Count number of triplets (a, b, c) from first N natural numbers such that a * b + c = N

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Given an integer N, the task is to count the triplets (a, b, c) from the first N natural numbers such that a * b + c = N.

Examples:

Input: N = 3
Output: 3
Explanation:
Triplets of the form a * b + c = N are { (1, 1, 2), (1, 2, 1), (2, 1, 1) }
Therefore, the required output is 3.

Input: N = 100
Output: 473

Approach: The problem can be solved based on the following observation:

For every possible pairs (a, b), If a * b < N, then only c exists. Therefore, count the pairs (a, b) whose product is less than N.

Follow the steps below to solve the problem:

Initialize a variable, say cntTriplets, to store the count of triplets of first N natural numbers that satisfy the given condition.
Iterate over the range [1, N – 1] using variable i and check if N % i == 0 or not. If found to be true, then update cntTriplets += (N / i) – 1.
Otherwise, update cntTriplets += (N / i).
Finally, print the value of cntTriplets.

Below is the implementation of the above approach.

C++

// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the count of
// triplets (a, b, c) with a * b + c = N
int findCntTriplet(int N)
{
    // Stores count of triplets of 1st
    // N natural numbers which are of
    // the form a * b + c = N
    int cntTriplet = 0;
 
    // Iterate over the range [1, N]
    for (int i = 1; i < N; i++) {
 
        // If N is divisible by i
        if (N % i != 0) {
 
            // Update cntTriplet
            cntTriplet += N / i;
        }
        else {
 
            // Update cntTriplet
            cntTriplet += (N / i) - 1;
        }
    }
    return cntTriplet;
}
 
// Driver Code
int main()
{
    int N = 3;
    cout << findCntTriplet(N);
    return 0;
}

Java

// Java program to implement 
// the above approach 
import java.util.*;
class GFG
{
 
  // Function to find the count of
  // triplets (a, b, c) with a * b + c = N
  static int findCntTriplet(int N)
  {
 
    // Stores count of triplets of 1st
    // N natural numbers which are of
    // the form a * b + c = N
    int cntTriplet = 0;
 
    // Iterate over the range [1, N]
    for (int i = 1; i < N; i++)
    {
 
      // If N is divisible by i
      if (N % i != 0)
      {
 
        // Update cntTriplet
        cntTriplet += N / i;
      }
      else
      {
 
        // Update cntTriplet
        cntTriplet += (N / i) - 1;
      }
    }
    return cntTriplet;
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int N = 3;
    System.out.println(findCntTriplet(N));
  }
}
 
// This code is contributed by susmitakundugoaldanga

Python3

# Python program to implement
# the above approach
 
# Function to find the count of
# triplets (a, b, c) with a * b + c = N
def findCntTriplet(N):
   
    # Stores count of triplets of 1st
    # N natural numbers which are of
    # the form a * b + c = N
    cntTriplet = 0;
 
    # Iterate over the range [1, N]
    for i in range(1, N):
 
        # If N is divisible by i
        if (N % i != 0):
 
            # Update cntTriplet
            cntTriplet += N // i;
        else:
 
            # Update cntTriplet
            cntTriplet += (N // i) - 1;
 
    return cntTriplet;
 
# Driver code
if __name__ == '__main__':
    N = 3;
    print(findCntTriplet(N));
 
# This code is contributed by 29AjayKumar

C#

// C# program to implement 
// the above approach 
using System;
class GFG
{
 
  // Function to find the count of
  // triplets (a, b, c) with a * b + c = N
  static int findCntTriplet(int N)
  {
 
    // Stores count of triplets of 1st
    // N natural numbers which are of
    // the form a * b + c = N
    int cntTriplet = 0;
 
    // Iterate over the range [1, N]
    for (int i = 1; i < N; i++)
    {
 
      // If N is divisible by i
      if (N % i != 0)
      {
 
        // Update cntTriplet
        cntTriplet += N / i;
      }
      else
      {
 
        // Update cntTriplet
        cntTriplet += (N / i) - 1;
      }
    }
    return cntTriplet;
  }
 
  // Driver code
  public static void Main(String[] args)
  {
    int N = 3;
    Console.WriteLine(findCntTriplet(N));
  }
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
 
// javascript program for the above approach
  
  // Function to find the count of
  // triplets (a, b, c) with a * b + c = N
  function findCntTriplet(N)
  {
 
    // Stores count of triplets of 1st
    // N natural numbers which are of
    // the form a * b + c = N
    let cntTriplet = 0;
 
    // Iterate over the range [1, N]
    for (let i = 1; i < N; i++)
    {
 
      // If N is divisible by i
      if (N % i != 0)
      {
 
        // Update cntTriplet
        cntTriplet += Math.floor(N / i);
      }
      else
      {
 
        // Update cntTriplet
        cntTriplet += Math.floor(N / i) - 1;
      }
    }
    return cntTriplet;
  }
 
// Driver Code
     
    let N = 3;
    document.write(findCntTriplet(N));
     
</script>

Output:

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

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