Program to find Median of two sorted arrays of equal size in C

October 12, 2022

Median of two sorted arrays of equal size in C

Here, in this page we will discuss the program to find median of two sorted arrays of equal size in C++ programming language. We are given with two arrays say arr1[] and arr2[] of the same size say n . We need to find the median after merging these arrays.

Median of two sorted arrays of equal size in C

Method 1:

  • Find the union of the given two arrays.
  • Sort both array 1 and array2.
  • Then the median element will be
    Median = (arr1[n-1]+arr2[0])/2
Run 
#include <stdio.h>

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++;
      }

    //sort ar1
    for(int i=0; i<n; i++){
        for(int j=i+1; j<n; j++){ if(ar1[i]>ar1[j]){
                int temp = ar1[i];
                ar1[i] = ar1[j];
                ar1[j] = temp;
            }
        }
    }
    
    //sort ar2
    for(int i=0; i<n; i++){
        for(int j=i+1; j<n; j++){ if(ar2[i]>ar2[j]){
                int temp = ar2[i];
                ar2[i] = ar2[j];
                ar2[j] = temp;
            }
        }
    }

    return (ar1[n - 1] + ar2[0]) / 2;
}

int main()
{
 
   int arr1[]={1, 12, 15, 26, 38}, arr2[]={2, 13, 17, 30, 45};
   int n = sizeof(arr1)/sizeof(arr1[0]);
   
   printf("%d",getMedian(arr1, arr2, n));

   return 0;
}

Output :

16

Given an array which consists of only 0, 1 and 2

Find the “Kth” max and min element of an array

Find the Union and Intersection of the two sorted arrays

Find Largest sum contiguous Subarray

Minimize the maximum difference between heights 

Method 2 :

  • Create variable count. Keep track of count while comparing elements of two arrays.
  • Run a loop that will terminate when count > n.
  • 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.

Time and Space Complexities :

  • Time-Complexity :O(n)
  • Space-Complexity : O(1)
Run 
#include <stdio.h>

int getMedian(int ar1[], int ar2[], int n)
{
    int i = 0;
    int j = 0; 
    int count;
    int m1 = -1, m2 = -1;

         for (count = 0; count <= n; count++){
              if (i == n){
                  m1 = m2;
                  m2 = ar2[0];
                  break;
              }
             else if (j == n){
                 m1 = m2;
                 m2 = ar1[0];
                 break;
             }
            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;
}

int main()
{
 
   int arr1[]={1, 12, 15, 26, 38}, arr2[]={2, 13, 17, 30, 45};
   int n = sizeof(arr1)/sizeof(arr1[0]);
   
   printf("%d",getMedian(arr1, arr2, n));

   return 0;
}

Output :

16

Method 3  :

  • Find the medians of the given two arrays and store them in variable say m1 and m2.
  • Now, if m1=m2, then that will be the required output.
  • If m1>m2, then we check in the subarrays, arr1[0 – middle element] and in arr2[middle element – last].
  • If m2>m1, then we check in the subarrays, arr1[middle element – last] and in arr2[0-middle element ].
  • This will be the recursive process and repeat this till size of both the arrays become 2.
  • As, when the size will be 2 , we can use the formula :

    Median = (max(ar1[0], ar2[0]) + min(ar1[1], ar2[1]))/2

Time and Space Complexities :

  • Time-Complexity :O(logn)
  • Space-Complexity : O(1)
Run 
#include<stdio.h> 

int max(int a, int b){
    if(a>b)
        return a;
    return b;
}

int min(int a, int b){
    if(a>b)
        return b;
    return a;
}

/* 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];
}

int getMedian(int ar1[], int ar2[], int n)
{

    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); 
    int m2 = median(ar2, n); 

    if (m1 == m2)
     return m1;

    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 (n % 2 == 0)
      return getMedian(ar2 + n/2 - 1, ar1, n - n/2 + 1);
    return getMedian(ar2 + n/2, ar1, n - n/2);
}

int main()
{
 
   int arr1[]={1, 12, 15, 26, 38}, arr2[]={2, 13, 17, 30, 45};
   int n = sizeof(arr1)/sizeof(arr1[0]);
   
   printf("%d",getMedian(arr1, arr2, n));

   return 0;
}

Output :

16

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