# Merge Sorted Array LeetCode Solution

Here, We see Merge Sorted Array LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

# List of all LeetCode Solution

## Problem Statement

You are given two integer arrays `nums1` and `nums2`, sorted in non-decreasing order, and two integers `m` and `n`, representing the number of elements in `nums1` and `nums2` respectively.

Merge `nums1` and `nums2` into a single array sorted in non-decreasing order.

The final sorted array should not be returned by the function, but instead be stored inside the array `nums1`. To accommodate this, `nums1` has a length of `m + n`, where the first `m` elements denote the elements that should be merged, and the last `n` elements are set to `0` and should be ignored. `nums2` has a length of `n`.

```Example 1:

Input: nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3
Output: [1,2,2,3,5,6]
Explanation: The arrays we are merging are [1,2,3] and [2,5,6].
The result of the merge is [1,2,2,3,5,6] with the underlined elements coming from nums1.

Example 2:

Input: nums1 = [1], m = 1, nums2 = [], n = 0
Output: [1]
Explanation: The arrays we are merging are [1] and [].
The result of the merge is [1].

Example 3:

Input: nums1 = [0], m = 0, nums2 = [1], n = 1
Output: [1]
Explanation: The arrays we are merging are [] and [1].
The result of the merge is [1].
Note that because m = 0, there are no elements in nums1. The 0 is only there to ensure the merge result can fit in nums1.
```

## Merge Sorted Array Leetcode Solution C++

``````class Solution {
public:
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
int i=m-1,j=n-1,k=m+n-1;
while(i>=0&&j>=0)
{
if(nums1[i]>nums2[j])
{
nums1[k]=nums1[i];
i--;
k--;
}
else
{
nums1[k]=nums2[j];
j--;
k--;
}
}
while(i>=0)
nums1[k--]=nums1[i--];
while(j>=0)
nums1[k--]=nums2[j--];
}
};```Code language: JavaScript (javascript)```

## Merge Sorted Array Leetcode Solution Java

``````class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
int i = m - 1 , j = n - 1;
int k = m + n - 1;
while(i >= 0 && j >= 0) {
if(nums1[i] >= nums2[j]) {
nums1[k] = nums1[i];
i--;
} else {
nums1[k] = nums2[j];
j--;
}
k--;
}
while(i >= 0)
nums1[k--] = nums1[i--];
while(j >= 0)
nums1[k--] = nums2[j--];
return;
}
}```Code language: JavaScript (javascript)```

## Merge Sorted Array Leetcode Solution JavaScript

``````var merge = function(nums1, m, nums2, n) {
nums1.splice(m, nums1.length - m);

var i = 0;
var j = 0;

while (j < nums2.length) {
if (nums1[i] === undefined || nums1[i] > nums2[j]) {
nums1.splice(i, 0, nums2[j]);
j++;
i++;
} else {
i++;
}
}
};```Code language: JavaScript (javascript)```

## Merge Sorted Array Leetcode Solution Python

``````class Solution(object):
def merge(self, nums1, m, nums2, n):
while m > 0 and n > 0:
if nums1[m - 1] > nums2[n - 1]:
nums1[m + n - 1] = nums1[m - 1]
m -= 1
else:
nums1[m + n - 1] = nums2[n - 1]
n -= 1
nums1[:n] = nums2[:n]``````
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