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

Given an array arr[] of length N, count the number of pairs (i, j) such that arr[i] * arr[j] = arr[i] + arr[j] and 0 <= i < j <= N. It is also given that elements of the array can be any positive integers including zero.

Examples:

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

Input : arr[] = {1, 2, 3, 4}
Output : 0

Simple solution:
We can generate all possible pairs of the array and count those pairs which satisfy the given condition.

Below is the implementation of the above approach:

CPP

// 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(i, j)
// 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++) {

            // Increment count if condition satisfy
            if (arr[i] * arr[j] == arr[i] + arr[j])
                count++;
        }
    }

    // Return count of pairs
    return count;
}

// Driver code
int main()
{

    int arr[] = { 2, 0, 3, 2, 0 };
    int n = sizeof(arr) / sizeof(arr[0]);

    // Get and print count of pairs
    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(i, j)
    // 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++) {

                // Increment count if condition satisfy
                if (arr[i] * arr[j] == arr[i] + arr[j])
                    count++;
            }
        }

        // Return count of pairs
        return count;
    }

    // Driver code
    public static void main(String[] args)
    {

        int arr[] = { 2, 0, 3, 2, 0 };
        int n = arr.length;

        // Get and print count of pairs
        System.out.println(countPairs(arr, n));
    }
}

Python3

# Python3 program to count pairs (i, j)
# such that arr[i] * arr[j] = arr[i] + arr[j]

# Function to return the count of pairs(i, j)
# such that arr[i] * arr[j] = arr[i] + arr[j]
def countPairs(arr, n) :

    count = 0;

    for i in range(n - 1) :
        for j in range(i + 1, n) :

            # Increment count if condition satisfy
            if (arr[i] * arr[j] == arr[i] + arr[j]) :
                count += 1;

    # Return count of pairs
    return count;

# Driver code
if __name__ == "__main__" :

    arr = [ 2, 0, 3, 2, 0 ];
    n = len(arr);

    # Get and print count of pairs
    print(countPairs(arr, n));

# This code is contributed by AnkitRai01

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(i, j)
    // 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++) {

                // Increment count if condition satisfy
                if (arr[i] * arr[j] == arr[i] + arr[j])
                    count++;
            }
        }

        // Return count of pairs
        return count;
    }

    // Driver code
    public static void Main(string[] args)
    {

        int[] arr = { 2, 0, 3, 2, 0 };
        int n = arr.Length;

        // Get and print count of pairs
        Console.WriteLine(countPairs(arr, n));
    }
}

Javascript

<script>
// Function to return the count of pairs(i, j)
// 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++) {


// Increment count if condition satisfy
if (arr[i] * arr[j] == arr[i] + arr[j])
count++;
}
}

// Return count of pairs
return count;
}


let arr = [ 2, 0, 3, 2, 0 ];
let n = arr.length;

document.write(countPairs(arr, n));
// This code is contributed by khatriharsh281</script>

Output:

2

Time Complexity: O(n²)
Auxiliary Space: O(1²)

Efficient Solution:
Taking arr[i] as x and arr[j] as y, we can rewrite the given condition as the following equation.

xy = x + y
xy - x - y = 0
xy - x - y + 1 = 1
x(y - 1) -(y - 1) = 1 
(x - 1)(y - 1) = 1

Case 1:
x - 1 = 1 i.e x = 2
y - 1 = 1 i.e y = 2

Case 2:
x - 1 = -1 i.e x = 0
y - 1 = -1 i.e y = 0

So, now we know that the condition arr[i] * arr[j] = arr[i] + arr[j] will satisfy only if either arr[i] = arr[j] = 0 or arr[i] = arr[j] = 2.
All we need to do is to count the occurrence of 2’s and 0’s. We can then get the number of pairs using formula

(count * (count - 1)) / 2

Below is the implementation of the above approach:

CPP

// 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(i, j)
// such that arr[i] * arr[j] = arr[i] + arr[j]
long countPairs(int arr[], int n)
{

    int countZero = 0;
    int countTwo = 0;

    // Count number of 0's and 2's in the array
    for (int i = 0; i < n; i++) {
        if (arr[i] == 0)
            countZero++;

        else if (arr[i] == 2)
            countTwo++;
    }

    // Total pairs due to occurrence of 0's
    long pair0 = (countZero * (countZero - 1)) / 2;

    // Total pairs due to occurrence of 2's
    long pair2 = (countTwo * (countTwo - 1)) / 2;

    // Return count of all pairs
    return pair0 + pair2;
}

// Driver code
int main()
{

    int arr[] = { 2, 0, 3, 2, 0 };
    int n = sizeof(arr) / sizeof(arr[0]);

    // Get and print count of pairs
    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(i, j)
    // such that arr[i] * arr[j] = arr[i] + arr[j]
    static long countPairs(int arr[], int n)
    {

        int countZero = 0;
        int countTwo = 0;

        // Count number of 0's and 2's in the array
        for (int i = 0; i < n; i++) {
            if (arr[i] == 0)
                countZero++;

            else if (arr[i] == 2)
                countTwo++;
        }

        // Total pairs due to occurrence of 0's
        long pair0 = (countZero * (countZero - 1)) / 2;

        // Total pairs due to occurrence of 2's
        long pair2 = (countTwo * (countTwo - 1)) / 2;

        // Return count of all pairs
        return pair0 + pair2;
    }

    // Driver code
    public static void main(String[] args)
    {

        int arr[] = { 2, 0, 3, 2, 0 };
        int n = arr.length;

        // Get and print count of pairs
        System.out.println(countPairs(arr, n));
    }
}

Python3

# Python3 program to count pairs (i, j)
# such that arr[i] * arr[j] = arr[i] + arr[j]

# Function to return the count of pairs(i, j)
# such that arr[i] * arr[j] = arr[i] + arr[j]
def countPairs(arr, n):

    countZero = 0;
    countTwo = 0;

    # Count number of 0's and 2's in the array
    for i in range(n) :
        if (arr[i] == 0) :
            countZero += 1;

        elif (arr[i] == 2) :
            countTwo += 1;

    # Total pairs due to occurrence of 0's
    pair0 = (countZero * (countZero - 1)) // 2;

    # Total pairs due to occurrence of 2's
    pair2 = (countTwo * (countTwo - 1)) // 2;

    # Return count of all pairs
    return pair0 + pair2;

# Driver code
if __name__ == "__main__" :

    arr = [ 2, 0, 3, 2, 0 ];
    n = len(arr);

    # Get and print count of pairs
    print(countPairs(arr, n));

# This code is contributed by AnkitRai01

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(i, j)
    // such that arr[i] * arr[j] = arr[i] + arr[j]
    static long countPairs(int[] arr, int n)
    {

        int countZero = 0;
        int countTwo = 0;

        // Count number of 0's and 2's in the array
        for (int i = 0; i < n; i++) {
            if (arr[i] == 0)
                countZero++;

            else if (arr[i] == 2)
                countTwo++;
        }

        // Total pairs due to occurrence of 0's
        long pair0 = (countZero * (countZero - 1)) / 2;

        // Total pairs due to occurrence of 2's
        long pair2 = (countTwo * (countTwo - 1)) / 2;

        // Return count of all pairs
        return pair0 + pair2;
    }

    // Driver code
    public static void Main(string[] args)
    {

        int[] arr = { 2, 0, 3, 2, 0 };
        int n = arr.Length;

        // Get and print count of pairs
        Console.WriteLine(countPairs(arr, n));
    }
}

Output:

2

Time Complexity: O(n)
Auxiliary Space: O(1)