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:

Solution

#include <vector>
#include <algorithm>
#include <climits>

class Solution {
public:
    double findMedianSortedArrays(std::vector<int>& nums1, std::vector<int>& nums2) {
        if (nums2.size() < nums1.size()) {
            return findMedianSortedArrays(nums2, nums1);
        }
        
        int n1 = nums1.size();
        int n2 = nums2.size();
        int low = 0;
        int high = n1;
        
        while (low <= high) {
            int cut1 = (low + high) / 2;
            int cut2 = (n1 + n2 + 1) / 2 - cut1;
            
            int l1 = (cut1 == 0) ? INT_MIN : nums1[cut1 - 1];
            int l2 = (cut2 == 0) ? INT_MIN : nums2[cut2 - 1];
            int r1 = (cut1 == n1) ? INT_MAX : nums1[cut1];
            int r2 = (cut2 == n2) ? INT_MAX : nums2[cut2];
            
            if (l1 <= r2 && l2 <= r1) {
                if ((n1 + n2) % 2 == 0) {
                    return (std::max(l1, l2) + std::min(r1, r2)) / 2.0;
                }
                return std::max(l1, l2);
            } else if (l1 > r2) {
                high = cut1 - 1;
            } else {
                low = cut1 + 1;
            }
        }
        
        return 0.0;
    }
};