of OpenAI's ChatGPT language model to simplify the extraction of insights and knowledge from PDF documents. With its user-friendly chat interface, natural language processing capabilities, and fast results, the AI PDF Query Master is an invaluable tool for researchers, students, and professionals working with extensive document repositories. By leveraging AI and NLP technologies, this application empowers users to unlock the potential of their PDF documents and streamline their information retrieval process. Experience the power of AI-assisted PDF querying today with the AI PDF Query Master!

Note: The code provided in the question is incomplete and may not function as expected. Additional code or modifications may be necessary for a fully functional application.

CODE:https://github.com/VAMSINADH2000/AI-PDF-Query-Master

Roman to Integer Leetcode Python Solution

2021-05-14T17:41:00.005+05:30

class Solution:
def romanToInt(self, s: str) -> int:
roman = {'I':1,
'V':5,
'X':10,
'L':50,
'C':100,
'D':500,
'M':1000}
res = 0
s = list(s)
for i in range(len(s)):
if res==0:
res+=roman[s[i]]
continue
if roman[s[i]]>roman[s[i-1]]:
res = res + roman[s[i]] - 2*roman[s[i-1]]
continue
res+=roman[s[i]]
return res

All Popular Sorting Algorithms in Python

2020-12-12T10:18:00.003+05:30

Here are Some Sorting Algorithms that are basically implemented in python

Bubble Sort:

def bubble_sort(lst):

length = len(lst) - 1

sorted = False

while not sorted:

sorted = True

for i in range(length):

if lst[i] > lst[i + 1]:

lst[i], lst[i + 1] = lst[i + 1], lst[i]

sorted = False

return lst

print(bubble_sort([7, 10, 4, 3, 20, 15]))

Insertion Sort:

def insertion_sort(lst):

for i in range(1,len(lst)):

val_to_sort = lst[i]

while lst[i-1] > val_to_sort and i > 0:

lst[i-1],lst[i] = lst[i],lst[i-1]

i-=1

return lst

print(insertion_sort([9,8,7,6,5,4,3,2,1]))

Merge Sort:

def merge_sort(arr):

if len(arr) <= 1:

return

mid = len(arr)//2

left = arr[:mid]

right = arr[mid:]

# print(left,right)

merge_sort(left)

merge_sort(right)

merge_two(left, right, arr)

def merge_two(a,b,arr):

len_a = len(a)

len_b = len(b)

i = j = k = 0

while i < len_a and j < len_b:

if a[i] <= b[j]:

arr[k] = a[i]

i+=1

else:

arr[k] = b[j]

j+=1

k+=1

while i < len_a:

arr[k] = a[i]

i+=1

k+=1

while j < len_b:

arr[k] = b[j]

j+=1

k+=1

array = [9,8,7,4,5,6,3,2,1]

merge_sort(array)

print(array)

Quick Sort

def quick_sort(sequence):

length = len(sequence)

if length <=1:

return sequence

else:

pivot = sequence.pop()

items_greater = []

items_lower = []

for item in sequence:

if item > pivot:

items_greater.append(item)

else:

items_lower.append(item)

return quick_sort(items_lower) + [pivot] + quick_sort(items_greater)

print(quick_sort([9,8,7,6,5,4,3,2,1]))

Selection Sort:

def selection_sort(lst):

n = len(lst)-1

for i in range(n):

mini_value = i

for j in range(i+1,n+1):

if lst[j] < lst[mini_value]:

mini_value = j

if mini_value !=i:

lst[mini_value],lst[i] = lst[i],lst[mini_value]

return lst

print(selection_sort([9,8,7,6,5,4,3,2,1]))

Everything You Need To Know To Become a Python Master

2020-10-04T09:18:00.004+05:30

Image credits: www.dreamstime.com

In this blog we would all be doing is discussing the topics and skills that you need to learn to become a master at the Python programming language. Now I've broken these skills up into three main categories. Those are

1.Basic

2.Intermediate

3.Advanced/ master level.

