# LeetCode: 88. Merge Sorted Array

**The 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`

.

**Examples**

```
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`

`-10`

^{9}<= nums1[i], nums2[j] <= 10^{9}

**Solution**

We opted for a simple approach using a **while** loop

```
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
int cur = nums1.size();
while(m > 0 && n>0){
if (nums1[m - 1] >= nums2[n - 1]) {
nums1[--cur] = nums1[--m];
} else {
nums1[--cur] = nums2[--n];
}
}
while(n > 0) nums1[--cur] = nums2[--n];
}
```

Here's what each part of the code does:

**Function Declaration:**

`void merge(vector<int>& nums1, int m, vector<int>& nums2, int n)`

: This function takes two sorted integer arrays`nums1`

and`nums2`

with sizes`m`

and`n`

respectively. It merges`nums2`

into`nums1`

while maintaining the sorted order.

**Initialization:**

`int cur = nums1.size();`

: Initialize the variable`cur`

to the size of`nums1`

. This variable will be used to index the elements being merged into`nums1`

.

**Merging Process:**

- The loop iterates as long as there are elements remaining in both
`nums1`

and`nums2`

: - It compares the last elements of
`nums1`

and`nums2`

(`nums1[m - 1]`

and`nums2[n - 1]`

respectively). - If the last element of
`nums1`

is greater than or equal to the last element of`nums2`

, it moves the last element of`nums1`

to the current index in`nums1`

and decrements`m`

. - Otherwise, it moves the last element of
`nums2`

to the current index in`nums1`

and decrements`n`

. - This process continues until one of the arrays is fully processed.

**Remaining Elements:**

- After the first loop, if there are remaining elements in
`nums2`

, the second loop copies them to`nums1`

. - This loop decrements
`n`

and continues until all elements in`nums2`

are copied to`nums1`

.