Count number of triplets with product not exceeding a given number
Given a positive integer N, the task is to find the number of triplets of positive integers (X, Y, Z), whose product is at most N.
Examples:
Input: N = 2
Output: 4
Explanation: Below are the triplets whose product is at most N(= 2):
- (1, 1, 1): Product is 1*1*1 = 1.
- (1, 1, 2): Product is 1*1*2 = 2.
- (1, 2, 1): Product is 1*2*1 = 2.
- (2, 1, 1): Product is 2*1*1 = 2.
Therefore, the total count is 4.
Input: 6
Output: 25
Naive Approach: The simplest approach to solve the given problem is to generate all possible triplets whose values lie over the range [0, N] and count those triplets whose product of values is at most N. After checking for all the triplets, print the total count obtained.
Time Complexity: O(N3)
Auxiliary Space: O(1)
Efficient Approach: The above approach can also be optimized by generating all possible pairs (i, j) over the range [1, N] and increment the count of all possible pairs by N / (i * j). Follow the steps below to solve the problem:
- Initialize a variable, say ans, that stores the count of all possible triplets.
- Generate all possible pairs (i, j) over the range [1, N] and if the product of the pairs is greater than N, then check for the next pairs. Otherwise, increment the count of all possible pairs by N/(i*j).
- After completing the above steps, print the value of ans as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <iostream> using namespace std; // Function to count the number of // triplets whose product is at most N int countTriplets( int N) { // Stores the count of triplets int ans = 0; // Iterate over the range [0, N] for ( int i = 1; i <= N; i++) { // Iterate over the range [0, N] for ( int j = 1; j <= N; j++) { // If the product of // pairs exceeds N if (i * j > N) break ; // Increment the count of // possible triplets ans += N / (i * j); } } // Return the total count return ans; } // Driver Code int main() { int N = 10; cout << countTriplets(N); return 0; } |
Java
// Java program for the above approach import java.util.*; class GFG{ // Function to count the number of // triplets whose product is at most N static int countTriplets( int N) { // Stores the count of triplets int ans = 0 ; // Iterate over the range [0, N] for ( int i = 1 ; i <= N; i++) { // Iterate over the range [0, N] for ( int j = 1 ; j <= N; j++) { // If the product of // pairs exceeds N if (i * j > N) break ; // Increment the count of // possible triplets ans += N / (i * j); } } // Return the total count return ans; } // Driver Code public static void main(String[] args) { int N = 10 ; System.out.print(countTriplets(N)); } } // This code is contributed by Amit Katiyar |
Python3
# Python3 program for the above approach # Function to count the number of # triplets whose product is at most N def countTriplets(N): # Stores the count of triplets ans = 0 # Iterate over the range [0, N] for i in range ( 1 , N + 1 ): # Iterate over the range [0, N] for j in range ( 1 , N + 1 ): # If the product of # pairs exceeds N if (i * j > N): break # Increment the count of # possible triplets ans + = N / / (i * j) # Return the total count return ans # Driver Code if __name__ = = "__main__" : N = 10 print (countTriplets(N)) # This code is contributed by ukasp. |
C#
// C# program for the above approach using System; class GFG{ // Function to count the number of // triplets whose product is at most N static int countTriplets( int N) { // Stores the count of triplets int ans = 0; // Iterate over the range [0, N] for ( int i = 1; i <= N; i++) { // Iterate over the range [0, N] for ( int j = 1; j <= N; j++) { // If the product of // pairs exceeds N if (i * j > N) break ; // Increment the count of // possible triplets ans += N / (i * j); } } // Return the total count return ans; } // Driver Code public static void Main(String[] args) { int N = 10; Console.Write(countTriplets(N)); } } // This code is contributed by Princi Singh |
Javascript
<script> // JavaScript program for the above approach // Function to count the number of // triplets whose product is at most N function countTriplets( N){ // Stores the count of triplets let ans = 0; // Iterate over the range [0, N] for (let i = 1; i <= N; i++) { // Iterate over the range [0, N] for (let j = 1; j <= N; j++) { // If the product of // pairs exceeds N if (i * j > N) break ; // Increment the count of // possible triplets ans += Math.floor(N / (i * j)); } } // Return the total count return ans; } // Driver Code let N = 10; document.write( countTriplets(N)); // This code is contributed by rohitsingh07052. </script> |
53
Time Complexity: O(N2)
Auxiliary Space: O(1)