Count triplets such that product of two numbers added with third number is N

Given a positive integer N, the task is to find the number of triplets (A, B, C) where A, B, C are positive integers such that the product of two numbers added with the third number is N i.e., A * B + C = N.

Examples:

Input: N = 3
Output: 3
Explanation:
Following are the possible triplets satisfying the given criteria:

1. (1, 1, 2): The value of 1*1 + 2 = 3.
2. (1, 2, 1): The value of 1*2 + 1 = 3.
3. (2, 1, 1): The value of 2*1 + 1 = 3.

Therefore, the total count of such triplets is 3.

Input: N = 5
Output: 8

Approach: The given problem can be solved by rearranging the equation A * B + C = N as A * B = N - C. Now, the only possible values A and B can have to satisfy the above equation is the divisors of N - C. For Example, if the value of N - C = 18, having 6 divisors that are 1, 2, 3, 6, 9, 18. So, values of A, B satisfying the above equation are: (1, 18), (2, 9), (3, 6), (6, 3), (9, 2), (18, 1). So, for the value of N - C = 18, possible values of A, B are 6, i.e., the number of divisors of N - C(= 18). Follow the steps below to solve the given problem:

• Initialize a variable, say count as 0 that stores the total count of possible triplets.
• Iterate a loop over the range [1, N - 1] using the variable i and for each value, i find the total count of divisors of (N - i) and add it to the variable count.
• After completing the above steps, print the value of count as the resultant number of triplets.

Below is the implementation of the above approach:

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the divisors of
// the number (N - i)
int countDivisors(int n)
{
    // Stores the resultant count of
    // divisors of (N - i)
    int divisors = 0;
    int i;

    // Iterate over range [1, sqrt(N)]
    for (i = 1; i * i < n; i++) {
        if (n % i == 0) {
            divisors++;
        }
    }
    if (i - (n / i) == 1) {
        i--;
    }
    for (; i >= 1; i--) {
        if (n % i == 0) {
            divisors++;
        }
    }
    // Return the total divisors
    return divisors;
}

// Function to find the number of triplets
// such that A * B - C = N
int possibleTriplets(int N)
{
    int count = 0;

    // Loop to fix the value of C
    for (int i = 1; i < N; i++) {

        // Adding the number of
        // divisors in count
        count += countDivisors(N - i);
    }

    // Return count of triplets
    return count;
}

// Driver Code
int main()
{
    int N = 10;
    cout << possibleTriplets(N);

    return 0;
}

// Java program for the above approach
import java.io.*;
class GFG
{
  
      // Function to find the divisors of
    // the number (N - i)
    static int countDivisors(int n)
    {
      
        // Stores the resultant count of
        // divisors of (N - i)
        int divisors = 0;
        int i;

        // Iterate over range [1, sqrt(N)]
        for (i = 1; i * i < n; i++) {
            if (n % i == 0) {
                divisors++;
            }
        }
        if (i - (n / i) == 1) {
            i--;
        }
        for (; i >= 1; i--) {
            if (n % i == 0) {
                divisors++;
            }
        }

        // Return the total divisors
        return divisors;
    }

    // Function to find the number of triplets
    // such that A * B - C = N
    static int possibleTriplets(int N)
    {
        int count = 0;

        // Loop to fix the value of C
        for (int i = 1; i < N; i++) {

            // Adding the number of
            // divisors in count
            count += countDivisors(N - i);
        }

        // Return count of triplets
        return count;
    }

    // Driver Code
    public static void main (String[] args) {
      
        int N = 10;
        System.out.println(possibleTriplets(N));
    }
}

// This code is contributed by Dharanendra L V.

# Python program for the above approach
import math

# function to find the divisors of
# the number (N - i)
def countDivisors(n):

    # Stores the resultant count of
    # divisors of (N - i)
    divisors = 0

    # Iterate over range [1, sqrt(N)]
    for i in range(1, math.ceil(math.sqrt(n))+1):
        if n % i == 0:
            divisors = divisors+1

        if (i - (n / i) == 1):
            i = i-1

    for i in range(math.ceil(math.sqrt(n))+1, 1, -1):
        if (n % i == 0):
            divisors = divisors+1

     # Return the total divisors
    return divisors

    # def to find the number of triplets
    # such that A * B - C = N


def possibleTriplets(N):
    count = 0

    # Loop to fix the value of C
    for i in range(1, N):

        # Adding the number of
        # divisors in count
        count = count + countDivisors(N - i)

        # Return count of triplets
    return count

    # Driver Code
# Driver Code
if __name__ == "__main__":
    N = 10
    print(possibleTriplets(N))

# This code is contributed by Potta Lokesh

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG{

// Function to find the divisors of
// the number (N - i)
static int countDivisors(int n)
{
    // Stores the resultant count of
    // divisors of (N - i)
    int divisors = 0;
    int i;

    // Iterate over range [1, sqrt(N)]
    for (i = 1; i * i < n; i++) {
        if (n % i == 0) {
            divisors++;
        }
    }
    if (i - (n / i) == 1) {
        i--;
    }
    for (; i >= 1; i--) {
        if (n % i == 0) {
            divisors++;
        }
    }
    // Return the total divisors
    return divisors;
}

// Function to find the number of triplets
// such that A * B - C = N
static int possibleTriplets(int N)
{
    int count = 0;

    // Loop to fix the value of C
    for (int i = 1; i < N; i++) {

        // Adding the number of
        // divisors in count
        count += countDivisors(N - i);
    }

    // Return count of triplets
    return count;
}

// Driver Code
public static void Main()
{
    int N = 10;
    Console.Write(possibleTriplets(N));
}
}

// This code is contributed by SURENDRA_GANGWAR.

<script>
// javascript program for the above approach 
      // Function to find the divisors of
    // the number (N - i)
    function countDivisors(n)
    {
      
        // Stores the resultant count of
        // divisors of (N - i)
        var divisors = 0;
        var i;

        // Iterate over range [1, sqrt(N)]
        for (i = 1; i * i < n; i++) {
            if (n % i == 0) {
                divisors++;
            }
        }
        if (i - (n / i) == 1) {
            i--;
        }
        for (; i >= 1; i--) {
            if (n % i == 0) {
                divisors++;
            }
        }

        // Return the total divisors
        return divisors;
    }

    // Function to find the number of triplets
    // such that A * B - C = N
    function possibleTriplets(N)
    {
        var count = 0;

        // Loop to fix the value of C
        for (var i = 1; i < N; i++) {

            // Adding the number of
            // divisors in count
            count += countDivisors(N - i);
        }

        // Return count of triplets
        return count;
    }

    // Driver Code
    var N = 10;
    document.write(possibleTriplets(N));

// This code is contributed by shikhasingrajput 
</script>

23

Time Complexity: O(N*sqrt(N))
Auxiliary Space: O(1)