# Merge Sorted Array LeetCode Solution | Easy Approach

Share:

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?

Time: OO(m+n))
Space: O(1)

### C++

`````` Will be updated Soonclass 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--];
}
};
Will be updated Soon``````

### Java

``````
Will be updated Soonclass 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--];
}
}

Will be updated Soon``````

### Python

``````  Will beclass 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

``````

#### Watch Tutorial

Checkout more Solutions here