Median of two sorted arrays of same size

06 Jul, 2022
There are 2 sorted arrays A and B of size n each. Write an algorithm to find the median of the array obtained after merging the above 2 arrays(i.e. array of length 2n). The complexity should be O(log(n)). 
Note: Since the size of the set for which we are looking for the median is even (2n), we need to take the average of the middle two numbers and return the floor of the average.Method 1 (Simply count while Merging) Use the 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. See the below implementation. 
C++
C
Java
Python3
C#
PHP
Javascript
// A Simple Merge based O(n) // solution to find median of// two sorted arrays#include <bits/stdc++.h>using namespace std;  /* This function returns median of ar1[] and ar2[].Assumptions in this function:Both ar1[] and ar2[] are sorted arraysBoth have n elements */int getMedian(int ar1[],              int ar2[], int n){    int i = 0; /* Current index of                   i/p array ar1[] */    int j = 0; /* Current index of                   i/p array ar2[] */    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 Codeint main(){    int ar1[] = {1, 12, 15, 26, 38};    int ar2[] = {2, 13, 17, 30, 45};      int n1 = sizeof(ar1) / sizeof(ar1[0]);    int n2 = sizeof(ar2) / sizeof(ar2[0]);    if (n1 == n2)        cout << "Median is "              << getMedian(ar1, ar2, n1) ;    else        cout << "Doesn't work for arrays"              << " of unequal size" ;    getchar();    return 0;}  // This code is contributed // by Shivi_Aggarwal
// A Simple Merge based O(n) solution to find median of// two sorted arrays#include <stdio.h>  /* This function returns median of ar1[] and ar2[].   Assumptions in this function:   Both ar1[] and ar2[] are sorted arrays   Both have n elements */int getMedian(int ar1[], int ar2[], int n){    int i = 0;  /* Current index of i/p array ar1[] */    int j = 0; /* Current index of i/p array ar2[] */    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])        {            m1 = m2;  /* Store the prev median */            m2 = ar1[i];            i++;        }        else        {            m1 = m2;  /* Store the prev median */            m2 = ar2[j];            j++;        }    }      return (m1 + m2)/2;}  /* Driver program to test above function */int main(){    int ar1[] = {1, 12, 15, 26, 38};    int ar2[] = {2, 13, 17, 30, 45};      int n1 = sizeof(ar1)/sizeof(ar1[0]);    int n2 = sizeof(ar2)/sizeof(ar2[0]);    if (n1 == n2)        printf("Median is %d", getMedian(ar1, ar2, n1));    else        printf("Doesn't work for arrays of unequal size");    getchar();    return 0;}
// 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");    }    }
# A Simple Merge based O(n) Python 3 solution # to find median of two sorted lists  # This function returns median of ar1[] and ar2[].# Assumptions in this function:# Both ar1[] and ar2[] are sorted arrays# Both have n elementsdef getMedian( ar1, ar2 , n):    i = 0 # Current index of i/p list ar1[]          j = 0 # Current index of i/p list ar2[]          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    count = 0    while count < n + 1:        count += 1                  # 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[]        elif j == n:            m1 = m2            m2 = ar1[0]            break        # equals sign because if two         # arrays have some common elements         if ar1[i] <= ar2[j]:            m1 = m2 # Store the prev median            m2 = ar1[i]            i += 1        else:            m1 = m2 # Store the prev median            m2 = ar2[j]            j += 1    return (m1 + m2)/2  # Driver code to test above functionar1 = [1, 12, 15, 26, 38]ar2 = [2, 13, 17, 30, 45]n1 = len(ar1)n2 = len(ar2)if n1 == n2:    print("Median is ", getMedian(ar1, ar2, n1))else:    print("Doesn't work for arrays of unequal size")  # This code is contributed by "Sharad_Bhardwaj".
