Problem Statement:-

*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.*

**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*

##
__Solution 1__

sort the `num1`

array by swapping num2 array elements, while swapping keep sorting num2, then add all num2 elements to num1 array.

lets understand this approach step by step.

**step-1** Initialise two varible `pointer1 = 0`

& `pointer2 = 0`

.

**step-2** if `n == 0`

, then return.

**step-3** run a loop from `i=1`

to `m`

,then

if

`num1[pointer1] > num2[pointer2]`

,then

1.swap(num1[pointer1],num2[pointer2]).

2.sort the`num2`

array.

3.increment`pointer1`

++else increment

`pointer1`

++

**step-4** again run a loop from i=1 to n,then

1.`num1[pointer1] = num2[pointer2]`

2.increment the`pointer2`

++.

**step-5** end.

Java

```
class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
int pointer1=0, pointer2=0;
if(n==0) return;
for(int i=0; i<m;i++){
if(nums1[pointer1] > nums2[pointer2]){
int temp = nums1[pointer1];
nums1[pointer1] = nums2[pointer2];
nums2[pointer2] = temp;
Arrays.sort(nums2);
pointer1++;
}else
pointer1++;
}
for(int i=1;i<=n;i++){
nums1[pointer1] = nums2[pointer2];
pointer1++;
pointer2++;
}
}
}
```

##
__Solution 2__

by reverse sorting, this problem can be solved in linear time complexity.

**step-1** initialise variables `i=m-1`

, `j=n-1`

, `arr1Len=nums1.length-1`

.

**step-2** run a loop if `i>=0 and j>=0`

, then

if nums1[i] > nums2[j], then

update the value from the last index.

`nums1[arr1Len] = nums[i];`

`i--`

`arr1Len--`

else

`nums1[arr1Len] = nums[j]`

`j--`

`arr1Len--`

**srep-3** run a loop if j >=0, then

`nums1[arr1Len] = nums[j]`

`arr1Len--`

`j--`

**step-4** end.

Java

```
class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
int i=m-1,j=n-1,arr1Len=nums1.length-1;
while(i>=0 && j>=0){
if(nums1[i] > nums2[j]){
nums1[arr1Len] = nums1[i];
i--;
arr1Len--;
}else{
nums1[arr1Len] = nums2[j];
j--;
arr1Len--;
}
}
while(j>=0){
nums1[arr1Len] = nums2[j];
j--;
arr1Len--;
}
}
}
```

**Time Complexity :** **O(m+n)**

**Space Complexity :** **O(1)**

Thank you for reading this article. if you find something wrong, let me know in the comment section.

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