Java Program for Median of two sorted arrays of same size

Last Updated : 24 Oct, 2023

Write a Java program for a given 2 sorted arrays A and B of size n each. the task is to find the median of the array obtained by merging the above 2 arrays(i.e. array of length 2n). The complexity should be O(log(n)).

Examples:

Input: ar1[] = {1, 12, 15, 26, 38}
ar2[] = {2, 13, 17, 30, 45}
Output: 16
Explanation:
After merging two arrays, we get {1, 2, 12, 13, 15, 17, 26, 30, 38, 45}
The middle two elements are 15 and 17
The average of middle elements is (15 + 17)/2 which is equal to 16

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Note : Since size of the set for which we are looking for median is even (2n), we need take average of middle two numbers and return floor of the average.

Java Program for Median of two sorted arrays of same size using Simply count while Merging:

Use merge procedure of merge sort. Keep track of count while comparing elements of two arrays. If count becomes n(For 2n elements), we have reached the median. Take the average of the elements at indexes n-1 and n in the merged array.

Below is the implementation of the above approach:

Java

// A Simple Merge based O(n) solution 
// to find median of two sorted arrays

class Main
{
    // function to calculate median
    static int getMedian(int ar1[], int ar2[], int n)
    { 
        int i = 0; 
        int j = 0; 
        int count;
        int m1 = -1, m2 = -1;
    
        /* Since there are 2n elements, median will 
        be average of elements at index n-1 and 
        n in the array obtained after merging ar1 
        and ar2 */
        for (count = 0; count <= n; count++)
        {
            /* Below is to handle case where all 
            elements of ar1[] are smaller than 
            smallest(or first) element of ar2[] */
            if (i == n)
            {
                m1 = m2;
                m2 = ar2[0];
                break;
            }
    
            /* Below is to handle case where all 
            elements of ar2[] are smaller than 
            smallest(or first) element of ar1[] */
            else if (j == n)
            {
                m1 = m2;
                m2 = ar1[0];
                break;
            }
            /* equals sign because if two 
            arrays have some common elements */
            if (ar1[i] <= ar2[j])
            { 
                /* Store the prev median */
                m1 = m2; 
                m2 = ar1[i];
                i++;
            }
            else
            {
                /* Store the prev median */
                m1 = m2; 
                m2 = ar2[j];
                j++;
            }
        }
    
        return (m1 + m2)/2;
    }
    
    /* Driver program to test above function */
    public static void main (String[] args)
    {
        int ar1[] = {1, 12, 15, 26, 38};
        int ar2[] = {2, 13, 17, 30, 45};
    
        int n1 = ar1.length;
        int n2 = ar2.length;
        if (n1 == n2)
            System.out.println("Median is " +
                        getMedian(ar1, ar2, n1));
        else
            System.out.println("arrays are of unequal size");
    } 
}

Output

Median is 16

Time Complexity: O(n)
Auxiliary Space: O(1)

Java Program for Median of two sorted arrays of same size (By comparing the medians of two arrays):

Step-by-step approach:

Merge the two input arrays ar1[] and ar2[].
Sort ar1[] and ar2[] respectively.
The median will be the last element of ar1[] + the first
element of ar2[] divided by 2. [(ar1[n-1] + ar2[0])/2].

Below is the implementation of the above approach:

Java

/*package whatever //do not write package name here */
import java.io.*;
import java.util.*;
class GFG
{

/* This function returns
    median of ar1[] and ar2[].
    Assumptions in this function:
    Both ar1[] and ar2[]
    are sorted arrays
    Both have n elements */
public static int getMedian(int ar1[],
                            int ar2[], int n)
{
    int j = 0;
    int i = n - 1;
    while (ar1[i] > ar2[j] && j < n && i > -1)
    {
    int temp = ar1[i];
    ar1[i] = ar2[j];
    ar2[j] = temp;
    i--; j++;
    }
    Arrays.sort(ar1);
    Arrays.sort(ar2);
    return (ar1[n - 1] + ar2[0]) / 2;
}

// Driver code
public static void main (String[] args)
{
    int ar1[] = { 1, 12, 15, 26, 38 };
    int ar2[] = { 2, 13, 17, 30, 45 };

    int n1 = 5;
    int n2 = 5;
    if (n1 == n2)
    System.out.println("Median is "+ getMedian(ar1, ar2, n1));
    else
    System.out.println("Doesn't work for arrays of unequal size");
}
}

// This code is contributed by Manu Pathria

Output

Median is 16

Time Complexity: O(nlogn)
Auxiliary Space: O(1)

Java Program for Median of two sorted arrays of same size using Binary Search:

Step-by-step approach:

We can find the kth element by using binary search on whole range of constraints of elements.

Initialize ans = 0.0
Initialize low = -10^9, high = 10^9 and pos = n
Run a loop while(low <= high):
Calculate mid = (low + (high – low)>>1)
Find total elements less or equal to mid in the given arrays
If the count is less or equal to pos
Update low = mid + 1
Else high = mid – 1
Store low in ans, i.e., ans = low.
Again follow step3 with pos as n – 1
Return (sum + low * 1.0)/2
Median of two sorted arrays of same size

Below is the implementation of the above approach:

Java

/*package whatever //do not write package name here */

import java.io.*;

class GFG {

    public static double getMedian(int[] nums1, int[] nums2,
                                int n)
    {
        // according to given constraints all numbers are in
        // this range
        int low = (int)-1e9, high = (int)1e9;

        int pos = n;
        double ans = 0.0;
        // binary search to find the element which will be
        // present at pos = totalLen/2 after merging two
        // arrays in sorted order
        while (low <= high) {
            int mid = low + ((high - low) >> 1);

            // total number of elements in arrays which are
            // less than mid
            int ub = upperBound(nums1, mid)
                    + upperBound(nums2, mid);

            if (ub <= pos)
                low = mid + 1;
            else
                high = mid - 1;
        }

        ans = low;

        // As there are even number of elements, we will
        // also have to find element at pos = totalLen/2 - 1
        pos--;
        low = (int)-1e9;
        high = (int)1e9;
        while (low <= high) {
            int mid = low + ((high - low) >> 1);
            int ub = upperBound(nums1, mid)
                    + upperBound(nums2, mid);

            if (ub <= pos)
                low = mid + 1;
            else
                high = mid - 1;
        }

        // average of two elements in case of even
        // number of elements
        ans = (ans + low * 1.0) / 2;

        return ans;
    }

    // a function which returns the index of smallest
    // element which is strictly greater than key (i.e. it
    // returns number of elements which are less than or
    // equal to key)
    public static int upperBound(int[] arr, int key)
    {
        int low = 0, high = arr.length;

        while (low < high) {
            int mid = low + ((high - low) >> 1);

            if (arr[mid] <= key)
                low = mid + 1;
            else
                high = mid;
        }

        return low;
    }

    public static void main(String[] args)
    {
        int[] arr = { 1, 4, 5, 6, 10 };
        int[] brr = { 2, 3, 4, 5, 7 };

        double median = getMedian(arr, brr, arr.length);

        System.out.println("Median is " + median);
    }
}