Merge Sorted Arrays - Leetcode Solution

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

• -10<sup>9</sup> <= nums1[i], nums2[j] <= 10<sup>9</sup>

• Approach: Two Pointers

• Set two pointers in nums1, one at end and another at m - 1.

• Start traversing the nums2 from end and check if nums2[k] < nums1[i]. because both arrays are sorted in ascending order then shift the ith element to the jth position and continue looking till you not reached the 0th index of nums1.

• Or if the kth element >= the ith of the nums1 then put at jth position.

• If you reached the 0ht index of nums1 check if there is any element remaining to put if yes then put them all in nums 1 from the jth location.

TC: O(N + M) SC: O(1)

void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
    int i = m - 1, j = (m + n - 1), k = n - 1;
    while(k >= 0 && i >= 0){
        if(nums2[k] >= nums1[i])
            nums1[j--] = nums2[k--];
        else
            nums1[j--] = nums1[i--];
    }
    while(k >= 0)
        nums1[j--] = nums2[k--];
}