Count number of pairs (i, j) such that arr[i] * arr[j] > arr[i] + arr[j]

Given an array arr[] of non-negative integers, the task is to count pairs of indices (i, j such that arr[i] * arr[j] > arr[i] + arr[j] where i < j.

Examples:

Input: arr[] = { 5, 0, 3, 1, 2 }
Output: 3

Input: arr[] = { 1, 1, 1 }
Output: 0



Naive Approach: Run two nested loops and check for every pair whether the condition is satisfied. If the condition is satisfied for any pair then update count = count + 1 and print the count in the end.

Below is the implementation of the above approach:

C++

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// C++ program to count pairs (i, j)
// such that arr[i] * arr[j] > arr[i] + arr[j]
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the count of pairs
// such that arr[i] * arr[j] > arr[i] + arr[j]
long countPairs(int arr[], int n)
{
    long count = 0;
    for (int i = 0; i < n - 1; i++) {
        for (int j = i + 1; j < n; j++) {
  
            // If condition is satisfied
            if (arr[i] * arr[j] > arr[i] + arr[j])
                count++;
        }
    }
    return count;
}
  
// Driver code
int main()
{
    int arr[] = { 5, 0, 3, 1, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countPairs(arr, n);
    return 0;
}

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Java














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// Java program to count pairs (i, j) 


// such that arr[i] * arr[j] > arr[i] + arr[j] 


import java.util.*; 


  


class solution 


{ 


  


// Function to return the count of pairs 


// such that arr[i] * arr[j] > arr[i] + arr[j] 


static long countPairs(int arr[], int n) 


{ 


    long count = 0; 


    for (int i = 0; i < n - 1; i++) { 


        for (int j = i + 1; j < n; j++) { 


  


            // If condition is satisfied 


            if (arr[i] * arr[j] > arr[i] + arr[j]) 


                count++; 


        } 


    } 


    return count; 


} 


  


// Driver code 


public static void main(String args[]) 


{ 


    int arr[] = { 5, 0, 3, 1, 2 }; 


    int n = arr.length; 


    System.out.println(countPairs(arr, n)); 


      


} 


} 


  


// This code is contributed by 


// Surendra_Gangwar 



















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Python3














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# Python3 program to count pairs(i,j) 


# such that arr[i]*arr[j]>arr[i]+arr[j] 


import math as mt 


  


# function to return the count of pairs  


# such that arr[i]*arr[j]>arr[i]+arr[j] 


def countPairs(arr, n): 


    count = 0


      


    for i in range(n): 


        for j in range(i + 1, n): 


              


            # if condition is satisified 


            if arr[i] * arr[j] > arr[i] + arr[j]: 


                count += 1


              


    return count 


  


# Driver code 


arr = [5, 0, 3, 1, 2] 


n = len(arr) 


  


print(countPairs(arr, n)) 


          


# This code is contributed  


# by Mohit Kumar 29 



















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C#














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// C# program to count pairs (i, j) 


// such that arr[i] * arr[j] > arr[i] + arr[j] 


  


using System; 


  


public class GFG{ 


      


// Function to return the count of pairs 


// such that arr[i] * arr[j] > arr[i] + arr[j] 


static long countPairs(int []arr, int n) 


{ 


    long count = 0; 


    for (int i = 0; i < n - 1; i++) { 


        for (int j = i + 1; j < n; j++) { 


  


            // If condition is satisfied 


            if (arr[i] * arr[j] > arr[i] + arr[j]) 


                count++; 


        } 


    } 


    return count; 


} 


  


// Driver code 


    static public void Main (){ 


    int []arr = { 5, 0, 3, 1, 2 }; 


    int n = arr.Length; 


    Console.WriteLine (countPairs(arr, n)); 


    } 


} 



















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PHP














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<?php 


// PHP program to count pairs (i, j)  


// such that arr[i] * arr[j] > arr[i] + arr[j]  


  


// Function to return the count of pairs  


// such that arr[i] * arr[j] > arr[i] + arr[j]  


function countPairs($arr, $n)  


{  


    $count = 0;  


    for ($i = 0; $i < $n - 1; $i++)  


    {  


        for ($j = $i + 1; $j < $n; $j++)  


        {  


  


            // If condition is satisfied  


            if ($arr[$i] *  


                $arr[$j] > $arr[$i] +  


                           $arr[$j])  


                $count++;  


        }  


    }  


    return $count;  


}  


  


// Driver code  


$arr = array( 5, 0, 3, 1, 2 );  


$n = sizeof($arr) ; 


  


echo countPairs($arr, $n); 


  


// This code is contributed by Ryuga 


?> 



















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Output:


3





Efficient Approach: Consider the following cases:



1) arr[i] = 0 or arr[i] = 1 and arr[j] = any element

In this case, arr[j] * arr[i] will always be less than arr[i] + arr[j].

Hence we can discard all pairs which have one element either 0 or 1.


2) arr[i] = 2 and arr[j] <= 2

In this case, arr[j] * arr[i] will always be less than or equal to arr[i] + arr[j].

Hence again we can discard all such pairs.


