# 88. Merge Sorted Array - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions
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LeetCode link: [88. Merge Sorted Array](https://leetcode.com/problems/merge-sorted-array), difficulty: **Easy**.
## LeetCode description of "88. 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 = [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`
Follow up: Can you come up with an algorithm that runs in `O(m + n)` time?
## Intuition
- This problem can be solved by using the built-in `sort()` method of a programming language to sort the array, which is very simple to implement. However, the time complexity of this approach is `O(n log n)`.
- That’s why this problem usually restricts the time complexity to `O(n)`. How can we achieve `O(n)`?
- We need to consider the given conditions. Both arrays are already sorted, and this condition must be leveraged.
- When reorganizing the arrays, we can either place the numbers in order from left to right, or from right to left. Which direction should we choose?
- If you don’t ask yourself this question, you’d normally just go left to right. But in that case, a problem arises: if the current number in `nums1` is not the smallest, it can’t be directly replaced and needs to be stored somewhere temporarily. A simple workaround is to create a new empty array to hold the final result, and then copy that array back into `nums1`. But this increases the amount of code you need to write.
- Therefore, it’s better to place the numbers in order from right to left.
## Complexity
- Time complexity: `m + n`.
- Space complexity: `m + n`.
## Python
```python
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.
"""
# Make a good example first.
# nums1: [1, 5, 7, 0, 0, 0, 0]
# nums2: [2, 3, 4, 8]
# result:[1, 2, 3, 4, 5, 7, 8]
i = m - 1
j = n - 1
for k in range(len(nums1) - 1, -1, -1):
if i < 0 or (j >= 0 and nums2[j] > nums1[i]):
nums1[k] = nums2[j]
j -= 1
else:
nums1[k] = nums1[i]
i -= 1
```
## Ruby
```ruby
# @param {Integer[]} nums1
# @param {Integer} m
# @param {Integer[]} nums2
# @param {Integer} n
# @return {Void} Do not return anything, modify nums1 in-place instead.
def merge(nums1, m, nums2, n)
# Make a good example first.
# nums1: [1, 5, 7, 0, 0, 0, 0]
# nums2: [2, 3, 4, 8]
# result:[1, 2, 3, 4, 5, 7, 8]
i = m - 1
j = n - 1
(nums1.size - 1).downto(0) do |k|
if i < 0 || (j >= 0 and nums2[j] > nums1[i])
nums1[k] = nums2[j]
j -= 1
else
nums1[k] = nums1[i]
i -= 1
end
end
end
```
## Java
```java
class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
// Make a good example first.
// nums1: [1, 5, 7, 0, 0, 0, 0]
// nums2: [2, 3, 4, 8]
// result:[1, 2, 3, 4, 5, 7, 8]
var i = m - 1;
var j = n - 1;
for (var k = nums1.length - 1; k >= 0; k--) {
if (i < 0 || (j >= 0 && nums2[j] > nums1[i])) {
nums1[k] = nums2[j];
j--;
} else {
nums1[k] = nums1[i];
i--;
}
}
}
}
```
## Other languages
```java
// Welcome to create a PR to complete the code of this language, thanks!
```
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Original link: [88. Merge Sorted Array - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions](https://leetcode.to/en/leetcode/88-merge-sorted-array).
GitHub repository: [leetcode-python-java](https://github.com/leetcode-python-java/leetcode-python-java).