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

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

 


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


class Solution {
    public void merge(int[] nums1, int m, int[] nums2, int n) {

     int [] result= new int [m+n];
     int i=0,j=0,k=0;

     while(i < m && j<n)
     {
     if(nums1[i] < nums2[j])
     {
result[k++]=nums1[i++];
     }
     else
     {
result[k++]=nums2[j++];
     }
     }

     while(i <m)
     {
        result[k++]=nums1[i++];
     }
     while(j<n)
     {
        result[k++]=nums2[j++];
     }

//return Arrays.toString(Arrays.copyOfRange(result, 0, result.length - 1));
for(int x=0;x<m+n;x++)
{
    nums1[x]=result[x];
}
//return nums1;

    }
}

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