# LeetCode 88. Merge Sorted Array

## Description

https://leetcode.com/problems/merge-sorted-array/

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

Example 3:

```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`
• `-109 <= nums1[i], nums2[j] <= 109`

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

## Explanation

The size of nums1 is m + n. nums1 is enough to store all numbers in nums1 and nums2. We only need to first shift all numbers in nums1 by n positions. Then we compare and put numbers in nums1 and nums2 into nums1 in ascending order.

## Python Solution

``````class Solution:
def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
"""
Do not return anything, modify nums1 in-place instead.
"""

for i in range(m - 1, -1, -1):
nums1[i + n] = nums1[i]

i = n
j = 0
k = 0
while i < m + n and j < n:
if nums1[i] < nums2[j]:
nums1[k] = nums1[i]
i += 1
else:
nums1[k] = nums2[j]
j += 1
k += 1

while j < n:
nums1[k] = nums2[j]
j += 1
k += 1``````
• Time complexity: O(m + n).
• Space complexity: O(m + n).

## 3 Thoughts to “LeetCode 88. Merge Sorted Array”

1. Vicky says:

class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
for (int i = m; i < (m+n); i++) {
nums1[i] = nums2[i – m];
}

Arrays.sort(nums1);
}
}

2. Vicky says:

/** Java solution */

public void merge(int[] nums1, int m, int[] nums2, int n) {
int insertP = m + n – 1;
int nums1P = m – 1;
int nums2P = n – 1;
while (nums1P >= 0 && nums2P >= 0){
if (nums1[nums1P] > nums2[nums2P]) {
nums1[insertP–] = nums1[nums1P–];
}
else {
nums1[insertP–] = nums2[nums2P–];
}
}
while (nums2P >= 0) {
nums1[insertP–] = nums2[nums2P–];
}
}