Median of Two Sorted Arrays

Problem Description

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

Solution (JavaScript)

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number}
 */
var findMedianSortedArrays = function(nums1, nums2) {
    if(nums1.length > nums2.length) return findMedianSortedArrays(nums2, nums1);
    
    const x = nums1.length, y = nums2.length;
    let lo = 0, hi = x;
    
    while(lo <= hi) {
        let partitionX = (lo + hi) / 2 | 0,
            partitionY = (x + y + 1) / 2 - partitionX | 0;
        
        let maxLeftX = partitionX === 0 ? -Infinity : nums1[partitionX - 1];
        let minRightX = partitionX === x ? Infinity : nums1[partitionX];
        
        let maxLeftY = partitionY === 0 ? -Infinity : nums2[partitionY - 1];
        let minRightY = partitionY === y ? Infinity : nums2[partitionY];
        
        if(maxLeftX <= minRightY && maxLeftY <= minRightX) {
            if((x + y) & 1) return Math.max(maxLeftX, maxLeftY);
            return (Math.max(maxLeftX, maxLeftY) + Math.min(minRightX, minRightY)) / 2;
        } else if(maxLeftX > minRightY) {
            hi = partitionX - 1;
        } else {
            lo = partitionX + 1;
        }
    }
};