Median of 2 Sorted Arrays of Different Sizes

Understanding the Problem

The goal is to find the median of two sorted arrays of different sizes.

Method 1: Simple Merge

This method merges the two arrays and finds the median.

import java.util.*;

public class MedianFinder {
    public static double findMedianSortedArrays(int[] arr1, int[] arr2) {
        int n1 = arr1.length, n2 = arr2.length;
        int[] merged = new int[n1 + n2];
        int i = 0, j = 0, k = 0;

        while (i < n1 && j < n2) {
            if (arr1[i] < arr2[j]) {
                merged[k++] = arr1[i++];
            } else {
                merged[k++] = arr2[j++];
            }
        }
        while (i < n1) merged[k++] = arr1[i++];
        while (j < n2) merged[k++] = arr2[j++];

        int total = n1 + n2;
        if (total % 2 == 0) {
            return (merged[total / 2 - 1] + merged[total / 2]) / 2.0;
        } else {
            return merged[total / 2];
        }
    }

    public static void main(String[] args) {
        int[] arr1 = {1, 3, 8};
        int[] arr2 = {7, 9, 10, 11};
        System.out.println("Median: " + findMedianSortedArrays(arr1, arr2));
    }
}
            

Output:

Median: 8.0

Method 2: Binary Search

This method uses binary search to find the median efficiently.

import java.util.*;

public class MedianFinderBinarySearch {
    public static double findMedianSortedArrays(int[] arr1, int[] arr2) {
        if (arr1.length > arr2.length) {
            return findMedianSortedArrays(arr2, arr1);
        }

        int x = arr1.length, y = arr2.length;
        int low = 0, high = x;
        while (low <= high) {
            int partitionX = (low + high) / 2;
            int partitionY = (x + y + 1) / 2 - partitionX;

            int maxLeftX = (partitionX == 0) ? Integer.MIN_VALUE : arr1[partitionX - 1];
            int minRightX = (partitionX == x) ? Integer.MAX_VALUE : arr1[partitionX];

            int maxLeftY = (partitionY == 0) ? Integer.MIN_VALUE : arr2[partitionY - 1];
            int minRightY = (partitionY == y) ? Integer.MAX_VALUE : arr2[partitionY];

            if (maxLeftX <= minRightY && maxLeftY <= minRightX) {
                if ((x + y) % 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.");
    }

    public static void main(String[] args) {
        int[] arr1 = {1, 3, 8};
        int[] arr2 = {7, 9, 10, 11};
        System.out.println("Median: " + findMedianSortedArrays(arr1, arr2));
    }
}
            

Output:

Median: 8.0

Method 3: Using a Combined Array