Median of two sorted arrays of different sizes | Set 1 (Linear)

This is an extension of the median of two sorted arrays of equal size problem. Here we handle arrays of unequal size also.

Examples:

Input: a[] = {1, 12, 15, 26, 38}
       b[] = {2, 13, 24}
Output: 14
Explanation:
After merging arrays the result is
1, 2, 12, 13, 15, 24, 26, 38
median = (13 + 15)/2 = 14.

Input: a[] = {1, 3, 5, 7}
       b[] = {2, 4, 6, 8}
Output: 5
Explanation : 
After merging arrays the result is
1, 2, 3, 4, 5, 6, 7, 8, 9
median = 5 

Approach:

1. The approach discussed in this post is similar to method 1 of equal size median. Use merge procedure of merge sort.
2. Keep a variable to track of count while comparing elements of two arrays.
3. Perform merge of two sorted arrays. Increase the value of count as a new element is inserted.
4. The most efficient way of merging two arrays is to compare the top of both arrays and pop the smaller value and increase the count.
5. Continue comparing the top/first elements of both arrays until there are no elements left in any arrays
6. Now to find the median there are two cases to be handled.
   • If the sum of length of array sizes is even then store the elements when the count is of value ((n1 + n2)/2)-1 and (n1 + n2)/2 in the merged array add the elements and divide by 2 return the value as median.
   • If the value(sum of sizes) is odd then return the (n1 + n2)/2 th element(when value of count is (n1 + n2)/2) as median.

C++

// A C++ program to find median of two sorted arrays of
// unequal sizes
#include <bits/stdc++.h>
using namespace std;

// A utility function to find median of two integers
/* This function returns median of a[] and b[].
Assumptions in this function: Both a[] and b[]
are sorted arrays  */
float findmedian(int a[], int n1, int b[], int n2)
{
    int i = 0; /* Current index of
             i/p array a[] */
    int j = 0; /* Current index of
                  i/p array b[] */
    int k;
    int m1 = -1, m2 = -1;
    for (k = 0; k <= (n1 + n2) / 2; k++) {

        if (i < n1 && j < n2) {
            if (a[i] < b[j]) {
                m2 = m1;
                m1 = a[i];
                i++;
            }
            else {
                m2 = m1;
                m1 = b[j];
                j++;
            }
        }

        /* Below is to handle the case where
           all elements of a[] are
           smaller than smallest(or first)
           element of b[] or a[] is empty*/
        else if (i == n1) {
            m2 = m1;
            m1 = b[j];
            j++;
        }

        /* Below is to handle case where
           all elements of b[] are
           smaller than smallest(or first)
           element of a[] or b[] is empty*/
        else if (j == n2) {
            m2 = m1;
            m1 = a[i];
            i++;
        }
    }

    /*Below is to handle the case where
     sum of number of elements
     of the arrays is even */
    if ((n1 + n2) % 2 == 0)
        return (m1 + m2) * 1.0 / 2;

    /* Below is to handle the case where
       sum of number of elements
       of the arrays is odd */
    return m1;
}

// Driver program to test above functions
int main()
{
    int a[] = { 1, 12, 15, 26, 38 };
    int b[] = { 2, 13, 24 };

    int n1 = sizeof(a) / sizeof(a[0]);
    int n2 = sizeof(b) / sizeof(b[0]);

    printf("%f", findmedian(a, n1, b, n2));

    return 0;
}

Java

// A Java program to find median of two sorted arrays of
// unequal sizes

import java.io.*;

class GFG {

    // A utility function to find median of two integers
    /* This function returns median of a[] and b[].
Assumptions in this function: Both a[] and b[]
are sorted arrays */
    static float findmedian(int a[], int n1, int b[], int n2)
    {
        int i = 0; /* Current index of
            i/p array a[] */
        int j = 0; /* Current index of
                i/p array b[] */
        int k;
        int m1 = -1, m2 = -1;
        for (k = 0; k <= (n1 + n2) / 2; k++) {

            if (i < n1 && j < n2) {
                if (a[i] < b[j]) {
                    m2 = m1;
                    m1 = a[i];
                    i++;
                }
                else {
                    m2 = m1;
                    m1 = b[j];
                    j++;
                }
            }

            /* Below is to handle the case where
        all elements of a[] are
        smaller than smallest(or first)
        element of b[] or a[] is empty*/
            else if (i == n1) {
                m2 = m1;
                m1 = b[j];
                j++;
            }