1.Basic

This is really focusing on just fundamental programming concepts and really understanding like the basic, basic building blocks of programming. So the core syntax, the way a program actually reads, the way the different lines of codes are going to be executed, conditions, variables, all that kind of stuff. So you want to understand how variables work. So, you know, X equals one, Y equals two X equals Y what does X equal? You know, just the basics like that and understanding why we even use variables in programming then probably moving forward into conditions. So Boolean conditions, you know, how do we compare two different variables together? How do we check if the user typed something in and it's equal to something else, chaining conditionals together. So, okay. How can we make a chain conditional that has more than one condition in it? How do we evaluate, you know, this condition and that condition at the same time.

Next that's going to be operators that you want to learn,how do we add two variables subtract two variables. What is the modulates operator do into your division? Just these basic tools that you're going to need to be using. As soon as you get out of this basic level, then after that, it's probably going to go to control flow. So it's things like if else, else, if you're going to want to master those, then just learning about loops and Iterable. So for loops, while loops, how do we loop through a list? How do we loop through a dictionary that leads me nicely into what else I would recommend learning, which is basic data structures in Python. So understanding lists, understanding, sets, understanding dictionaries, what about strings booleans ins floats. That's all stuff that you really need to master after that. And you've learned kind of the core basic syntax and language features of Python.

Then you're probably going to move into functions. Functions are really important, a huge part of Python, and really will teach you a lot if you can understand how those work. Then after you've done that, try to learn about some of the common methods in Python. So for example, if you have a string, how do you turn that string into all uppercase or all lowercase? How do you add a new element to a list or add a new element to a set, just kind of these basics, operations and methods, which are really easy to learn and find it, but that you kinda need to know before moving much,much further. And then finally I would recommend having a basic knowledge of IO operations.

After learning all these I would strongly suggest you to practice these things in any one of the competitive programming website.(HackerRank,LeetCode,HackerEarth..etc)

2.Intermediate

The first thing that I'm going to focus on here and intermediate skills is object oriented, programming, understanding classes and understanding objects. So this goes a really long way, and this is something that you're going to want to understand, especially if you're using other languages other than Python, uh, but super important. It's used pretty much everywhere. And if you only get one thing from this intermediate section, understand that you have to have a really good foundation and object oriented programming to really understand anything above this level.

And then after you've learned that I would recommend you move into data structures. So once you have a solid foundation of object oriented, you want to learn about data structures understanding the ques, you know, hash maps, the stacks priority cues, whatever it may be, you just need to understand those topics cause they will come up and understanding the efficiency and time complexity in big O notation is actually really important when it comes to becoming a good programmer.If you see them next, I would recommend you learn about Lambda functions. So these are anonymous functions. They're actually super useful.

3.Advanced

I want to be clear here that a lot of the topics in this section you don't really need to know to be good at Python, right? To consider yourself a Python developer to go work as a software engineer, you don't need to know most of what I'm going to describe here. if you know the basic and intermediate stuff, you can learn this when you need to learn it kind of thing. So yeah, just don't be discouraged if you don't already know these things.The first thing I have is decorators. if you have like a class method, a static method and UCL at static method at class method at login required, those are examples of decorators, things that decorate a function or a method. I think you can decorate a class as well. Generators are a way to kind of utilize memory efficiently in Python. Let's say you have a really long collection or you're generating a collection rather than just generating the entire thing. You can generate one item at a time. If you only need access to one item from that collection at a time, or maybe two items or three items or something like that, you can use a generator to do that.

Next we have medic classes. These are too advanced, even really give a basic description, but they are pretty much how Python classes actually operate and allow you to do some really advanced things with classes, hook into some things that you can't usually hook into unless you know about them. And then next after you learn about decorators, generators, context, pantries, and yeah, medic classes, I would recommend learning about concurrency and parallelism. So say you have a CPU core that has four processing cores on it. So for actual CPU, how can you use Python to utilize all of those cores? Well, that's when you're going to learn about something called a global interpreter lock, and then you can talk about multiprocessing and multithreading so multithreading is essentially how do we let one part of our program kind of wait while the other program is doing something and how do we not block pieces of code, when they don't need to be blocked. So that's a really vague definition of it, but that should hopefully give you a little bit of an idea.

It's pretty much revolving around speed within the Python language, then I would get into testing. So once you've learned all of those, I would learn about unit testing and how you can actually do test driven development in Python or in whatever programming language, just understanding testing pipelines. And all of that is very important. Then I would learn about how you actually build and manipulate packages. So packages are different than modules, but I throw this in the advanced section because if you're writing enough code that you could fill a module, a package you're probably advanced.

