### Merge Two sorted array Without Extra Space

Problem statement-

Given two sorted arrays arr1[] and arr2[] of sizes N and M in non-decreasing order. Merge them in sorted order without using any extra space. Modify arr1 so that it contains the first N elements and modify arr2 so that it contains the last M elements.

Example 1:

Input:

N = 4, arr1[] = [1 3 5 7]

M = 5, arr2[] = [0 2 6 8 9]

Output:

arr1[] = [0 1 2 3]

arr2[] = [5 6 7 8 9]

Explanation: After merging the two

non-decreasing arrays, we get,

0 1 2 3 5 6 7 8 9.

Example 2:

Input:

N = 2, arr1[] = [10, 12]

M = 3, arr2[] = [5 18 20]

Output:

arr1[] = [5 10]

arr2[] = [12 18 20]

Explanation: After merging two sorted arrays

we get 5 10 12 18 20.

You don't need to read input or print anything. You only need to complete the function merge() that takes arr1, arr2, N and M as input parameters and modifies them in-place so that they look like the sorted merged array when concatenated.

Expected Time Complexity:  O((n+m) log(n+m))

Expected Auxilliary Space: O(1)

Constraints:

1 <= X, Y <= 5*104

0 <= arr1i, arr2i <= 109

### Solution-

There are 3 approach to solve this question.

1st the brute force approach in this case you will use one extra array of size(m+n ) and insert which element is smaller.

2nd-

you will iterate over array 1 and every time you will compare ith element with array2[0] element and if arr1[i] > arr2[0] you will swap them and apply insertion sort in

array 2.

3rd-

This is most efficient method

The idea: We start comparing elements that are far from each other rather than adjacent.

For every pass, we calculate the gap and compare the elements towards the right of the gap. Every pass, the gap reduces to the ceiling value of dividing by 2

credit - gfg

code-

### codeforces rating system | Codeforces rating Newbie to Legendary Grandmaster

Codeforces rating system | Codeforces rating Newbie to Legendary Grandmaster- Codeforces is one of the most popular platforms for competitive programmers and  codeforces rating matters a lot  .  Competitive Programming  teaches you to find the easiest solution in the quickest possible way. CP enhances your problem-solving and debugging skills giving you real-time fun. It's brain-sport. As you start solving harder and harder problems in live-contests your analytical and rational thinking intensifies. To have a good codeforces profile makes a good impression on the interviewer. If you have a good  codeforces profile so it is very easy to get a referral for product base company like amazon, google , facebook etc.So in this blog I have explained everything about codeforces rating system. What are different titles on codeforces- based on rating codeforces divide rating into 10 part. Newbie Pupil Specialist Expert Candidate Codemaster Master International Master Grandmaster Internat

### Apple Division CSES Problem Set Solution | CSES Problem Set Solution Apple division with code

Apple Division CSES Problem Set Solution | CSES Problem Set Solution Apple division with code - Apple Division CSES Problem Solution Easy Explanation. Apple division is problem is taken form cses introductory problem set.Let's Read Problem statement first. Problem Statement- Time limit:  1.00 s   Memory limit:  512 MB There are  n n  apples with known weights. Your task is to divide the apples into two groups so that the difference between the weights of the groups is minimal. Input The first input line has an integer  n n : the number of apples. The next line has  n n  integers  p 1 , p 2 , … , p n p 1 , p 2 , … , p n : the weight of each apple. Output Print one integer: the minimum difference between the weights of the groups. Constraints 1 ≤ n ≤ 20 1 ≤ n ≤ 20 1 ≤ p i ≤ 10 9 1 ≤ p i ≤ 10 9 Example Input: 5 3 2 7 4 1 Output: 1 Explanation: Group 1 has weights 2, 3 and 4 (total weight 9), and group 2 has weights 1 and 7 (total weight 8). Join Telegram channel for code discussi

### Movie Festival CSES problem set solution

Movie Festival CSES problem set solution - Problem statement-  Time limit:  1.00 s   Memory limit:  512 MB In a movie festival  n n  movies will be shown. You know the starting and ending time of each movie. What is the maximum number of movies you can watch entirely? Input The first input line has an integer  n n : the number of movies. After this, there are  n n  lines that describe the movies. Each line has two integers  a a  and  b b : the starting and ending times of a movie. Output Print one integer: the maximum number of movies. Constraints 1 ≤ n ≤ 2 ⋅ 10 5 1 ≤ n ≤ 2 ⋅ 10 5 1 ≤ a < b ≤ 10 9 1 ≤ a < b ≤ 10 9 Example Input: 3 3 5 4 9 5 8 Output: 2 Solution - Step -1 take input in a vector as a pair first element is ending time ans second element is starting time Step -2 sort the vector in increasing order because we will watch that movie which is ending first. Step -3 iterate over vector and calculate ans .           follow code below-