            /* Below is to handle case where
        all elements of b[] are
        smaller than smallest(or first)
        element of a[] or b[] is empty*/
            else if (j == n2) {
                m2 = m1;
                m1 = a[i];
                i++;
            }
        }

        /*Below is to handle the case where
    sum of number of elements
    of the arrays is even */
        if ((n1 + n2) % 2 == 0) {
            return (m1 + m2) * (float)1.0 / 2;
        }
        /* Below is to handle the case where
    sum of number of elements
    of the arrays is odd */
        return m1;
    }

    // Driver program to test above functions
    public static void main(String[] args)
    {
        int a[] = { 1, 12, 15, 26, 38 };
        int b[] = { 2, 13, 24 };

        int n1 = a.length;
        int n2 = b.length;

        System.out.println(findmedian(a, n1, b, n2));
    }
}

// This code has been contributed by inder_verma.

Python3

# Python3 program to find median of
# two sorted arrays of unequal sizes

# A utility function to find median
# of two integers
''' This function returns median of
a[] and b[]. Assumptions in this
function: Both a[] and b[] are sorted
arrays '''

def findmedian(a, n1, b, n2):

    i = 0 # Current index of i / p array a[]
    j = 0 # Current index of i / p array b[]

    m1 = -1
    m2 = -1
    for k in range(((n1 + n2) // 2) + 1) :

        if (i < n1 and j < n2) :
            if (a[i] < b[j]) :
                m2 = m1
                m1 = a[i]
                i += 1

            else :
                m2 = m1
                m1 = b[j]
                j += 1

        # Below is to handle the case where
        # all elements of a[] are
        # smaller than smallest(or first)
        # element of b[] or a[] is empty

        elif(i == n1) :
            m2 = m1
            m1 = b[j]
            j += 1

        # Below is to handle case where
        # all elements of b[] are
        # smaller than smallest(or first)
        # element of a[] or b[] is empty
        elif (j == n2) :
            m2 = m1
            m1 = a[i]
            i += 1

    '''Below is to handle the case
    where sum of number of elements
    of the arrays is even '''
    if ((n1 + n2) % 2 == 0):
        return (m1 + m2) * 1.0 / 2

    ''' Below is to handle the case
    where sum of number of elements
    of the arrays is odd '''
    return m1

# Driver Code
if __name__ == "__main__":

    a = [ 1, 12, 15, 26, 38 ]
    b = [ 2, 13, 24 ]

    n1 = len(a)
    n2 = len(b)

    print(findmedian(a, n1, b, n2))

# This code is contributed
# by ChitraNayal

C#

// A C# program to find median
// of two sorted arrays of
// unequal sizes
using System;

class GFG {

    // A utility function to find
    // median of two integers
    /* This function returns
   median of a[] and b[].
   Assumptions in this
   function: Both a[] and b[]
   are sorted arrays */
    static float findmedian(int[] a, int n1,
                            int[] b, int n2)
    {
        int i = 0; /* Current index of
              i/p array a[] */
        int j = 0; /* Current index of
              i/p array b[] */
        int k;
        int m1 = -1, m2 = -1;
        for (k = 0; k <= (n1 + n2) / 2; k++) {
            if (i < n1 && j < n2) {
                if (a[i] < b[j]) {
                    m2 = m1;
                    m1 = a[i];
                    i++;
                }
                else {
                    m2 = m1;
                    m1 = b[j];
                    j++;
                }
            }

            /* Below is to handle the case where
    all elements of a[] are
    smaller than smallest(or first)
    element of b[] or a[] is empty*/
            else if (i == n1) {
                m2 = m1;
                m1 = b[j];
                j++;
            }

            /* Below is to handle case where
    all elements of b[] are
    smaller than smallest(or first)
    element of a[] or b[] is empty*/
            else if (j == n2) {
                m2 = m1;
                m1 = a[i];
                i++;
            }
        }

        /*Below is to handle the case where
sum of number of elements
of the arrays is even */
        if ((n1 + n2) % 2 == 0) {
            return (m1 + m2) * (float)1.0 / 2;
        }
        /* Below is to handle the case
where sum of number of elements
of the arrays is odd */
        return m1;
    }

