# Merge Sorted Array Problem

## Description

LeetCode Problem 88.

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:

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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:

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

Example 3:

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

## Sample C++ Code

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