3) arr[i] = 2 and arr[j] > 2

This case will produce valid pairs. If count_2 is the count of ‘2’s and count_others

is the count of elements greater than 2,

then number of pairs will be count_2 * count_others.


4) arr[i] > 2 and arr[j] > 2

This case will also produce valid pairs. Let count_others be the number of elements

greater than 2, then every two elements among these count_others elements

will form a valid pair. Hence the number of pairs will be




Therefore, total count = (count_2 * count_others) + (count_others * (count_others – 1)) / 2.



Below is the implementation of the above approach:




C++















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// C++ program to count pairs (i, j) 


// such that arr[i] * arr[j] > arr[i] + arr[j] 


#include <bits/stdc++.h> 


using namespace std; 


  


// Function to return the count of pairs 


// such that arr[i] * arr[j] > arr[i] + arr[j] 


long countPairs(const int* arr, int n) 


{ 


    int count_2 = 0, count_others = 0; 


    for (int i = 0; i < n; i++) { 


        if (arr[i] == 2) 


            count_2++; 


        else if (arr[i] > 2) 


            count_others++; 


    } 


    long ans 


        = 1L * count_2 * count_others 


          + (1L * count_others * (count_others - 1)) / 2; 


    return ans; 


} 


  


// Driver code 


int main() 


{ 


    int arr[] = { 5, 0, 3, 1, 2 }; 


    int n = sizeof(arr) / sizeof(arr[0]); 


    cout << countPairs(arr, n); 


    return 0; 


} 



















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Java














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// Java program to count pairs (i, j)  


// such that arr[i] * arr[j] > arr[i] + arr[j]  


  


class GFG  


{ 


    // Function to return the count of pairs  


    // such that arr[i] * arr[j] > arr[i] + arr[j]  


    static long countPairs(int[] arr, int n)  


    { 


        int count_2 = 0, count_others = 0; 


        for (int i = 0; i < n; i++)  


        { 


            if (arr[i] == 2)  


            { 


                count_2++; 


            }  


            else if (arr[i] > 2)  


            { 


                count_others++; 


            } 


        } 


          


        long ans = 1L * count_2 * count_others + 


                (1L * count_others * (count_others - 1)) / 2; 


        return ans; 


    } 


  


    // Driver code  


    public static void main(String[] args)  


    { 


        int arr[] = {5, 0, 3, 1, 2}; 


        int n = arr.length; 


        System.out.println(countPairs(arr, n)); 


    } 


} 


  


// This code is contributed by  


// 29AjayKumar  



















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Python3














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# Python3 program to count pairs(i,j) 


# such that arr[i]*arr[j]>arr[i]+arr[j] 


import math as mt 


  


# function to return the count of pairs  


# such that arr[i]*arr[j]>arr[i]+arr[j] 


def countPairs(arr, n): 


    count_2, count_others = 0, 0


      


    for i in range(n): 


        if arr[i] == 2: 


            count_2 += 1


        elif arr[i] > 2: 


            count_others += 1


    ans = (count_2 * count_others +


          (count_others * 


          (count_others - 1)) // 2) 


    return ans 


  


# Driver code 


arr = [5, 0, 3, 1, 2] 


n = len(arr) 


  


print(countPairs(arr, n)) 


          


# This code is contributed  


# by Mohit Kumar 



















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C#














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// C# program to count pairs (i, j) such  


// that arr[i] * arr[j] > arr[i] + arr[j]  


using System; 


  


class GFG  


{ 


    // Function to return the count of pairs  


    // such that arr[i] * arr[j] > arr[i] + arr[j]  


    static long countPairs(int[] arr, int n)  


    { 


        int count_2 = 0, count_others = 0; 


        for (int i = 0; i < n; i++)  


        { 


            if (arr[i] == 2)  


            { 


                count_2++; 


            }  


            else if (arr[i] > 2)  


            { 


                count_others++; 


            } 


        } 


          


        long ans = 1L * count_2 * count_others + 


                  (1L * count_others *  


                       (count_others - 1)) / 2; 


        return ans; 


    } 


  


    // Driver code  


    public static void Main()  


    { 


        int[] arr = {5, 0, 3, 1, 2}; 


        int n = arr.Length; 


        Console.WriteLine(countPairs(arr, n)); 


    } 


} 


  


// This code is contributed by  


// Mukul Singh 



















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PHP





 arr[i] + arr[j]


// Function to return the count of pairs

// such that arr[i] * arr[j] > arr[i] + arr[j]

function countPairs($arr, $n)

{

	$count_2 = 0; $count_others = 0;

	for ($i = 0; $i < $n; $i++) 
	{
		if ($arr[$i] == 2)
			$count_2++;
		else if ($arr[$i] > 2)

			$count_others++;

	}

	$ans = $count_2 * $count_others +

		             ($count_others *

		             ($count_others – 1)) / 2;

	return $ans;

}


// Driver code

$arr = array( 5, 0, 3, 1, 2 );

$n = sizeof($arr);

echo countPairs($arr, $n);


// This code is contributed

// by Akanksha Rai

?>





Output:


3






          
          
          
          

            

    

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