Resources that I recommend to learn Python for free:

1.Telusko Youtube Channel

2.Tech with Tim Youtube Channel

3.corey schafer Youtube Channel

I think you should know if you're working towards becoming a master in Python. So if this Blog helped you out.

Stay Tuned!

😎 HAPPY CODING😎

How to Prepare for Placement Aptitude

2020-07-21T08:34:00.002+05:30

I will discuss the preparation strategies for the aptitude test of campus placements this strategies has been prepared with the combined expertise of trainers who have helped more than 15 lakh students get placed in the company of their choice.

Cracking the aptitude test is usually the toughest part of any recruitment process quite a few to do get rejected at this stage despite having good technical knowledge and communication skills. In aptitude test you compete with thousands of students where the cutoff is not usually pre decided so your performance along with the performance of your competitors decides whether you get selected for the next round or not that is why it is absolutely essential to be amongst the top scorers in the exam. Let us help you understand how you can achieve this.

The aptitude test consists of three sections quantitative aptitude, logical reasoning and verbal ability over all the major chapters that you have to cover are

1.numbers

2.ages

3.problems on trains

4.time and work

5.permutations and combinations

6.probability

7.relationship based questions

8.seating arrangements

9.para jumbles

10.sentence Corrections and

11.grammar

These chapters will definitely help in answering the bulk of questions asked for quantitative aptitude.Test yourself along with that be a regular on websites like India Bix and Lofoya from where you can get hundreds of free practice questions all questions do not carry equal important to emphasize on important topics like

1. number systems

2.time and work

3.permutations and combinations

4.probability

5.time speed and distance

6.percentage and ratios to name a few you don't have to practice hundreds of questions but it is necessary to practice questions of all types and all difficulty.

If a topic has 50 questions be sure to try and solve all 50 without looking at solutions mark those which you could not solve and give them a try after it it if you are not able to solve even then look at the solutions and understand why you got them wrong and what is the concept behind them.

Logical reasoning chapters like

1.series completion

2.blood relations

3.directions

4.coding and decoding are important follow the same strategy as Kwan speed is really important in logical reasoning and only practice will help you achieve the desired score.

verbal ability is a different ballgame because you require a good grasp on the language communicating in English is the easiest way to practice grammar reading newspapers will also help you some important topics here are

1.Reading comprehension

2.para jumbles

3.grammar based sentence Corrections and

4.fill in the blanks along with synonyms antonyms and patter completion arrangement eyes questions.

In conclusion start from concepts and practice questions from websites thus mastering all sections topic by topic those who aim for cracking the cutoff you will fall short of your goal aim to be amongst the top 2% and you will easily clear the cutoff.

Don't learn shortcuts to solve problems when you are a beginner learn the concepts behind them so that you can able to crack tough problems.

Prefered Books:1.Rs Agarwal

2.Aptipedia

👍ALL THE BEST👍

Subsets LeetCode Solution

2020-07-12T09:17:00.002+05:30

Subsets LeetCode

Quoestion: For the given set find all unique subsets i.e,powerset

For Example:

set = [1,2,3]

powerset=[[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]]

Approach:

1.intialize powerset with empty List

2.Add 1st element ([[],[1]])

3.Add 2nd element and add possible unique combinations of 1st and 2nd elements.([[],[1],[2],[1,2]])

4.repeat above step until size of powerset is 2^n

5.return Powerset

Python Code:

class Solution(object):

def subsets(self, nums):

powerset=[[]]

n = len(nums)

if n == 0:

return powerset

for i in nums:

for j in range(len(powerset)):

print(len(powerset))

temp = powerset[j]+[i]

powerset.append(temp)

return powerset

Python One Liner:

from itertools import chain, combinations

class Solution(object):

def subsets(self,nums):

temp = list(nums)

return chain.from_iterable(combinations(temp, i) for i in range(len(nums)+1))

Try Problem in LeetCode

Maximum Width of Binary Tree LeetCode Solution

2020-07-10T07:32:00.004+05:30

Quoestion:Click Here

Approach Using BFS:

we use BFS or level order traversal.we keep tarck of level,index of each node in queue for calculating max width.

