Number of pairs in an array such that product is greater than sum

Given a array a[] of non-negative integers. Count the number of pairs (i, j) in the array such that a[i] + a[j] < a[i]*a[j]. (the pair (i, j) and (j, i) are considered same and i should not be equal to j)

Examples:

Input : a[] = {3, 4, 5}
Output : 3
Pairs are (3, 4) , (4, 5) and (3,5)

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

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Naive approach
For each value a[i] count the number of a[j] (i > j) such that a[i]*a[j] > a[i] + a[j]

C++

// Naive C++ program to count number of pairs 
// such that their sum is more than product. 
#include<bits/stdc++.h> 
using namespace std; 
  
// Returns the number of valid pairs 
int countPairs (int arr[], int n) 
{     
    int ans = 0;  // initializing answer 
  
    // Traversing the array. For each array 
    // element, checking its predecessors that 
    // follow the condition 
    for (int i = 0; i<n; i++) 
        for (int j = i-1; j>= 0; j--) 
            if (arr[i]*arr[j] > arr[i] + arr[j]) 
                ans++; 
    return ans; 
} 
  
// Driver function 
int main() 
{ 
    int arr[] = {3, 4, 5}; 
    int n = sizeof(arr)/sizeof(arr[0]); 
    cout << countPairs(arr, n); 
    return 0; 
} 

Java

// Naive java program to count number of pairs 
// such that their sum is more than product. 
import java.*; 
  
public class GFG  
{ 
      
    // Returns the number of valid pairs 
    static int countPairs (int arr[], int n) 
    {  
        int ans = 0; // initializing answer 
      
        // Traversing the array. For each array 
        // element, checking its predecessors that 
        // follow the condition 
        for (int i = 0; i<n; i++) 
            for (int j = i-1; j>= 0; j--) 
                if (arr[i]*arr[j] > arr[i] + arr[j]) 
                    ans++; 
        return ans; 
    } 
      
    // Driver code 
    public static void main(String args[]) 
    { 
        int arr[] = {3, 4, 5}; 
        int n = arr.length; 
        System.out.println(countPairs(arr, n)); 
    } 
} 
  
// This code is contributed by Sam007 

Python3

# Naive Python program to count number 
# of pairs such that their sum is more 
# than product. 
  
# Returns the number of valid pairs 
def countPairs(arr, n): 
      
    # initializing answer 
    ans = 0
      
    # Traversing the array. For each 
    # array element, checking its  
    # predecessors that follow the 
    # condition 
    for i in range(0, n): 
        j = i-1
        while(j >= 0): 
            if (arr[i] * arr[j] >  
                     arr[i] + arr[j]): 
                ans = ans + 1
            j = j - 1
    return ans 
      
# Driver program to test above function. 
arr = [3, 4, 5] 
n = len(arr)  
k = countPairs(arr, n) 
print(k) 
      
# This code is contributed by Sam007. 

C#

// Naive C# program to count number of pairs 
// such that their sum is more than product. 
using System; 
          
public class GFG  
{ 
    // Returns the number of valid pairs 
    static int countPairs (int []arr, int n) 
    {  
        int ans = 0; // initializing answer 
      
        // Traversing the array. For each array 
        // element, checking its predecessors that 
        // follow the condition 
        for (int i = 0; i<n; i++) 
            for (int j = i-1; j>= 0; j--) 
                if (arr[i]*arr[j] > arr[i] + arr[j]) 
                    ans++; 
        return ans; 
    } 
      
    // driver program 
    public static void Main()  
    { 
        int []arr = {3, 4, 5}; 
        int n = arr.Length; 
        Console.Write( countPairs(arr, n)); 
    } 
} 
  
// This code is contributed by Sam007 

PHP

<?php 
// Naive PHP program to  
// count number of pairs 
// such that their sum  
// is more than product. 
  
// Returns the number 
// of valid pairs 
function countPairs ($arr, $n) 
{  
    // initializing answer 
    $ans = 0;  
  
    // Traversing the array. 
    // For each array 
    // element, checking  
    // its predecessors that 
    // follow the condition 
    for ($i = 0; $i < $n; $i++) 
        for ($j = $i - 1; $j >= 0; $j--) 
            if ($arr[$i] * $arr[$j] >  
                $arr[$i] + $arr[$j]) 
                $ans++; 
    return $ans; 
} 
  
// Driver Code 
$arr = array(3, 4, 5); 
$n = sizeof($arr); 
echo(countPairs($arr, $n)); 
  
// This code is contributed by Ajit. 
?> 

Output:

Efficient approach
When a[i] = 0 : a[i]*a[j] = 0 and a[i] + a[j] >= 0 so if a[i] = 0 no pairs can be found.
When a[i] = 1 : a[i]*a[j] = a[j] and a[i] + a[j] = 1 + a[j], so no pairs can be found when a[i] = 1
When a[i] = 2 and a[j] = 2 : a[i]*a[j] = a[i] + a[j] = 4
When a[i] = 2 and a[j] > 2 or a[i] > 2 and a[j] >= 2 : All such pairs are valid.

