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# 88. Merge Sorted Array

## Problem Statement

<br>

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`.

&#x20;

**Example 1:**

<pre><code><strong>Input: nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3
</strong><strong>Output: [1,2,2,3,5,6]
</strong><strong>Explanation: The arrays we are merging are [1,2,3] and [2,5,6].
</strong>The result of the merge is [1,2,2,3,5,6] with the underlined elements coming from nums1.
</code></pre>

**Example 2:**

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

**Example 3:**

<pre><code><strong>Input: nums1 = [0], m = 0, nums2 = [1], n = 1
</strong><strong>Output: [1]
</strong><strong>Explanation: The arrays we are merging are [] and [1].
</strong>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.
</code></pre>

&#x20;

**Constraints:**

* `nums1.length == m + n`
* `nums2.length == n`
* `0 <= m, n <= 200`
* `1 <= m + n <= 200`
* `-109 <= nums1[i], nums2[j] <= 109`

## Intuition

```
Approach:

One can do in extra space, sorting the array

But we want no extra space and O(m+n)

So we start the pointers from behind i,j
We want to exhaust j pointer of second array, J exhausted means, All went to correct 
Position


```

### Links

<https://leetcode.com/problems/merge-sorted-array/description/>

### Video Links

### Approach 1:

```
Pointers
```

{% code title="C++" lineNumbers="true" %}

```cpp
class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
        int i=m-1, j=n-1, k=m+n-1;

        while(i>=0 and j>=0){
            if(nums1[i] > nums2[j])
                nums1[k--] = nums1[i--];
            else
                nums1[k--] = nums2[j--];
        }

        while(j>=0)
            nums1[k--] = nums2[j--];
    }
};
```

{% endcode %}

### Approach 2:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 3:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 4:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Similar Problems

###