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 = Nint 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 Codeint 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 = Ndef 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 codeif __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:
3
Time Complexity: O(N)
Auxiliary Space: O(1)
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