Algorithm:

1.if root=null;return 0

2.create a queue to store the index,level,TreeNode

3.we fill indexes according to the following

indexof_leftchild = (2*(index_of_root))

indexof_rightchild = (2*(index_of_root)+1)

4.we pop the first item in queue and check for level if cur_level > prev_level we update them

5.maximum width will be max(prev_width,(cur_idx-idx_old)+1)

6.return maxwidth

TimeComplexity:O(N)

SpaceComplexity:O(N)

Python Code:

# Definition for a binary tree node.

# class TreeNode(object):

# def __init__(self, val=0, left=None, right=None):

# self.val = val

# self.left = left

# self.right = right

class Solution:

def widthOfBinaryTree(self, root):

prev_level, idx_old, result = 1, 1, 0

queue = [[prev_level,idx_old,root]]

while queue:

[cur_idx, cur_level, node] = queue.pop(0)

if cur_level > prev_level:

prev_level, idx_old = cur_level, cur_idx

result = max(result, cur_idx - idx_old + 1)

if node.left: queue.append([cur_idx*2, cur_level+1, node.left])

if node.right: queue.append([cur_idx*2+1,cur_level+1, node.right])

return result

Try Problem In LeetCode

3Sum LeetCode Solution

2020-07-09T10:30:00.010+05:30

Quiestion:Click Here

Approach:

since we need to find triplet,we fix one nums[i] in given array then we traverse through array for the other two numbers such that their sum is equal to complement of nums[i](i.e,-(nums[i])) using two pointer technique.

we fix low pointer next to the nums[i] and high pointer to the last element.

if nums[i]+nums[low]+nums[high] <0

increase low pointer

if nums[i]+nums[low]+nums[high] >0

decrement high pointer

if nums[i]+nums[low]+nums[high] ==0

add nums[i],nums[low],nums[high] to result list

Python Code:

class Solution(object):

def threeSum(self, nums):

res = []

nums.sort()

for i in range(len(nums)-2):

if nums[i]>0: break #[7]

if i>0 and nums[i]==nums[i-1]: continue

low,high = i+1,len(nums)-1

while low < high:

if nums[i]+nums[low]+nums[high]==0:

res.append([nums[i],nums[low],nums[high]])

while low<high and nums[low]==nums[low+1]:

low+=1

while low<high and nums[high]==nums[high-1]:

high-=1

low+=1

high-=1

elif nums[i]+nums[low]+nums[high]<0:

low+=1

else:

high-=1

return res

click here for code

Try Problem in leetcode

Happy Coding 😃!

Island Perimeter LeetCode Solution

2020-07-08T09:02:00.000+05:30

Quoestion:ClickHere

Approach:

if value in the box is 1 then add 4 to result and check for adjacent boxes if they are also 1 if yes then decrement 2 in the result.because boxes connect using the perimeter and each box contibutes 1 so we reduce 1 + 1 in the result.

0	1	0
1	1	0
0	0	1
1	0	1

Algorithm:

1.Intialize res variable

2.Loop from i:1 to rows

.loop from j:1 to columns

.if grid[i][j] =1 then,

.res= res + 4

.if i >0 and grid[i-1][j]=1 then,

.res= res- 2

.if j >0 and grid[i][j-1]=1 then,

.res= res- 2

3.return res

Python Code:

class Solution(object):
    def islandPerimeter(self, grid):
        res = 0
        for i in range(len(grid)):
            for j in range(len(grid[i])):
                if grid[i][j]==1:
                    res +=4
                    if i>0 and grid[i-1][j]==1:
                        res-=2
                    if j>0 and grid[i][j-1]==1:
                        res-=2
        return res

Shortened Python Code:

class Solution(object):

def islandPerimeter(self, g):

res = 0

for i, p in enumerate(g):

for j, q in enumerate(p):

if q:

res += 4

if i > 0 and g[i-1][j]: res -= 2

if j > 0 and g[i][j-1]: res -= 2

return res

Try Problem in LeetCode

Plus One LeetCode Solution

2020-07-07T09:48:00.000+05:30

Quoestion:Click here

Python Code:

class Solution(object):
    def plusOne(self, digits):
        for i in range(len(digits)-1,-1,-1):
            
            if digits[i]==9:
                digits[i]=0
            else:
                digits[i]=digits[i]+ 1
                return digits
        newarr=[0]*(len(digits)+1)
        newarr[0]=1
        return newarr

Hamming Distance LeetCode Solution Brian Kernighan's Algorithm

2020-07-05T21:26:00.003+05:30

Quoestion:Click Here

Approach:

we know that XOR operation return 1 when both bits are different,So we use it in finding hamming distance.