    // Driver Code
    public static void Main()
    {
        int[] a = { 1, 12, 15, 26, 38 };
        int[] b = { 2, 13, 24 };

        int n1 = a.Length;
        int n2 = b.Length;

        Console.WriteLine(findmedian(a, n1, b, n2));
    }
}

// This code is contributed
// by Subhadeep Gupta

PHP

<?php
// A PHP program to find median of
// two sorted arrays of unequal sizes

// A utility function to find
// median of two integers
/* This function returns median
of a[] and b[]. Assumptions in this
function: Both a[] and b[] are
sorted arrays */
function findmedian($a, $n1, $b, $n2)
{
    $i = 0; /* Current index of
               i/p array a[] */
    $j = 0; /* Current index of
               i/p array b[] */
    $k;
    $m1 = -1; $m2 = -1;
    for ($k = 0;
         $k <= ($n1 + $n2) / 2; $k++)
    {

        if ($i < $n1 and $j < $n2)
        {
            if ($a[$i] < $b[$j])
            {
                $m2 = $m1;
                $m1 = $a[$i];
                $i++;
            }
            else
            {
                $m2 = $m1;
                $m1 = $b[$j];
                $j++;
            }
        }

        /* Below is to handle the case
        where all elements of a[] are
        smaller than smallest(or first)
        element of b[] or a[] is empty*/
        else if (i == n1)
        {
            $m2 = $m1;
            $m1 = $b[j];
            $j++;
        }

        /* Below is to handle case
        where all elements of b[] are
        smaller than smallest(or first)
        element of a[] or b[] is empty*/
        else if ($j == $n2)
        {
            $m2 = $m1;
            $m1 = $a[$i];
            $i++;
        }
    }

    /*Below is to handle the case
    where sum of number of elements
    of the arrays is even */
    if (($n1 + $n2) % 2 == 0)
        return ($m1 + $m2) * 1.0 / 2;

    /* Below is to handle the case
    where sum of number of elements
    of the arrays is odd */
    return m1;
}

// Driver Code
$a = array( 1, 12, 15, 26, 38 );
$b = array( 2, 13, 24 );

$n1 = count($a);
$n2 = count($b);

echo(findmedian($a, $n1, $b, $n2));

// This code is contributed
// by inder_verma.
?>

Javascript

<script>

// A Javascript program to find median
// of two sorted arrays of unequal sizes

// A utility function to find median of two integers
// This function returns median of a[] and b[].
// Assumptions in this function: Both a[] and b[]
// are sorted arrays
function findmedian(a, n1, b, n2)
{
    let i = 0; /* Current index of
                  i/p array a[] */
    let j = 0; /* Current index of
                  i/p array b[] */
    let k;
    let m1 = -1, m2 = -1;

    for(k = 0; k <= (n1 + n2) / 2; k++)
    {
        if (i < n1 && j < n2)
        {
            if (a[i] < b[j])
            {
                m2 = m1;
                m1 = a[i];
                i++;
            }
            else
            {
                m2 = m1;
                m1 = b[j];
                j++;
            }
        }

        /* Below is to handle the case where
        all elements of a[] are
        smaller than smallest(or first)
        element of b[] or a[] is empty*/
        else if (i == n1)
        {
            m2 = m1;
            m1 = b[j];
            j++;
        }

        /* Below is to handle case where
        all elements of b[] are
        smaller than smallest(or first)
        element of a[] or b[] is empty*/
        else if (j == n2)
        {
            m2 = m1;
            m1 = a[i];
            i++;
        }
    }

    /*Below is to handle the case where
    sum of number of elements
    of the arrays is even */
    if ((n1 + n2) % 2 == 0)
    {
        return (m1 + m2) * 1.0 / 2;
    }

    /* Below is to handle the case where
    sum of number of elements
    of the arrays is odd */
    return m1;
}

// Driver code
let a = [ 1, 12, 15, 26, 38 ];
let b = [ 2, 13, 24 ];
let n1 = a.length;
let n2 = b.length;

document.write(findmedian(a, n1, b, n2));

// This code is contributed by rag2127

</script>

Output:

14.000000

Complexity Analysis:

• Time Complexity: O(n), Both the lists need to be traversed so the time complexity is O(n).
• Auxiliary Space: O(1), no extra space is required.

More Efficient (Log time complexity methods)

1. Median of two sorted arrays of different sizes.
2. Median of two sorted arrays of different sizes in min(Log (n1, n2))