Merge Sorted Array | LeetCode # 88

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.

Constraints

• nums1.length == m + n
• nums2.length == n
• 0 <= m, n <= 200
• 1 <= m + n <= 200
• -109 <= nums1[i], nums2[j] <= 109

Example

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

We will solve this coding problem using multiple pointers.

• Set a pointer i to the last element of the first array.
• Set a pointer j to the last element of the second array.
• Set a pointer k to the last position of the merged array.

• While there are elements remaining in the second array (j >= 0):
  • If there are elements remaining in the first array (i >= 0) and the current element of the first array is greater than the current element of the second array:
    • Copy the current element from the first array to the current position of the merged array.
    • Move the pointer of the first array one step to the left.
    • Move the pointer of the merged array one step to the left.
  • else:
    • Copy the current element from the second array to the current position of the merged array.
    • Move the pointer of the second array one step to the left.
    • Move the pointer of the merged array one step to the left.

Step by step walkthrough of solution using given example is as follows.

Here is C++ implementation of solution.

class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
        int i = m - 1;
        int j = n - 1;
        int k = nums1.size() - 1;

        while(j >= 0) {
            if(i >= 0 && nums1[i] > nums2[j]) {
                nums1[k] = nums1[i];
                i--;
                k--;
            }
            else {
                nums1[k] = nums2[j];
                j--;
                k--;
            }
        }
    }
};

Time Complexity

O(m + n)

Space Complexity

O(1)

Thanks for reading and have a nice day.