Example: x = 1,y = 4 if we perform XOR operation on this we get 0 1 0 1

Output:No of 1's in XOR result

So if we count No of 1's in XOR result then it will be the Hamming distance.

There are many methods to count no of 1's in a binary number out of these most optimized method is Brian Kernighan's Algorithm

Brian Kernighan's Algorithm:

Step1:Intialize integer variable,distance

Step2:while number !=0

---Turn-off the rightmost 1 bit by performing below operation

--num = num&(num-1)

---increment distance

Step3:return distance

Optimization:Rather than performing iterations for all the bits,we reduced it to only the small number of bits.

Time Complexity:O(1)

Space Complexity:O(1)

Python Code:

#Brian Kernighan's Algorithm

class Solution:

def hammingDistance(self, x: int, y: int) -> int:

num = x^y

if num==0:

return 0

dist = 0

while True:

num = num & (num-1)

dist+=1

if num==0:

return dist

Python One Liner:

class Solution:

def hammingDistance(self, x: int, y: int) -> int:

return bin(x^y).count('1')

Python Naive Approach:

class Solution(object):

def hammingDistance(self, x, y):

x = [i for i in str(bin(x))]

x = x[2:]

y = [i for i in str(bin(y))]

y = y[2:]

diff = abs(len(x)-len(y))

if len(x) > len(y):

for i in range(diff):

y.insert(i,'0')

else:

for i in range(diff):

x.insert(i,'0')

count = 0

for i in range(len(x)):

if x[i]==y[i]:

continue

else:

count+=1

return count

Try Problem in LeetCode

Ugly Number II LeetCode Solution

2020-07-04T19:31:00.013+05:30

Quoestion: Click Here

If we pick a naive approach we will end up getting time exceeded error.So,we will solve this problem more efficiently Using Dynamic Programming

Algorithm:

1.Intialize an array result of size n

2.Update result[0] = 1

3.Intialize pointers p2,p3 and p5 to 0 (pointer for value 2,3 and 5)

4.Loop from i:1 to N

=>find the minimum of three value (2*result[p2],3*result[p3] and 5*result[p5])

=>put the value at the index i

=>if minimum = 2*result[p2], then increment p2.

=>if minimum = 3*result[p3], then increment p3.

=>if minimum = 5*result[p5], then increment p5.

5.return last value of result array.

Python Code:

class Solution(object):

def nthUglyNumber(self, n):

res = []

res.append(1)

p2 = 0

p3 = 0

p5 = 0

for i in range(1,n+1):

minimum = min(res[p2]*2,res[p3]*3,res[p5]*5)

res.append(minimum)

if minimum == res[p2]*2:p2+=1

if minimum == res[p3]*3:p3+=1

if minimum == res[p5]*5:p5+=1

return res[n-1]

Try Problem In LeetCode

Prison Cells After N Days LeetCode Solution

2020-07-04T10:34:00.003+05:30

Quoestion: Click Here

Input: cells = [0,1,0,1,1,0,0,1], N = 7

Output: [0,0,1,1,0,0,0,0]

Given 1<= N <= 10^9

Approach:

If we repeat the process for atmost 10^9 times we get time limit exceeded error.

so,we need to optimize

If we Look at the above picture It is known that it's repeating after every 14 days and it is making a cycle.So we change N by N = N%14.So that no of iterations can be decreased and code runs faster.

Algorithm:

=>Update N to N % 14,if N==0 then update N = 14.

