Count of triplets till N whose product is at most N

Given a positive integer N, the task is to find the number of triplets (A, B, C) from the first N Natural Numbers such that A * B * C ≤ N.

Examples:

Input: N = 3
Output: 4
Explanation:
Following are the triplets that satisfy the given criteria:

1. ( 1, 1, 1 ) => 1 * 1 * 1 = 1 ≤  2.
2. ( 1, 1, 2 ) => 1 * 1 * 2 = 2 ≤  2.
3. ( 1, 2, 1 ) => 1 * 2 * 1 = 2 ≤ 2.
4. ( 2, 1, 1 ) => 2 * 1 * 1 = 2 ≤ 2.

Therefore, the total count of triplets is 4.

Input: N = 10
Output: 53

Naive Approach: The simplest approach to solve the given problem is to generate all possible triplets from the first N natural numbers and count those triplets that satisfy the given criteria. After checking for all the triplets, print the total count obtained.

Java

/*package whatever //do not write package name here */

import java.io.*;

class GFG {
    public static void main(String[] args)
    {
        int N = 4;
        System.out.println(countTriplets(N));
    }
    public static int countTriplets(int n)
    {
        int count = 0;

        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                for (int k = 1; k <= n; k++) {
                    if (i * j * k <= n) {
                        count++;
                    }
                }
            }
        }

        return count;
    }
}
//This code is contributed by aeroabrar_31

Output

4

Time Complexity: O(N³)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized by the observation that if A and B are fixed, then it is possible to calculate all the possible choices for C by doing N/(A*B) because N/(A*B) will give the maximum value, say X, which on multiplication with (A*B) results in a value less than or equal to N. So, all possible choices of C will be from 1 to X. Now, A and B can be fixed by trying A for every possible number till N, and B for every possible number till (N/A). Follow the steps below to solve the given problem:

• Initialize variable, say cnt as 0 that stores the count of triplets possible.
• Iterate a loop over the range [1, N] using the variable i and nested iterate over the range [1, N] using the variable j and increment the value of cnt by the value of cnt/(i*j).
• After completing the above steps, print the value of cnt as the result.

Below is the implementation of the approach:

C++

// C++ program for the above approach

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

// Function to find number of triplets
// (A, B, C) having A * B * C <= N
int countTriplets(int N)
{
    // Stores the count of triplets
    int cnt = 0;

    // Iterate a loop fixing the value
    // of A
    for (int A = 1; A <= N; ++A) {

        // Iterate a loop fixing the
        // value of A
        for (int B = 1; B <= N / A; ++B) {

            // Find the total count of
            // triplets and add it to cnt
            cnt += N / (A * B);
        }
    }

    // Return the total triplets formed
    return cnt;
}

// Driver Code
int main()
{
    int N = 2;
    cout << countTriplets(N);

    return 0;
}

Java

// Java program for the above approach
import java.io.*;
class GFG {

    // Function to find number of triplets
    // (A, B, C) having A * B * C <= N
    static int countTriplets(int N)
    {

        // Stores the count of triplets
        int cnt = 0;

        // Iterate a loop fixing the value
        // of A
        for (int A = 1; A <= N; ++A) {

            // Iterate a loop fixing the
            // value of A
            for (int B = 1; B <= N / A; ++B) {

                // Find the total count of
                // triplets and add it to cnt
                cnt += N / (A * B);
            }
        }

        // Return the total triplets formed
        return cnt;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int N = 2;

        System.out.println(countTriplets(N));
    }
}

// This code is contributed by dwivediyash

Python3

# Python3 program for the above approach

# Function to find number of triplets
# (A, B, C) having A * B * C <= N
def countTriplets(N) :

    # Stores the count of triplets
    cnt = 0;

    # Iterate a loop fixing the value
    # of A
    for A in range( 1, N + 1) :

        # Iterate a loop fixing the
        # value of A
        for B in range(1, N // A + 1) :

            # Find the total count of
            # triplets and add it to cnt
            cnt += N // (A * B);

    # Return the total triplets formed
    return cnt;

# Driver Code
if __name__ == "__main__" :

    N = 2;
    print(countTriplets(N));

    # This code is contributed by AnkThon

C#

// C# program for the above approach
using System;

public class GFG {

    // Function to find number of triplets
    // (A, B, C) having A * B * C <= N
    static int countTriplets(int N)
    {

        // Stores the count of triplets
        int cnt = 0;

        // Iterate a loop fixing the value
        // of A
        for (int A = 1; A <= N; ++A) {

            // Iterate a loop fixing the
            // value of A
            for (int B = 1; B <= N / A; ++B) {

                // Find the total count of
                // triplets and add it to cnt
                cnt += N / (A * B);
            }
        }

        // Return the total triplets formed
        return cnt;
    }

    // Driver Code
    public static void Main(string[] args)
    {
        int N = 2;

        Console.WriteLine(countTriplets(N));
    }
}

// This code is contributed by AnkThon

Javascript

<script>
        // JavaScript Program to implement
        // the above approach


        // Function to find number of triplets
        // (A, B, C) having A * B * C <= N
        function countTriplets(N) {
            // Stores the count of triplets
            let cnt = 0;

            // Iterate a loop fixing the value
            // of A
            for (let A = 1; A <= N; ++A) {

                // Iterate a loop fixing the
                // value of A
                for (let B = 1; B <= N / A; ++B) {

                    // Find the total count of
                    // triplets and add it to cnt
                    cnt += N / (A * B);
                }
            }

            // Return the total triplets formed
            return cnt;
        }

        // Driver Code

        let N = 2;
        document.write(countTriplets(N));


// This code is contributed by Potta Lokesh
    </script>

Output:

4

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