LeeCode 4. Median of Two Sorted Arrays

4. Median of Two Sorted Arrays

There are two sorted arrays nums1 and nums2 of size m and n respectively.

Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

You may assume nums1 and nums2 cannot be both empty.

Example 1:

nums1 = [1, 3]
nums2 = [2]

The median is 2.0

Example 2:

nums1 = [1, 2]
nums2 = [3, 4]

The median is (2 + 3)/2 = 2.5

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        //the length effects the logic. It's a good method
        if(nums1.capacity() > nums2.capacity()) {
            return findMedianSortedArrays(nums2, nums1);
        }
        
        int x = nums1.capacity();
        int y = nums2.capacity();
        
        int low = 0;
        int high = x;
        while(low<=high) {
            int partitionX = (low + high) / 2;
            int partitionY = (x + y) / 2 - partitionX;
            //Solve the problem of array left and right overflow
            int xMaxLeft = (partitionX == 0) ?  INT_MIN : nums1[partitionX - 1];
            int xMInRight = (partitionX == x) ? INT_MAX : nums1[partitionX];
            int yMaxLeft = (partitionY == 0) ? INT_MIN : nums2[partitionY - 1];
            int yMinRight = (partitionY == y) ? INT_MAX : nums2[partitionY];

            if(xMaxLeft<=yMinRight && yMaxLeft<=xMInRight) {
                if((x+y)%2 == 0) {
                    //When the denominator is an integer, the result must be an integer.
                    return ((double)(max(xMaxLeft, yMaxLeft) + min(xMInRight, yMinRight))) / 2;
                }
                else {
                    return (double)min(xMInRight, yMinRight);
                }
            }  //binary search
            else if(xMaxLeft > yMinRight) {
                high = partitionX - 1;
            }
            else {
                low = partitionX + 1;
            }
            
        }
        return 0;
    }
};