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.

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

Solution Approach

• We start from the end of both arrays to avoid overwriting the contents of nums1.
• nums1 has enough space (size m + n), so we start filling it from the back.
• Compare elements of nums1 and nums2, and place the larger one at the end.
• If nums2 still has elements left, copy them to the front of nums1.
• No need to copy remaining nums1 elements since they are already in place.

Example

nums1 = [1, 2, 3, 0, 0, 0, 0], m = 3
nums2 = [2, 4, 5, 6], n = 4

We need to merge both arrays into nums1, which has enough space (size = m + n = 7).

Final nums1

[1, 2, 2, 3, 4, 5, 6]

Step-by-Step Iterations

We initialize:

• i = m - 1 = 2 → points to 3 in nums1
• j = n - 1 = 3 → points to 6 in nums2
• k = m + n - 1 = 6 → last position in nums1

Iteration	i	j	k	nums1[i]	nums2[j]	Action	Updated nums1
1	2	3	6	3	6	6 > 3 → nums1[6] = 6	[1, 2, 3, 0, 0, 0, 6]
2	2	2	5	3	5	5 > 3 → nums1[5] = 5	[1, 2, 3, 0, 0, 5, 6]
3	2	1	4	3	4	4 > 3 → nums1[4] = 4	[1, 2, 3, 0, 4, 5, 6]
4	2	0	3	3	2	3 > 2 → nums1[3] = 3	[1, 2, 3, 3, 4, 5, 6]
5	1	0	2	2	2	2 == 2 → nums1[2] = 2	[1, 2, 2, 3, 4, 5, 6]
6	0	-1	1	1	—	j < 0, done	[1, 2, 2, 3, 4, 5, 6]

Java Implementation

import java.util.Arrays;

public class MergeSortedArray {

    public void merge(int[] nums1, int m, int[] nums2, int n) {
        int i = m - 1;            // Last index of actual nums1 elements
        int j = n - 1;            // Last index of nums2
        int k = m + n - 1;        // Last index of nums1

        // Merge from back to front
        while (i >= 0 && j >= 0) {
            if (nums1[i] > nums2[j]) {
                nums1[k--] = nums1[i--];
            } else {
                nums1[k--] = nums2[j--];
            }
        }

        // Copy remaining nums2 elements
        while (j >= 0) {
            nums1[k--] = nums2[j--];
        }
    }

    public static void main(String[] args) {
        MergeSortedArray merger = new MergeSortedArray();

        // Updated Test Case: nums2 has 4 elements
        int[] nums1 = {1, 2, 3, 0, 0, 0, 0};
        int[] nums2 = {2, 4, 5, 6};
        merger.merge(nums1, 3, nums2, 4);
        System.out.println("Merged Array: " + Arrays.toString(nums1)); // [1, 2, 2, 3, 4, 5, 6]
    }
}