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.

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My Personal Notes

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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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