Skip to content
Related Articles

Related Articles

Count of pairs (x, y) in an array such that x < y
  • Last Updated : 11 Feb, 2020

Given an array of N distinct integers, the task is to find the number of pairs (x, y) such that x < y.

Example:

Input: arr[] = {2, 4, 3, 1}
Output: 6
Possible pairs are (1, 2), (1, 3), (1, 4), (2, 3), (2, 4) and (3, 4)

Input: arr[] = {5, 10}
Output: 1
Only possible pair is (5, 10)

Naive approach: Find every possible pair and check whether it satisfies the given condition or not.



C++




// C++ implementation of the approach 
#include <iostream>
using namespace std;
  
// Function to return the number of 
// pairs (x, y) such that x < y 
int getPairs(int a[],int n) 
    // To store the number of valid pairs 
    int count = 0; 
    for (int i = 0; i < n; i++) 
    
        for (int j = 0; j < n; j++) 
        
  
            // If a valid pair is found 
            if (a[i] < a[j]) 
                count++; 
        
    
  
    // Return the count of valid pairs 
    return count; 
  
// Driver code 
int main() 
    int a[] = { 2, 4, 3, 1 }; 
    int n = sizeof(a) / sizeof(a[0]);
    cout << getPairs(a, n); 
    return 0;
      
// This code is contributed by SHUBHAMSINGH10

Java




// Java implementation of the approach
class GFG {
  
    // Function to return the number of
    // pairs (x, y) such that x < y
    static int getPairs(int a[])
    {
        // To store the number of valid pairs
        int count = 0;
        for (int i = 0; i < a.length; i++) {
            for (int j = 0; j < a.length; j++) {
  
                // If a valid pair is found
                if (a[i] < a[j])
                    count++;
            }
        }
  
        // Return the count of valid pairs
        return count;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int a[] = { 2, 4, 3, 1 };
        System.out.println(getPairs(a));
    }
}

Python 3




# Python3 implementation of the approach
  
# Function to return the number of
# pairs (x, y) such that x < y
def getPairs(a):
      
    # To store the number of valid pairs
    count = 0
    for i in range(len(a)):
        for j in range(len(a)):
  
            # If a valid pair is found
            if (a[i] < a[j]):
                count += 1
  
    # Return the count of valid pairs
    return count
  
# Driver code
if __name__ == "__main__":
  
    a = [ 2, 4, 3, 1 ]
    print(getPairs(a))
  
# This code is contributed by ita_c

C#




// C# implementation of the approach 
using System;
  
class GFG 
  
    // Function to return the number of 
    // pairs (x, y) such that x < y 
    static int getPairs(int []a) 
    
        // To store the number of valid pairs 
        int count = 0; 
        for (int i = 0; i < a.Length; i++) 
        
            for (int j = 0; j < a.Length; j++) 
            
  
                // If a valid pair is found 
                if (a[i] < a[j]) 
                    count++; 
            
        
  
        // Return the count of valid pairs 
        return count; 
    
  
    // Driver code 
    public static void Main() 
    
        int []a = { 2, 4, 3, 1 }; 
        Console.WriteLine(getPairs(a)); 
    
  
// This code is contributed by Ryuga

PHP




<?php
// PHP implementation of the approach
  
// Function to return the number of
// pairs (x, y) such that x < y
function getPairs($a)
{
    // To store the number of valid pairs
    $count = 0;
    for ($i = 0; $i < sizeof($a); $i++) 
    {
        for ($j = 0; $j < sizeof($a); $j++) 
        {
  
            // If a valid pair is found
            if ($a[$i] < $a[$j])
                $count++;
        }
    }
  
    // Return the count of valid pairs
    return $count;
}
  
// Driver code
$a = array(2, 4, 3, 1);
echo getPairs($a);
  
// This code is contributed 
// by Akanksha Rai
?>
Output:
6

Time Complexity: O(n2)

Efficient approach: For an element x. In order to find the count of valid pairs of the form (x, y1), (x, y2), …, (x, yn), we need to count the elements which are greater than x. For the smallest element, there will be n – 1 elements greater than it. Similarly, the second smallest element can form n – 2 pairs and so on. Therefore, the desired count of valid pairs will be (n – 1) + (n – 2) + …. + 1 = n * (n – 1) / 2 where n is the length of the array.

Below is the implementation of the above approach:

C++




// C++ implementation of the approach
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to return the number of
// pairs (x, y) such that x < y
int getPairs(int a[])
{
    // Length of the array
    int n = sizeof(a[0]);
      
    // Calculate the number valid pairs
    int count = (n * (n - 1)) / 2;
  
    // Return the count of valid pairs
    return count;
}
  
// Driver code
int main()
{
    int a[] = { 2, 4, 3, 1 };
    cout << getPairs(a);
    return 0;
}
  
// This code is contributed 
// by SHUBHAMSINGH10

Java




// Java implementation of the approach
class GFG {
  
    // Function to return the number of
    // pairs (x, y) such that x < y
    static int getPairs(int a[])
    {
        // Length of the array
        int n = a.length;
  
        // Calculate the number of valid pairs
        int count = (n * (n - 1)) / 2;
  
        // Return the count of valid pairs
        return count;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int a[] = { 2, 4, 3, 1 };
        System.out.println(getPairs(a));
    }
}

Python




# Python implementation of the approach 
  
# Function to return the number of 
# pairs (x, y) such that x < y 
def getPairs(a):
      
    # Length of the array 
    n = len(a)
      
    # Calculate the number of valid pairs 
    count = (n * (n - 1)) // 2
      
    # Return the count of valid pairs 
    return count
      
# Driver code 
a = [2, 4, 3, 1]
print(getPairs(a)) 
  
# This code is contributed by SHUBHAMSINGH10

C#




// C# implementation of the approach
using System;
class GFG 
{
  
    // Function to return the number of
    // pairs (x, y) such that x < y
    static int getPairs(int []a)
    {
        // Length of the array
        int n = a.Length;
  
        // Calculate the number of valid pairs
        int count = (n * (n - 1)) / 2;
  
        // Return the count of valid pairs
        return count;
    }
  
    // Driver code
    public static void Main()
    {
        int []a = { 2, 4, 3, 1 };
        Console.Write(getPairs(a));
    }
}
  
// This code is contributed 
// by Akanksha Rai
Output:
6

Time Complexity: O(1)

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

My Personal Notes arrow_drop_up
Recommended Articles
Page :