Median of Two Sorted Arrays LeetCode Solution Best & Easiest

Posted by admin February 15, 2023

In this post, we will solve the Median of Two Sorted Arrays LeetCode Solution.

Problem Statement:

Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays.

The overall run time complexity should be O(log (m+n)).

 

Example 1:

Input: nums1 = [1,3], nums2 = [2]
Output: 2.00000
Explanation: merged array = [1,2,3] and median is 2.

Example 2:

Input: nums1 = [1,2], nums2 = [3,4]
Output: 2.50000
Explanation: merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5.

 

Constraints:

  • nums1.length == m
  • nums2.length == n
  • 0 <= m <= 1000
  • 0 <= n <= 1000
  • 1 <= m + n <= 2000
  • -106 <= nums1[i], nums2[i] <= 106

Median of Two Sorted Arrays LeetCode Solution

Problem solution in JavaScript programming:

function findMedianSortedArrays(nums1, nums2) {
  const merged = nums1.concat(nums2).sort((a, b) => a - b);
  const length = merged.length;
  if (length % 2 === 0) {
    return (merged[length / 2 - 1] + merged[length / 2]) / 2;
  } else {
    return merged[Math.floor(length / 2)];
  }
}

Problem solution in Python programming:

class Solution:
    def findMedianSortedArrays(self, nums1, nums2):
        merged = sorted(nums1 + nums2)
        length = len(merged)
        if length % 2 == 0:
            return (merged[length // 2 - 1] + merged[length // 2]) / 2
        else:
            return merged[length // 2]

Problem solution in C++ programming:

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
       vector<int> merged;
        int i = 0, j = 0;
        while (i < nums1.size() && j < nums2.size()) {
            if (nums1[i] <= nums2[j]) {
                merged.push_back(nums1[i]);
                i++;
            } else {
                merged.push_back(nums2[j]);
                j++;
            }
        }
        while (i < nums1.size()) {
            merged.push_back(nums1[i]);
            i++;
        }
        while (j < nums2.size()) {
            merged.push_back(nums2[j]);
            j++;
        }
        int length = merged.size();
        if (length % 2 == 0) {
            return (merged[length / 2 - 1] + merged[length / 2]) / 2.0;
        } else {
            return merged[length / 2];
        }
     
    }
};