LeetCode-in-All

4. Median of Two Sorted Arrays

Hard

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.

Example 3:

Input: nums1 = [0,0], nums2 = [0,0]

Output: 0.00000

Example 4:

Input: nums1 = [], nums2 = [1]

Output: 1.00000

Example 5:

Input: nums1 = [2], nums2 = []

Output: 2.00000

Constraints:

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

Solution

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number}
 */
var findMedianSortedArrays = function (nums1, nums2) {
    if (nums2.length < nums1.length) {
        return findMedianSortedArrays(nums2, nums1)
    }

    let n1 = nums1.length,
        n2 = nums2.length
    let low = 0,
        high = n1

    while (low <= high) {
        let cut1 = Math.floor((low + high) / 2)
        let cut2 = Math.floor((n1 + n2 + 1) / 2) - cut1

        let l1 = cut1 === 0 ? -Infinity : nums1[cut1 - 1]
        let l2 = cut2 === 0 ? -Infinity : nums2[cut2 - 1]
        let r1 = cut1 === n1 ? Infinity : nums1[cut1]
        let r2 = cut2 === n2 ? Infinity : nums2[cut2]

        if (l1 <= r2 && l2 <= r1) {
            if ((n1 + n2) % 2 === 0) {
                return (Math.max(l1, l2) + Math.min(r1, r2)) / 2.0
            }
            return Math.max(l1, l2)
        } else if (l1 > r2) {
            high = cut1 - 1
        } else {
            low = cut1 + 1
        }
    }

    return 0.0
}

export { findMedianSortedArrays }

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