LeetCode 4. Median of Two Sorted Arrays

Binary Search

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

Follow up: 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

• -10^6 <= nums1[i], nums2[i] <= 10^6

Solution:

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class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        int m = nums1.size(), n = nums2.size();
        if ((m + n) % 2 == 1) {
            return findKthElement((m + n) / 2 + 1, nums1, 0, nums2, 0);
        } else {
            return (findKthElement((m + n) / 2, nums1, 0 ,nums2, 0) + findKthElement((m + n) / 2 + 1, nums1, 0 ,nums2, 0)) / 2;
        }
    }
    
    double findKthElement(int k, const vector<int>& nums1, int idx1, const vector<int>& nums2, int idx2) {
        int l1 = nums1.size() - idx1;
        int l2 = nums2.size() - idx2;
        if (l1 > l2) {
            return findKthElement(k, nums2, idx2, nums1, idx1);
        }
        if (l1 == 0) {
            return nums2[idx2 + k - 1];
        }
        if (k == 1) {
            return min(nums1[idx1], nums2[idx2]);
        }
        int cut1 = min(k/2, l1);
        int cut2 = k - cut1;
        if (nums1[idx1 + cut1 - 1] > nums2[idx2 + cut2 - 1]) {
            return findKthElement(k - cut2, nums1, idx1, nums2, idx2 + cut2);
        } else {
            return findKthElement(k - cut1, nums1, idx1 + cut1, nums2, idx2);
        }
    }
};