=>Loop from 1 to N

=>Intialize a temp array of size 8

=>Loop from 1 to 6

=>if cells[i-1]== cells[i+1]:

=>add 1 to temp

=>else:

=>add 0 to temp

=>add 0 to temp at end

=> update cell = temp

=>return cell

Python Code:

class Solution(object):
    def prisonAfterNDays(self, cells, N):
        N = N % 14
        if N==0:
            N = 14
        while N:
            lst = []
            lst.append(0)
            for i in range(1,len(cells)-1):
                if cells[i-1]== cells[i+1]:
                    lst.append(1)
                else:
                    lst.append(0)
            lst.append(0)
            cells = lst
            N-=1
        return cells

Try Problem in LeetCode

Binary Tree Level Order Traversal II Solution

2020-07-03T08:38:00.003+05:30

For example:
Given binary tree
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[[15,7],[9,20],[3] ]

Approach:
1.Add root to queue
2.pop item in queue and add left and right nodes of that item to queue.now add value of popped to a temporary array
3.repeat the above step len(queue) no of times
4.Add that temporary to result array
5.if size of queue !=0 go to step 2
6.return reversed result array

Python Code:
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution(object):
def levelOrderBottom(self, root):
if not root :
return []
queue = []
result = []
queue.append(root)
while len(queue)>0:
temp = []
for i in range(len(queue)):
node = queue.pop(0)
temp.append(node.val)
if node.left!=None: queue.append(node.left)
if node.right!=None: queue.append(node.right)
result.insert(0,temp)

return result

Try Quoeston Here

Arranging Coins LeetCode Solution

2020-07-02T09:16:00.003+05:30

Quoestion: Click Here

Approach:

If we take a close look at problem we need to find the greatest m,such that m(m+1)/2 <= n or m^2 + m - 2n <= 0. This is quadriatic inequality,to solve we should find roots of m^2 + m - 2n = 0.

_m1,2=⟨-1±√1+8n⟩/2 or m >= (2n +0.25)½ - 0.5

Complexity:

Time:O(1)
Space:O(1)

Pyhton Code:

Class Solution:

def arrangeCoins(self,n):

return int(sqrt(2*n + 0.25) - 0.5)

Second Approach:

we add arrange i coins in kth row and decrease n by k (n=n-i) until i >= n and finally we return i

Python Code:

def arrangeCoins(self, n):

if n==0:

return n

i =0

while i<n:

n-=(i+1)

i+=1

return i

Try Problem here

Best YouTube Channels To Learn Data Science and Machine Learning For Free

2020-06-27T18:53:00.011+05:30

There are Several Online Platforms to learn Data Science.But for those who are looking to learn Data Science for free YouTube is the best platform.I strongly recommend to follow these channels if you are heading towards Data Science.

The Channels that I mentioned here also contains mathematics channels because it is also a part of Data science.
("Below Red Coloured words are Links to channels")

1. CodeBasics
This Channel has all playlists that are required for data science.He also included some real world data science projects.The way he teaches is very interactive. Even if you are noob at programming he has python playlist from scratch.

2. DataSchool
If you want to understand functionality behind the library functions,what they will do and how do they work then this channel is a very good choice.

3. Data Professor
If you are completely from a non technical background then this channel is for you.

4. Ken Jee
This channel mainly focuses on real world data science projects

5. Sentdex

Best channel to learn Data Science in depth.It includes almost all stuff that you can do with python

6.Cory Schafer
Best for learning Python Libraries
7.Andreas Kretz

Building up big data platforms. Figuring out how to ingest, process and store data. Enabling Data Scientists to do awesome stuff for customers .That's what data engineering, the plumbing of data science is all about. This channel help you get into this awesome job.

8.StatQuest with Josh Starmer

StatQuest breaks down complicated Statistics and Machine Learning methods into small, bite-sized pieces that are easy to understand.

9.Krish Naik
He is self motivated to learn data science.He explains the concepts very clearly so that a beginner can easily understand.

10.3Blue1Brown

All tutorials are made in animated format so that we can understand even much better .This channel contain mathematics vedeos that make math at your finger tips,you will definitely love math if you watch these vedoes and also it is best channel to learn Neural networks,Deep Learning.

.11. Python Programmer

12. Lex Fridman

13.Abhishek_Thakur

14. RichardOnData

15.Data Science Dojo

Do Share Some of your best Channels that you found.

How to concatenate multiple CSV files in one single CSV with Python

2020-06-24T22:51:00.000+05:30

We will Come across this case when we have to analyze data from different csv files.
The task is to read and concatenate all csv files in same directory.

