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.