4. Median of Two Sorted Arrays

LeetCode

My Solution

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class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        int m = nums1.length;
        int n = nums2.length;

        int p1 = 0;
        int p2 = 0;

        int last = 0;
        int last_next = 0;
        double median = 0;

        if ((m + n) % 2 != 0) {
            while(p1 + p2 < (m + n + 1) / 2){
                if (p1 < m && (p2 >= n || nums1[p1] < nums2[p2])) {
                    last = nums1[p1];
                    p1++;
                } else {
                    last = nums2[p2];
                    p2++;
                }
            }
            median = last;
        } else {
            while(p1 + p2 < (m + n) / 2){
                if (p1 < m && (p2 >= n || nums1[p1] < nums2[p2])) {
                    last = nums1[p1];
                    p1++;
                } else {
                    last = nums2[p2];
                    p2++;
                }
            }
            if (p1 < m && (p2 >= n || nums1[p1] < nums2[p2])) {
                last_next = nums1[p1];
            } else {
                last_next = nums2[p2];
            }

            median = (last + last_next) / 2.0;
        }

    return median;
    }
}

Remarks:

  1. check the border of arrays! p1 < m && (p2 >= n || nums1[p1] < nums2[p2])
  2. Use double but not float
  3. be careful of division results: /2.0 -> double; /2 -> integer
  4. TC $O(m+n)$

Binary Search, Recursive

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public class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        // Ensure nums1 is the smaller array
        if (nums1.length > nums2.length) {
            return findMedianSortedArrays(nums2, nums1);
        }

        int m = nums1.length;
        int n = nums2.length;
        int low = 0, high = m;

        while (low <= high) {
            int partitionX = (low + high) / 2;
            int partitionY = (m + n + 1) / 2 - partitionX;

            // Edge cases where partition is at the boundaries
            int maxLeftX = (partitionX == 0) ? Integer.MIN_VALUE : nums1[partitionX - 1];
            int minRightX = (partitionX == m) ? Integer.MAX_VALUE : nums1[partitionX];

            int maxLeftY = (partitionY == 0) ? Integer.MIN_VALUE : nums2[partitionY - 1];
            int minRightY = (partitionY == n) ? Integer.MAX_VALUE : nums2[partitionY];

            // Check for correct partition
            if (maxLeftX <= minRightY && maxLeftY <= minRightX) {
                // If total length is even
                if ((m + n) % 2 == 0) {
                    return (Math.max(maxLeftX, maxLeftY) + Math.min(minRightX, minRightY)) / 2.0;
                } else {
                    return Math.max(maxLeftX, maxLeftY);
                }
            } else if (maxLeftX > minRightY) {
                high = partitionX - 1;
            } else {
                low = partitionX + 1;
            }
        }

        throw new IllegalArgumentException("Input arrays are not sorted properly.");
    }
}