First we should import numpy,pandas,os libraries

The Following code will list the all the csv file iin the directory

import pandas as pd
import numpy as np
import os
files = [file for file in os.listdir(r'C:\Users\sarad\1DataScience\Pandas-Data-Science-Tasks-master\SalesAnalysis\Sales_Data')]
for i in files:
print(i)
with the following code we get final concatenated dataframe

for file in files:
df = pd.read_csv('./Pandas-Data-Science-Tasks-master/SalesAnalysis/Sales_Data/'+file) #adding file name at the end of the path
result_df = pd.concat([result_df,df]) #concatenating previous and current data Frame

result_df.head()

Final Output:

Code:

Unique Binary Search Trees LeetCode Solution

2020-06-24T20:00:00.004+05:30

1 3 3 2 1
\ / / / \ \
3 2 1 1 3 2
/ / \ \

2 1 2 3
Approach:
Let us consider result[i] be the number of unique binary search trees for i. The number of trees are determined by the number of subtrees which have different root node. For example,

i=0, result[0]=1 //empty tree

i=1, result[1]=1 //one tree

i=2, result[2]=result[0]*result[1] // 0 is root
+ result[1]*result[0] // 1 is root

i=3, result[3]=result[0]*result[2] // 1 is root
+ result[1]*result[1] // 2 is root
+ result[2]*result[0] // 3 is root

i=4, result[4]=result[0]*result[3] // 1 is root
+ result[1]*result[2] // 2 is root
+ result[2]*result[1] // 3 is root
+ result[3]*result[0] // 4 is root
..
..
..

i=n, result[n] = sum(result[0..k]*result[k+1...n]) 0 <= k < n-1

Python Code:

class Solution(object):
def numTrees(self, n):
result = [0]*(n+1)
result[0],result[1] = 1,1
for i in range(2,n+1):
for j in range(0,i):
result[i] = result[i]+result[j]*result[i - j - 1]

return result[n]
Java Code:

class Solution {
public int numTrees(int n) {
int[] result = new int[n + 1];

result[0] = 1;
result[1] = 1;

for (int i = 2; i <= n; i++) {
for (int j = 0; j <= i - 1; j++) {
result[i] = result[i] + result[j] * result[i - j - 1];
}
}

return result[n];
}

}

Count Complete Tree Nodes Solution LeetCode

2020-06-23T18:03:00.001+05:30

Quoestion:Given a complete binary tree, count the number of nodes

Example1:
input:[1,2,3,4,5,6]
output:6
Example2:
input:[]
output:0

Algorithm:
Traverse through BST after visiting each node increase the count by 1 when we reach leaf node return count

You can follow any of the traversal algorithm.inorder is recomended

Python:

# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def __init__(self):
self.c = 0
def inorder(self,root):
if not root:
return self.c
if root.left:
self.inorder(root.left)
self.c+=1
# self.result.append(root.val)
if root.right:
self.inorder(root.right)
def countNodes(self, root: TreeNode) -> int:
self.inorder(root)
return self.c
One Liner Python Code:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def countNodes(self, root: TreeNode) -> int:
if not root:
return 0

return 1+self.countNodes(root.left)+self.countNodes(root.right)

Single Number II Solution LeetCode June Day 22

2020-06-22T17:06:00.001+05:30

Quoestion:Given a non-empty array of integers, every element appears three times except for one, which appears exactly once. Find that single one.

Python Code:
from collections import defaultdict
class Solution:
def singleNumber(self, nums: List[int]) -> int:
d = defaultdict(int)
for i in nums:
d[i]+=1
for i,j in d.items():
if j==1:

return i

Dungeon Game Solution Using DP LeetCode June Day 21

2020-06-21T14:56:00.002+05:30

Quoestion: Click Here

Approach:
After observing the problem, we can find that there is one requirement. knight's health must >= 1. since the goal is to find the minimum health. A intuitive thought is to traverse from destination, from each cell compute the minimum health required to enter such cell. then back propagate to top will get us muimum health required

Python Code:(Using DP):

class Solution:

def calculateMinimumHP(self, dungeon):

