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

  • Difficulty Level : Basic
  • Last Updated : 03 Jun, 2021

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:
 

 

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++




// 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;
}

Java




// 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

Python3




# 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

C#




// 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));
    }
}

PHP




<?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
?>

Javascript




<script>
 
    // JavaScript 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)
    {
        let count = 0;
        for (let i = 0; i < n - 1; i++) {
            for (let j = i + 1; j < n; j++) {
 
                // If condition is satisfied
                if (arr[i] * arr[j] > arr[i] + arr[j])
                    count++;
            }
        }
        return count;
    }
     
    let arr = [ 5, 0, 3, 1, 2 ];
    let n = arr.length;
    document.write(countPairs(arr, n));
 
</script>
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++




// 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;
}

Java




// 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

Python3




# 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

C#




// 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

PHP




<?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_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
?>

Javascript




<script>
 
    // JavaScript 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)
    {
        let count_2 = 0, count_others = 0;
        for (let i = 0; i < n; i++)
        {
            if (arr[i] == 2)
            {
                count_2++;
            }
            else if (arr[i] > 2)
            {
                count_others++;
            }
        }
          
        let ans = count_2 * count_others +
        (count_others * (count_others - 1)) / 2;
        return ans;
    }
     
    let arr = [5, 0, 3, 1, 2];
    let n = arr.length;
    document.write(countPairs(arr, n));
 
</script>
Output: 
3

 

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