// A Simple Merge based O(n) solution // to find median of two sorted arraysusing System;class GFG{    // 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 Code    public static void Main ()    {        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)            Console.Write("Median is " +                        getMedian(ar1, ar2, n1));        else            Console.Write("arrays are of unequal size");    } }
<?php// A Simple Merge based O(n) solution // to find median of two sorted arrays  // This function returns median of // ar1[] and ar2[]. Assumptions in // this function: Both ar1[] and ar2[] // are sorted arrays Both have n elements function getMedian($ar1, $ar2, $n){    // Current index of i/p array ar1[]    $i = 0;           // Current index of i/p array ar2[]    $j = 0;     $count;    $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 + $m2) / 2;}  // Driver Code$ar1 = array(1, 12, 15, 26, 38);$ar2 = array(2, 13, 17, 30, 45);  $n1 = sizeof($ar1);$n2 = sizeof($ar2);if ($n1 == $n2)    echo("Median is " .           getMedian($ar1, $ar2, $n1));else    echo("Doesn't work for arrays".          "of unequal size");  // This code is contributed by Ajit.?>
<script>  // A Simple Merge based O(n) solution to find median of// two sorted arrays  /* This function returns median of ar1[] and ar2[].Assumptions in this function:Both ar1[] and ar2[] are sorted arraysBoth have n elements */function getMedian(ar1, ar2, n){    var i = 0; /* Current index of i/p array ar1[] */    var j = 0; /* Current index of i/p array ar2[] */    var count;    var 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])        {            m1 = m2; /* Store the prev median */            m2 = ar1[i];            i++;        }        else        {            m1 = m2; /* Store the prev median */            m2 = ar2[j];            j++;        }    }      return (m1 + m2)/2;}  /* Driver program to test above function */var ar1 = [1, 12, 15, 26, 38];var ar2 = [2, 13, 17, 30, 45];var n1 = ar1.length;var n2 = ar2.length;if (n1 == n2)    document.write("Median is "+ getMedian(ar1, ar2, n1));else    document.write("Doesn't work for arrays of unequal size");  </script>
Median is 16
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Time Complexity : O(n)Auxiliary Space: O(1)
Method 2 (By comparing the medians of two arrays) This method works by first getting medians of the two sorted arrays and then comparing them.Let ar1 and ar2 be the input arrays. Algorithm :  
1) Calculate the medians m1 and m2 of the input arrays ar1[] 
   and ar2[] respectively.
2) If m1 and m2 both are equal then we are done.
     return m1 (or m2)
3) If m1 is greater than m2, then median is present in one 
   of the below two subarrays.
    a)  From first element of ar1 to m1 (ar1[0...|_n/2_|])
    b)  From m2 to last element of ar2  (ar2[|_n/2_|...n-1])
4) If m2 is greater than m1, then median is present in one    
   of the below two subarrays.
   a)  From m1 to last element of ar1  (ar1[|_n/2_|...n-1])
   b)  From first element of ar2 to m2 (ar2[0...|_n/2_|])
5) Repeat the above process until size of both the subarrays 
   becomes 2.
6) If size of the two arrays is 2 then use below formula to get 
  the median.
    Median = (max(ar1[0], ar2[0]) + min(ar1[1], ar2[1]))/2
Examples :  
   ar1[] = {1, 12, 15, 26, 38}
   ar2[] = {2, 13, 17, 30, 45}
For above two arrays m1 = 15 and m2 = 17For the above ar1[] and ar2[], m1 is smaller than m2. So median is present in one of the following two subarrays. 
   [15, 26, 38] and [2, 13, 17]
Let us repeat the process for above two subarrays:  
    m1 = 26 m2 = 13.
m1 is greater than m2. So the subarrays become  
  [15, 26] and [13, 17]
Now size is 2, so median = (max(ar1[0], ar2[0]) + min(ar1[1], ar2[1]))/2
                       = (max(15, 13) + min(26, 17))/2 
                       = (15 + 17)/2
                       = 16
Implementation : 
C
C++
Java
Python
C#
Javascript
