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Maximum and minimum of an array using minimum number of comparisons

Write a C function to return minimum and maximum in an array. You program should make minimum number of comparisons.


First of all, how do we return multiple values from a C function? We can do it either using structures or pointers.

We have created a structure named pair (which contains min and max) to return multiple values.

struct pair 
{
  int min;
  int max;
};  

And the function declaration becomes: struct pair getMinMax(int arr[], int n) where arr[] is the array of size n whose minimum and maximum are needed.

METHOD 1 (Simple Linear Search)
Initialize values of min and max as minimum and maximum of the first two elements respectively. Starting from 3rd, compare each element with max and min, and change max and min accordingly (i.e., if the element is smaller than min then change min, else if the element is greater than max then change max, else ignore the element)

C



/* structure is used to return two values from minMax() */
#include<stdio.h>
struct pair 
{
  int min;
  int max;
};  
  
struct pair getMinMax(int arr[], int n)
{
  struct pair minmax;     
  int i;
    
  /*If there is only one element then return it as min and max both*/
  if (n == 1)
  {
     minmax.max = arr[0];
     minmax.min = arr[0];     
     return minmax;
  }    
  
  /* If there are more than one elements, then initialize min 
      and max*/
  if (arr[0] > arr[1])  
  {
      minmax.max = arr[0];
      minmax.min = arr[1];
  }  
  else
  {
      minmax.max = arr[1];
      minmax.min = arr[0];
  }    
  
  for (i = 2; i<n; i++)
  {
    if (arr[i] >  minmax.max)      
      minmax.max = arr[i];
    
    else if (arr[i] <  minmax.min)      
      minmax.min = arr[i];
  }
    
  return minmax;
}
  
/* Driver program to test above function */
int main()
{
  int arr[] = {1000, 11, 445, 1, 330, 3000};
  int arr_size = 6;
  struct pair minmax = getMinMax (arr, arr_size);
  printf("nMinimum element is %d", minmax.min);
  printf("nMaximum element is %d", minmax.max);
  getchar();
}  

Java

// Java program of above implementation
public class GFG {
/* Class Pair is used to return two values from getMinMax() */
    static class Pair {
  
        int min;
        int max;
    }
  
    static Pair getMinMax(int arr[], int n) {
        Pair minmax = new  Pair();
        int i;
  
        /*If there is only one element then return it as min and max both*/
        if (n == 1) {
            minmax.max = arr[0];
            minmax.min = arr[0];
            return minmax;
        }
  
        /* If there are more than one elements, then initialize min 
    and max*/
        if (arr[0] > arr[1]) {
            minmax.max = arr[0];
            minmax.min = arr[1];
        } else {
            minmax.max = arr[1];
            minmax.min = arr[0];
        }
  
        for (i = 2; i < n; i++) {
            if (arr[i] > minmax.max) {
                minmax.max = arr[i];
            } else if (arr[i] < minmax.min) {
                minmax.min = arr[i];
            }
        }
  
        return minmax;
    }
  
    /* Driver program to test above function */
    public static void main(String args[]) {
        int arr[] = {1000, 11, 445, 1, 330, 3000};
        int arr_size = 6;
        Pair minmax = getMinMax(arr, arr_size);
        System.out.printf(" Minimum element is %d", minmax.min);
        System.out.printf(" Maximum element is %d", minmax.max);
  
    }
  
}

This article is attributed to GeeksforGeeks.org

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