# Merge Sorted Array Leetcode Solution

In this post, we are going to solve the Merge Sorted Array Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

## Problem

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.
```

Constraints:

• `nums1.length == m + n`
• `nums2.length == n`
• `0 <= m, n <= 200`
• `1 <= m + n <= 200`
• `-109 <= nums1[i], nums2[j] <= 109`

Follow up: Can you come up with an algorithm that runs in `O(m + n)` time?

Now, lets see the leetcode solution of Merge Sorted Array Leetcode Solution.

### Merge Sorted ArrayLeetcode Solution in Python

```class Solution:
def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
i = m - 1      # nums1's index (actual nums)
j = n - 1      # nums2's index
k = m + n - 1  # nums1's index (next filled position)

while j >= 0:
if i >= 0 and nums1[i] > nums2[j]:
nums1[k] = nums1[i]
k -= 1
i -= 1
else:
nums1[k] = nums2[j]
k -= 1
j -= 1
```

### Merge Sorted Array Leetcode Solutionin CPP

```class Solution {
public:
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
int i = m - 1;      // nums1's index (actual nums)
int j = n - 1;      // nums2's index
int k = m + n - 1;  // nums1's index (next filled position)

while (j >= 0)
if (i >= 0 && nums1[i] > nums2[j])
nums1[k--] = nums1[i--];
else
nums1[k--] = nums2[j--];
}
};
```

### Merge Sorted Array Leetcode Solution in Java

```class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
int i = m - 1;     // nums1's index (actual nums)
int j = n - 1;     // nums2's index
int k = m + n - 1; // nums1's index (next filled position)

while (j >= 0)
if (i >= 0 && nums1[i] > nums2[j])
nums1[k--] = nums1[i--];
else
nums1[k--] = nums2[j--];
}
}
```

Note: This problem Merge Sorted Array is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.

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