Count ways of choosing a pair with maximum difference

Given an array of n integers, we need to find the no. of ways of choosing pairs with maximum difference.

Examples:

Input : a[] = {3, 2, 1, 1, 3}
Output : 4
Explanation:- Here, the maximum difference 
you can find is 2 which is from (1, 3).
No. of ways of choosing it:
 1) Choosing the first and third elements,
 2) Choosing the first and fourth elements,
 3) Choosing the third and fifth elements,
 4) Choosing the fourth and fifth elements.
Hence ans is 4.

Input : a[] = {2, 4, 1, 1}
Output : 2
Explanation:- Here, the maximum difference 
is 3 from (1, 4). No. of ways choosing it:
1) Choosing the second and third elements,
2) Choosing the second and fourth elements.
   Hence ans is 2.

Naive Approach : A Simple solution is to find the minimum element and maximum element to find the maximum difference. Then we can find the no. of ways of choosing a pair by running two loops. In the inner loop, check if the two elements(one in outer loop and other in inner loop) are making maximum difference, if yes increase the count.at last output the count.
Time Complexity: O(n^2)
Auxiliary Space: O(1)

Efficient approach:
An efficient approach will be:

  • Case I (if all the elements are equal): The ans is no. of ways of choosing 2 elements from a set of n elements nC2 which is n(n-1)/2.
  • Case II (If all the elements are not equal) : The answer is product of count of no. of minimum elements(c1) and count of no. of maximum elements(c2), i.e., c1*c2

C++

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// CPP Code to find no. of Ways of choosing
// a pair with maximum difference
#include <bits/stdc++.h>
using namespace std;
  
int countPairs(int a[], int n)
{
    // To find minimum and maximum of
    // the array
    int mn = INT_MAX;
    int mx = INT_MIN;
    for (int i = 0; i < n; i++) {
        mn = min(mn, a[i]);
        mx = max(mx, a[i]);
    }
  
    // to find the count of minimum and
    // maximum elements
    int c1 = 0;
    int c2 = 0; // Count variables
    for (int i = 0; i < n; i++) {
        if (a[i] == mn)
            c1++;
        if (a[i] == mx)
            c2++;
    }
  
    // condition for all elements equal
    if (mn == mx)
        return n * (n - 1) / 2;
    else
        return c1 * c2;
}
  
// Driver code
int main()
{
    int a[] = { 3, 2, 1, 1, 3 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << countPairs(a, n);
    return 0;
}

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Java

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// Java Code to find no. of Ways of choosing
// a pair with maximum difference
import java.util.*;
  
class GFG {
  
    static int countPairs(int a[], int n)
    {
  
        // To find minimum and maximum of
        // the array
        int mn = Integer.MAX_VALUE;
        int mx = Integer.MIN_VALUE;
        for (int i = 0; i < n; i++) {
            mn = Math.min(mn, a[i]);
            mx = Math.max(mx, a[i]);
        }
  
        // to find the count of minimum and
        // maximum elements
        int c1 = 0;
        int c2 = 0; // Count variables
        for (int i = 0; i < n; i++) {
            if (a[i] == mn)
                c1++;
            if (a[i] == mx)
                c2++;
        }
  
        // condition for all elements equal
        if (mn == mx)
            return n * (n - 1) / 2;
        else
            return c1 * c2;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int a[] = { 3, 2, 1, 1, 3 };
        int n = a.length;
        System.out.print(countPairs(a, n));
    }
}
  
// This code is contributed by Anant Agarwal.

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Python3

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# Python Code to find no.
# of Ways of choosing
# a pair with maximum difference
  
def countPairs(a, n):
  
    # To find minimum and maximum of 
    # the array 
    mn = +2147483647
    mx = -2147483648
    for i in range(n):
        mn = min(mn, a[i])
        mx = max(mx, a[i])
      
       
    # to find the count of minimum and 
    # maximum elements
    c1 = 0
    c2 = 0 # Count variables
    for i in range(n):
        if (a[i] == mn):
            c1+= 1
        if (a[i] == mx):
            c2+= 1
      
   
    # condition for all elements equal
    if (mn == mx): 
        return  n*(n - 1) // 2
    else:
        return c1 * c2
  
# Driver code
  
a = [ 3, 2, 1, 1, 3]
n = len(a)
  
print(countPairs(a, n))
  
# This code is contributed
# by Anant Agarwal.

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C#

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// C# Code to find no. of Ways of choosing
// a pair with maximum difference
using System;
  
class GFG {
  
    static int countPairs(int[] a, int n)
    {
  
        // To find minimum and maximum of
        // the array
        int mn = int.MaxValue;
        int mx = int.MinValue;
        for (int i = 0; i < n; i++) {
            mn = Math.Min(mn, a[i]);
            mx = Math.Max(mx, a[i]);
        }
  
        // to find the count of minimum and
        // maximum elements
        int c1 = 0;
        int c2 = 0; // Count variables
        for (int i = 0; i < n; i++) {
            if (a[i] == mn)
                c1++;
            if (a[i] == mx)
                c2++;
        }
  
        // condition for all elements equal
        if (mn == mx)
            return n * (n - 1) / 2;
        else
            return c1 * c2;
    }
  
    // Driver code
    public static void Main()
    {
          
        int[] a = { 3, 2, 1, 1, 3 };
        int n = a.Length;
          
        Console.WriteLine(countPairs(a, n));
    }
}
  
// This code is contributed by vt_m.

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PHP

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<?php
// PHP Code to find no. of Ways of choosing
// a pair with maximum difference
  
function countPairs($a, $n)
{
    // To find minimum and maximum of
    // the array
    $mn = PHP_INT_MAX;
    $mx = PHP_INT_MIN;
    for ($i = 0; $i < $n; $i++) {
        $mn = min($mn, $a[$i]);
        $mx = max($mx, $a[$i]);
    }
  
    // to find the count of minimum and
    // maximum elements
    $c1 = 0;
    $c2 = 0; // Count variables
    for ($i = 0; $i < $n; $i++) {
        if ($a[$i] == $mn)
            $c1++;
        if ($a[$i] == $mx)
            $c2++;
    }
  
    // condition for all elements equal
    if ($mn == $mx)
        return $n * ($n - 1) / 2;
    else
        return $c1 * $c2;
}
  
// Driver code
  
    $a = array( 3, 2, 1, 1, 3 );
    $n = count($a);
    echo countPairs($a, $n);
  
// This code is contributed by anuj_67.
?>

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Output:

  4

Time Complexity: Time complexity to find minimum and maximum is O(n) and Time Complexity to find count of minimum and maximum is O(n)
Overall Time complexity : O(n)
Auxiliary Space : O(1)

This article is contributed by Harsha mogali. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : vt_m