R, C = len(dungeon), len(dungeon[0])

dp = [[0] * C for _ in range(R)]

dp[-1][-1] = max(1 - dungeon[-1][-1], 1)

for i in range(R - 2, -1, -1):

dp[i][C - 1] = max(dp[i + 1][C - 1] - dungeon[i][C - 1], 1)

for i in range(C - 2, -1, -1):

dp[R - 1][i] = max(dp[R - 1][i + 1] - dungeon[R - 1][i], 1)

for i in range(R - 2, -1, -1):

for j in range(C - 2, -1, -1):

# if adding health is much larger than health required, set it to 1

dp[i][j] = max(min(dp[i + 1][j], dp[i][j + 1]) - dungeon[i][j], 1)

return dp[0][0]

Longest Duplicate Substring Python Solution LeetCode June Day 19

2020-06-19T19:06:00.005+05:30

Quoestion: ClickHere

Python Code:

from collections import defaultdict

prime = 10 ** 9 + 7
base = 26

class Solution(object):
def longestDupSubstring(self, S):
"""
:type S: str
:rtype: str
"""

def helper(s, n):
max_base = pow(base, n - 1, prime)
cur = 0
for i in s[:n]:
cur = (cur * base + ord(i)) % prime
dic = defaultdict(list)
dic[cur].append(n - 1)

for j, i in enumerate(s[n:]):
cur -= ord(s[j]) * max_base
cur = (cur * base + ord(i)) % prime
if cur in dic:
st = s[j + 1:j + n + 1]
for v in dic[cur]:
if st == s[v - n + 1:v + 1]:
return st
dic[cur].append(j + n)
return ""

low = 0
high = len(S)

res = ""
while low + 1 < high:
mid = (low + high) // 2
v = helper(S, mid)
if v:
res = v
low = mid
else:
high = mid

return res

Search in a Binary Search Tree LeetCode Challenge June Day 15

2020-06-15T16:25:00.003+05:30

Quoestion: ClickHere

Example:
Given Tree

You Should Return This:

Recursive Approach:
1.Check if root's value is equal to search value or not if yes return root
2.if search value is > root's value update root as root.right
3.if search value is < root's value update root as root.left
4.Repeat above steps until leaf node
5.If search value is not found then return None

Python Code:
class Solution:

def searchBST(self, root: TreeNode, value: int) -> TreeNode:
print(root.val)
if root.val == value:
return root
if root.val < value:
if root.right:
return self.searchBST(root.right,value)
if root.val > value:
if root.left:
return self.searchBST(root.left,value)

return None

Remove Nth Node From End of List LeetCode Solution

2020-06-15T15:29:00.002+05:30

Quoestion: ClickHere

Algorithm:
step1:intialize slow and fast pointers with head
step2:Move forward fast pointer by n positions
step3:if fast.next is None then return head.next
step4:Move forward fast and slow pointers by one step until fast.next == None.
step5:slow.next = slow.next.next

Python Code
class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
slow = fast = head
for i in range(n):
print(fast.val)
fast = fast.next
if fast is None:
return head.next
while fast.next:
print(fast.val,slow.val)
fast= fast.next
slow = slow.next
slow.next = slow.next.next
return head

Intersection of Two Linked Lists LeetCode And GeeksForGeeks Solution

2020-06-15T10:35:00.002+05:30

Quoestion:Find the point of intersection of two singly linked lists begins.
Algorithm:
1.Calculate the size of both linked lists as L1,L2
2.Find the absolute difference of L1,L2 as d = L1-L2
3.Move d nodes in longer linked list
4.Then move by one step in both Linked Lists and Check if p==q if they are eaqual return p

Python Code:
# Definition for singly-linked list.
class ListNode:
def __init__(self, x):
self.val = x
self.next = None

class Solution:
def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
l1size,l2size = 0,0
node = headA
while node:
l1size+=1
node = node.next
node = headB
while node:
node = node.next
l2size+=1
d = abs(l1size-l2size)
# print(d,l1size,l2size)
if l1size>l2size:
for i in range(d):
headA = headA.next
# print(headA.val)
else:
for i in range(d):
headB = headB.next

while True:
if headA == headB:
return headA

if headA:
headA = headA.next
else:
break
if headB:
headB = headB.next
else:
break
# print(headA.val,headB.val)
return