Last updated on January 19th, 2025 at 10:46 pm
Here, we see a Merge Sorted Array LeetCode Solution. This Leetcode problem is solved using different approaches in many programming languages, such as C++, Java, JavaScript, Python, etc.
List of all LeetCode Solution
Topics
Array, Two-Pointers
Companies
Bloomberg, Facebook, Microsoft
Level of Question
Easy

Merge Sorted Array LeetCode Solution
Table of Contents
1. 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.
2. Coding Pattern Used in Solution
The coding pattern used in the provided code is “Two Pointers”. This pattern is commonly used when dealing with sorted arrays or lists, where two pointers are used to traverse the arrays from different directions (e.g., from the end or the beginning) to perform operations like merging, comparing, or finding specific elements.
3. Code Implementation in Different Languages
3.1 Merge Sorted Array C++
class Solution { public: void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) { int i=m-1,j=n-1,k=m+n-1; while(i>=0&&j>=0) { if(nums1[i]>nums2[j]) { nums1[k]=nums1[i]; i--; k--; } else { nums1[k]=nums2[j]; j--; k--; } } while(i>=0) nums1[k--]=nums1[i--]; while(j>=0) nums1[k--]=nums2[j--]; } };
3.2 Merge Sorted Array Java
class Solution { public void merge(int[] nums1, int m, int[] nums2, int n) { int i = m - 1 , j = n - 1; int k = m + n - 1; while(i >= 0 && j >= 0) { if(nums1[i] >= nums2[j]) { nums1[k] = nums1[i]; i--; } else { nums1[k] = nums2[j]; j--; } k--; } while(i >= 0) nums1[k--] = nums1[i--]; while(j >= 0) nums1[k--] = nums2[j--]; return; } }
3.3 Merge Sorted Array JavaScript
var merge = function(nums1, m, nums2, n) { nums1.splice(m, nums1.length - m); var i = 0; var j = 0; while (j < nums2.length) { if (nums1[i] === undefined || nums1[i] > nums2[j]) { nums1.splice(i, 0, nums2[j]); j++; i++; } else { i++; } } };
3.4 Merge Sorted Array Python
class Solution(object): def merge(self, nums1, m, nums2, n): while m > 0 and n > 0: if nums1[m - 1] > nums2[n - 1]: nums1[m + n - 1] = nums1[m - 1] m -= 1 else: nums1[m + n - 1] = nums2[n - 1] n -= 1 nums1[:n] = nums2[:n]
4. Time and Space Complexity
Time Complexity | Space Complexity | |
C++ | O(m + n) | O(1) |
Java | O(m + n) | O(1) |
JavaScript | O(m + n) | O(n) |
Python | O(m + n) | O(1) |
- The C++, Java, and Python implementations are efficient with O(m + n) time complexity and O(1) space complexity.
- However, the JavaScript implementation is less efficient due to the use of
splice
, which increases the time complexity to O(m * n) and space complexity to O(n).