// A divide and conquer based efficient solution to find median// of two sorted arrays of same size.#include<bits/stdc++.h>using namespace std;  int median(int [], int); /* to get median of a sorted array */  /* This function returns median of ar1[] and ar2[].   Assumptions in this function:   Both ar1[] and ar2[] are sorted arrays   Both have n elements */int getMedian(int ar1[], int ar2[], int n){    /* return -1  for invalid input */    if (n <= 0)        return -1;    if (n == 1)        return (ar1[0] + ar2[0])/2;    if (n == 2)        return (max(ar1[0], ar2[0]) + min(ar1[1], ar2[1])) / 2;      int m1 = median(ar1, n); /* get the median of the first array */    int m2 = median(ar2, n); /* get the median of the second array */      /* If medians are equal then return either m1 or m2 */    if (m1 == m2)        return m1;      /* if m1 < m2 then median must exist in ar1[m1....] and        ar2[....m2] */    if (m1 < m2)    {        if (n % 2 == 0)            return getMedian(ar1 + n/2 - 1, ar2, n - n/2 +1);        return getMedian(ar1 + n/2, ar2, n - n/2);    }      /* if m1 > m2 then median must exist in ar1[....m1] and        ar2[m2...] */    if (n % 2 == 0)        return getMedian(ar2 + n/2 - 1, ar1, n - n/2 + 1);    return getMedian(ar2 + n/2, ar1, n - n/2);}  /* Function to get median of a sorted array */int median(int arr[], int n){    if (n%2 == 0)        return (arr[n/2] + arr[n/2-1])/2;    else        return arr[n/2];}  /* Driver program to test above function */int main(){    int ar1[] = {1, 2, 3, 6};    int ar2[] = {4, 6, 8, 10};    int n1 = sizeof(ar1)/sizeof(ar1[0]);    int n2 = sizeof(ar2)/sizeof(ar2[0]);    if (n1 == n2)        printf("Median is %d", getMedian(ar1, ar2, n1));    else        printf("Doesn't work for arrays of unequal size");    return 0;}
// A divide and conquer based // efficient solution to find // median of two sorted arrays // of same size.#include<bits/stdc++.h>using namespace std;  /* to get median of a   sorted array */int median(int [], int);   /* This function returns median    of ar1[] and ar2[].Assumptions in this function:    Both ar1[] and ar2[] are     sorted arrays    Both have n elements */int getMedian(int ar1[],               int ar2[], int n){    /* return -1 for        invalid input */    if (n <= 0)        return -1;    if (n == 1)        return (ar1[0] +                 ar2[0]) / 2;    if (n == 2)        return (max(ar1[0], ar2[0]) +                 min(ar1[1], ar2[1])) / 2;      /* get the median of        the first array */    int m1 = median(ar1, n);           /* get the median of        the second array */    int m2 = median(ar2, n);       /* If medians are equal then        return either m1 or m2 */    if (m1 == m2)        return m1;      /* if m1 < m2 then median must        exist in ar1[m1....] and                ar2[....m2] */    if (m1 < m2)    {        if (n % 2 == 0)            return getMedian(ar1 + n / 2 - 1,                              ar2, n - n / 2 + 1);        return getMedian(ar1 + n / 2,                          ar2, n - n / 2);    }      /* if m1 > m2 then median must        exist in ar1[....m1] and                 ar2[m2...] */    if (n % 2 == 0)        return getMedian(ar2 + n / 2 - 1,                          ar1, n - n / 2 + 1);    return getMedian(ar2 + n / 2,                      ar1, n - n / 2);}  /* Function to get median    of a sorted array */int median(int arr[], int n){    if (n % 2 == 0)        return (arr[n / 2] +                 arr[n / 2 - 1]) / 2;    else        return arr[n / 2];}  // Driver codeint main(){    int ar1[] = {1, 2, 3, 6};    int ar2[] = {4, 6, 8, 10};    int n1 = sizeof(ar1) /              sizeof(ar1[0]);    int n2 = sizeof(ar2) /              sizeof(ar2[0]);    if (n1 == n2)        cout << "Median is "             << getMedian(ar1, ar2, n1);    else        cout << "Doesn't work for arrays "              << "of unequal size";    return 0;}  // This code is contributed// by Shivi_Aggarwal
// A Java program to divide and conquer based// efficient solution to find// median of two sorted arrays// of same size.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 */      static int getMedian(        int[] a, int[] b, int startA,        int startB, int endA, int endB)    {        if (endA - startA == 1) {            return (                       Math.max(a[startA],                                b[startB])                       + Math.min(a[endA], b[endB]))                / 2;        }        /* get the median of    the first array */        int m1 = median(a, startA, endA);          /* get the median of    the second array */        int m2 = median(b, startB, endB);          /* If medians are equal then    return either m1 or m2 */        if (m1 == m2) {            return m1;        }          /* if m1 < m2 then median must    exist in ar1[m1....] and                ar2[....m2] */        else if (m1 < m2) {            return getMedian(                a, b, (endA + startA + 1) / 2,                startB, endA,                (endB + startB + 1) / 2);        }          /* if m1 > m2 then median must    exist in ar1[....m1] and    ar2[m2...] */        else {            return getMedian(                a, b, startA,                (endB + startB + 1) / 2,                (endA + startA + 1) / 2, endB);        }    }      /* Function to get median    of a sorted array */    static int median(        int[] arr, int start, int end)    {        int n = end - start + 1;        if (n % 2 == 0) {            return (                       arr[start + (n / 2)]                       + arr[start + (n / 2 - 1)])                / 2;        }        else {            return arr[start + n / 2];        }    }      // Driver code    public static void main(String[] args)    {        int ar1[] = { 1, 2, 3, 6 };        int ar2[] = { 4, 6, 8, 10 };        int n1 = ar1.length;        int n2 = ar2.length;        if (n1 != n2) {            System.out.println(                "Doesn't work for arrays "                + "of unequal size");        }        else if (n1 == 0) {            System.out.println("Arrays are empty.");        }        else if (n1 == 1) {            System.out.println((ar1[0] + ar2[0]) / 2);        }        else {            System.out.println(                "Median is "                + getMedian(                      ar1, ar2, 0, 0,                      ar1.length - 1, ar2.length - 1));        }    }}
# using divide and conquer we divide# the 2 arrays accordingly recursively# till we get two elements in each # array, hence then we calculate median  #condition len(arr1)=len(arr2)=ndef getMedian(arr1, arr2, n):           # there is no element in any array    if n == 0:         return -1              # 1 element in each => median of     # sorted arr made of two arrays will        elif n == 1:         # be sum of both elements by 2        return (arr1[0]+arr2[0])/2              # Eg. [1,4] , [6,10] => [1, 4, 6, 10]    # median = (6+4)/2        elif n == 2:         # which implies median = (max(arr1[0],        # arr2[0])+min(arr1[1],arr2[1]))/2        return (max(arr1[0], arr2[0]) +                 min(arr1[1], arr2[1])) / 2          else:        #calculating medians             m1 = median(arr1, n)        m2 = median(arr2, n)                  # then the elements at median         # position must be between the         # greater median and the first         # element of respective array and         # between the other median and         # the last element in its respective array.        if m1 > m2:                          if n % 2 == 0:                return getMedian(arr1[:int(n / 2) + 1],                        arr2[int(n / 2) - 1:], int(n / 2) + 1)            else:                return getMedian(arr1[:int(n / 2) + 1],                         arr2[int(n / 2):], int(n / 2) + 1)                  else:            if n % 2 == 0:                return getMedian(arr1[int(n / 2 - 1):],                        arr2[:int(n / 2 + 1)], int(n / 2) + 1)            else:                return getMedian(arr1[int(n / 2):],                         arr2[0:int(n / 2) + 1], int(n / 2) + 1)   # function to find median of arraydef median(arr, n):    if n % 2 == 0:        return (arr[int(n / 2)] +                arr[int(n / 2) - 1]) / 2    else:        return arr[int(n/2)]        # Driver codearr1 = [1, 2, 3, 6]arr2 = [4, 6, 8, 10]n = len(arr1)print(int(getMedian(arr1,arr2,n)))  # This code is contributed by# baby_gog9800
// A C# program to divide and conquer based// efficient solution to find// median of two sorted arrays// of same size.using System;class GfG{  /* This function returns medianof ar1[] and ar2[].Assumptions in this function:Both ar1[] and ar2[] aresorted arraysBoth have n elements */  static int getMedian(int[] a, int[] b,                      int startA, int startB,                      int endA, int endB){  if (endA - startA == 1)   {    return (Math.Max(a[startA],                      b[startB]) +             Math.Min(a[endA], b[endB])) / 2;  }      /* get the median of    the first array */  int m1 = median(a, startA, endA);    /* get the median of    the second array */  int m2 = median(b, startB, endB);    /* If medians are equal then    return either m1 or m2 */  if (m1 == m2)   {    return m1;  }    /*if m1 < m2 then median must    exist in ar1[m1....] and    ar2[....m2] */  else if (m1 < m2)   {    return getMedian(a, b,                     (endA + startA + 1) / 2,                     startB, endA,                     (endB + startB + 1) / 2);  }    /*if m1 > m2 then median must    exist in ar1[....m1] and    ar2[m2...] */  else   {    return getMedian(a, b, startA,                    (endB + startB + 1) / 2,                    (endA + startA + 1) / 2, endB);  }}  /* Function to get medianof a sorted array */static int median(int[] arr,                   int start, int end){  int n = end - start + 1;  if (n % 2 == 0)   {    return (arr[start + (n / 2)] +             arr[start + (n / 2 - 1)]) / 2;  }  else   {    return arr[start + n / 2];  }}  // Driver codepublic static void Main(String[] args){  int []ar1 = {1, 2, 3, 6};  int []ar2 = {4, 6, 8, 10};  int n1 = ar1.Length;  int n2 = ar2.Length;  if (n1 != n2)   {    Console.WriteLine("Doesn't work for arrays " +                       "of unequal size");  }  else if (n1 == 0)   {    Console.WriteLine("Arrays are empty.");  }  else if (n1 == 1)   {    Console.WriteLine((ar1[0] + ar2[0]) / 2);  }  else   {    Console.WriteLine("Median is " +                        getMedian(ar1, ar2, 0, 0,                                  ar1.Length - 1,                                  ar2.Length - 1));  }}}  // This code is contributed by gauravrajput1
<script>// A Javascript program to divide and conquer based// efficient solution to find// median of two sorted arrays// of same size.   /* This function returns median    of ar1[] and ar2[].    Assumptions in this function:        Both ar1[] and ar2[] are        sorted arrays        Both have n elements */function getMedian(a,b,startA,startB,endA,endB){    if (endA - startA == 1) {            return (                       Math.max(a[startA],                                b[startB])                       + Math.min(a[endA], b[endB]))                / 2;        }        /* get the median of    the first array */        let m1 = median(a, startA, endA);           /* get the median of    the second array */        let m2 = median(b, startB, endB);           /* If medians are equal then    return either m1 or m2 */        if (m1 == m2) {            return m1;        }           /* if m1 < m2 then median must    exist in ar1[m1....] and                ar2[....m2] */        else if (m1 < m2) {            return getMedian(                a, b, (endA + startA + 1) / 2,                startB, endA,                (endB + startB + 1) / 2);        }           /* if m1 > m2 then median must    exist in ar1[....m1] and    ar2[m2...] */        else {            return getMedian(                a, b, startA,                (endB + startB + 1) / 2,                (endA + startA + 1) / 2, endB);        }}  /* Function to get median    of a sorted array */function median(arr,start,end){    let n = end - start + 1;        if (n % 2 == 0) {            return (                       arr[start + (n / 2)]                       + arr[start + (n / 2 - 1)])                / 2;        }        else {            return arr[start + n / 2];        }}  // Driver codelet ar1 = [ 1, 2, 3, 6 ];        let ar2 = [ 4, 6, 8, 10 ];        let n1 = ar1.length;        let n2 = ar2.length;        if (n1 != n2) {            document.write(                "Doesn't work for arrays "                + "of unequal size<br>");        }        else if (n1 == 0) {            document.write("Arrays are empty.<br>");        }        else if (n1 == 1) {            document.write((ar1[0] + ar2[0]) / 2+"<br>");        }        else {            document.write(                "Median is "                + getMedian(                      ar1, ar2, 0, 0,                      ar1.length - 1, ar2.length - 1)+"<br>");        }  // This code is contributed by avanitrachhadiya2155</script>
Median is 5
Time Complexity : O(logn) Auxiliary Space: O(1)Algorithmic Paradigm: Divide and Conquer 
Method 3 (By Merging two arrays w/o extra space)This method works by merging two arrays without extra space and then sorting them.
Algorithm : 
1) Merge the two input arrays ar1[] and ar2[].
2) Sort ar1[] and ar2[] respectively.
