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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 
?> 

Output:

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

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:

My Personal Notes