To solve this problem, count the number of 2s in the array say twoCount. Count the numbers greater than 2 in the array say twoGreaterCount. Answer will be twoCount * twoGreaterCount + twoGreaterCount * (twoGreaterCount-1)/2

C++

// C++ implementation of efficient approach 
// to count valid pairs. 
#include<bits/stdc++.h> 
using namespace std; 
  
// returns the number of valid pairs 
int CountPairs (int arr[], int n) 
{ 
    // traversing the array, counting the 
    // number of 2s and greater than 2 
    // in array 
    int twoCount = 0, twoGrCount = 0; 
    for (int i = 0; i<n; i++) 
    { 
        if (arr[i] == 2) 
            twoCount++; 
        else if (arr[i] > 2) 
            twoGrCount++; 
    } 
    return twoCount*twoGrCount + 
          (twoGrCount*(twoGrCount-1))/2; 
} 
  
// Driver function 
int main() 
{ 
    int arr[] = {3, 4, 5}; 
    int n = sizeof(arr)/sizeof(arr[0]); 
    cout << CountPairs(arr, n); 
    return 0; 
} 

Java

// Java implementation of efficient approach 
// to count valid pairs. 
import java.*; 
  
public class GFG  
{ 
    // Returns the number of valid pairs 
    static int countPairs (int arr[], int n) 
    {  
        // traversing the array, counting the 
        // number of 2s and greater than 2 
        // in array 
        int twoCount = 0, twoGrCount = 0; 
        for (int i = 0; i<n; i++) 
        { 
          if (arr[i] == 2) 
            twoCount++; 
          else if (arr[i] > 2) 
            twoGrCount++; 
        } 
        return twoCount*twoGrCount + 
        (twoGrCount*(twoGrCount-1))/2; 
    } 
      
    // Driver code 
    public static void main(String args[]) 
    { 
        int arr[] = {3, 4, 5}; 
        int n = arr.length; 
        System.out.println(countPairs(arr, n)); 
    } 
} 
  
// This code is contributed by Sam007 

Python3

# python implementation of efficient approach 
# to count valid pairs. 
  
# returns the number of valid pairs 
def CountPairs (arr,n): 
      
    # traversing the array, counting the 
    # number of 2s and greater than 2 
    # in array 
    twoCount = 0
    twoGrCount = 0
    for i in range(0, n): 
          
        if (arr[i] == 2): 
            twoCount += 1
        elif (arr[i] > 2): 
            twoGrCount += 1
      
    return ((twoCount * twoGrCount)  
      + (twoGrCount * (twoGrCount - 1)) / 2) 
  
# Driver function 
arr = [3, 4, 5] 
n = len(arr) 
print( CountPairs(arr, n)) 
  
# This code is contributed by Sam007. 

C#

// C# implementation of efficient approach 
// to count valid pairs. 
using System; 
          
public class GFG  
{ 
    // Returns the number of valid pairs 
    static int countPairs (int []arr, int n) { 
          
    // traversing the array, counting the 
    // number of 2s and greater than 2 
    // in array 
    int twoCount = 0, twoGrCount = 0; 
    for (int i = 0; i<n; i++) 
    { 
        if (arr[i] == 2) 
            twoCount++; 
        else if (arr[i] > 2) 
            twoGrCount++; 
    } 
    return twoCount*twoGrCount + 
           (twoGrCount*(twoGrCount-1))/2; 
    } 
      
    // driver program 
    public static void Main()  
    { 
        int []arr = {3, 4, 5}; 
        int n = arr.Length; 
        Console.Write( countPairs(arr, n)); 
    } 
} 
  
// This code is contributed by Sam007 

PHP

<?php 
// PHP implementation of 
// efficient approach 
// to count valid pairs. 
  
// returns the number 
// of valid pairs 
function CountPairs ($arr, $n) 
{ 
      
    // traversing the array, counting  
    // the number of 2s and greater  
    // than 2 in array 
    $twoCount = 0; $twoGrCount = 0; 
      
    for ($i = 0; $i < $n; $i++) 
    { 
        if ($arr[$i] == 2) 
            $twoCount++; 
        else if ($arr[$i] > 2) 
            $twoGrCount++; 
    } 
    return $twoCount * $twoGrCount +  
          ($twoGrCount * ($twoGrCount -  
                               1)) / 2; 
} 
  
// Driver Code 
$arr = array(3, 4, 5); 
$n = sizeof($arr); 
echo(CountPairs($arr, $n)); 
  
// This code is contributed by Ajit. 
?> 

Output:

This article is contributed by Ayush Jha. 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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

My Personal Notes

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

PHP

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