3) 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:
C++
C
Java
Python3
C#
Javascript
// CPP program for the above approach#include <bits/stdc++.h>using namespace std;  /* This function returnsmedian of ar1[] and ar2[].Assumptions in this function:Both ar1[] and ar2[]are sorted arraysBoth have n elements */int getMedian(int ar1[], int ar2[], int n){    int j = 0;    int i = n - 1;    while (ar1[i] > ar2[j] && j < n && i > -1)        swap(ar1[i--], ar2[j++]);    sort(ar1, ar1 + n);    sort(ar2, ar2 + n);    return (ar1[n - 1] + ar2[0]) / 2;}  // Driver Codeint main(){    int ar1[] = { 1, 12, 15, 26, 38 };    int ar2[] = { 2, 13, 17, 30, 45 };      int n1 = sizeof(ar1) / sizeof(ar1[0]);    int n2 = sizeof(ar2) / sizeof(ar2[0]);    if (n1 == n2)        cout << "Median is " << getMedian(ar1, ar2, n1);    else        cout << "Doesn't work for arrays"            << " of unequal size";    getchar();    return 0;}  // This code is contributed// by Lakshay
// C program for the above approach#include <stdio.h>#include <stdlib.h>  /* This function returnsmedian of ar1[] and ar2[].Assumptions in this function:Both ar1[] and ar2[]are sorted arraysBoth have n elements */  // compare function, compares two elementsint compare(const void* num1, const void* num2){    if (*(int*)num1 > *(int*)num2)        return 1;    else        return -1;}  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++;    }      qsort(ar1, n, sizeof(int), compare);    qsort(ar2, n, sizeof(int), compare);      return (ar1[n - 1] + ar2[0]) / 2;}  // Driver Codeint main(){    int ar1[] = { 1, 12, 15, 26, 38 };    int ar2[] = { 2, 13, 17, 30, 45 };      int n1 = sizeof(ar1) / sizeof(ar1[0]);    int n2 = sizeof(ar2) / sizeof(ar2[0]);    if (n1 == n2)        printf("Median is %d ", getMedian(ar1, ar2, n1));    else        printf("Doesn't work for arrays of unequal size");    return 0;}  // This code is contributed by Deepthi
/*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 codepublic 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
# Python program for above approach  # function to return median of the arrays# both are sorted & of same sizedef getMedian(ar1, ar2, n):    i, j = n - 1, 0      # while loop to swap all smaller numbers to arr1    while(ar1[i] > ar2[j] and i > -1 and j < n):        ar1[i], ar2[j] = ar2[j], ar1[i]        i -= 1        j += 1      ar1.sort()    ar2.sort()      return (ar1[-1] + ar2[0]) >> 1    # Driver programif __name__ == '__main__':    ar1 = [1, 12, 15, 26, 38]    ar2 = [2, 13, 17, 30, 45]      n1, n2 = len(ar1), len(ar2)      if(n1 == n2):        print('Median is', getMedian(ar1, ar2, n1))    else:        print("Doesn't work for arrays of unequal size")  # This code is contributed by saitejagampala
/*package whatever //do not write package name here */using System;public 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++;    }    Array.Sort(ar1);    Array.Sort(ar2);    return (ar1[n - 1] + ar2[0]) / 2;}  // Driver codepublic 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)    Console.WriteLine("Median is "+ getMedian(ar1, ar2, n1));    else    Console.WriteLine("Doesn't work for arrays of unequal size");}}  // This code is contributed by aashish1995
<script>      /* This function returns    median of ar1[] and ar2[].    Assumptions in this function:    Both ar1[] and ar2[]    are sorted arrays    Both have n elements */    function getMedian(ar1, ar2, n)    {    let j = 0;    let i = n - 1;    while (ar1[i] > ar2[j] && j < n && i > -1)    {        let temp = ar1[i];        ar1[i] = ar2[j];        ar2[j] = temp;        i--; j++;    }    ar1.sort(function(a, b){return a - b});    ar2.sort(function(a, b){return a - b});    return parseInt((ar1[n - 1] + ar2[0]) / 2, 10);    }          let ar1 = [ 1, 12, 15, 26, 38 ];    let ar2 = [ 2, 13, 17, 30, 45 ];      let n1 = 5;    let n2 = 5;    if (n1 == n2)    document.write("Median is "+ getMedian(ar1, ar2, n1));    else    document.write("Doesn't work for arrays of unequal size");      </script>
Median is 16
Time Complexity: O(nlogn)Auxiliary Space: O(1)
Median of two sorted arrays of different sizesReferences: http://en.wikipedia.org/wiki/Medianhttp://ocw.alfaisal.edu/NR/rdonlyres/Electrical-Engineering-and-Computer-Science/6-046JFall-2005/30C68118-E436-4FE3-8C79-6BAFBB07D935/0/ps9sol.pdf ds3etph5wnPlease write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.
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manupathria
saitejagampala
aashish1995
rrrtnx
2435shivkant
vaibhavrabadiya117
avanitrachhadiya2155
subhammahato348
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deepthiinduri
Code_r
sanyamjain10168
Accolite
Amazon
D-E-Shaw
FactSet
median-finding
Samsung
statistical-algorithms
Arrays
Divide and Conquer
Mathematical
Searching
Accolite
Amazon
Samsung
D-E-Shaw
FactSet
Arrays
Searching
Mathematical
Divide and Conquer
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