[LeetCode] 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)).
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

Thought process:
If the arrays are combined into one sorted array, the number we're looking for will be at index (m + n) / 2, or ((m + n) / 2 - 1 + (m + n) / 2) / 2. The numbers before the median in the combined array will consist of some numbers from nums1, and some numbers from nums2.
Assume half of the (m + n) / 2 numbers are from nums1, and the other half is from nums2. Say i = (m + n) / 2, we compare nums1[i / 2] with nums2[i / 2]. If nums1[i / 2] < nums2[i / 2], then we're sure that the numbers before nums1[i / 2] will not contain the median that we're looking for.

To avoid off-by-one errors and stack overflow, we'll make i represent the ith largest element, instead of the index.

Solution:
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public class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        int m = nums1.length;
        int n = nums2.length;
        
        if ((m + n) % 2 == 0) {
            return (findMedianSortedArrays(nums1, 0, nums2, 0, (m + n) / 2) + 
                    findMedianSortedArrays(nums1, 0, nums2, 0, (m + n) / 2 + 1)) / 2.0;
        } else {
            return findMedianSortedArrays(nums1, 0, nums2, 0, (m + n) / 2 + 1);
        }
    }
    
    private int findMedianSortedArrays(int[] nums1, int start1, int[] nums2, int start2, int target) {
        if (start1 >= nums1.length) {
            return nums2[start2 + target - 1];
        }
        if (start2 >= nums2.length) {
            return nums1[start1 + target - 1];
        }
        if (target == 1) {
            return Math.min(nums1[start1], nums2[start2]);
        }
        
        int num1 = start1 + target / 2 > nums1.length ? Integer.MAX_VALUE : nums1[start1 + target / 2 - 1];
        int num2 = start2 + target / 2 > nums2.length ? Integer.MAX_VALUE : nums2[start2 + target / 2 - 1];
        
        int remain = target - target / 2;
        if (num1 < num2) {
            return findMedianSortedArrays(nums1, start1 + target / 2, nums2, start2, remain);
        } else {
            return findMedianSortedArrays(nums1, start1, nums2, start2 + target / 2, remain);
        }
    }
}

Time complexity: O(log(m + n)).
Location: Philadelphia, PA, USA

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