# Count of sorted triplets (a, b, c) whose product is at most N

Given an integer **N**, the task is to find the count of triplets (a, b, c) such that a <= b <= c and a * b * c <= N.

**Examples:**

Input:N = 5Output:6Explanation:The triplets that follow the required conditions are (1, 1, 1), (1, 1, 2), (1, 1, 3), (1, 1, 4), (1, 1, 5) and (1, 2, 2). Hence the count of value triplets is 6.

Input:N = 20Output:40

**Approach: **The given problem can be solved based on the following observations:

- Since
**a <= b <=c**, it can be observed that the value of**a**must lie in the range**[1, N**.^{1/3}] - Similarly, for a given
**a**the value of**b**must lie in the range**[a, (N/a)**.^{1/2}] - Similarly, when the value of
**a**and**b**is fixed, the value of**c**must lie in the range**[b, N/(a*b)]**. Hence the number of possible values of**c**is the count of integers in the range**[b, N/(a*b)]**. Therefore, for each valid**a**and**b**, the number of possible values of**c**are**N/(a*b) – b + 1**.

Therefore, using the above observations below is the implementation of the above approach:

## C++

`// C++ implementation of the above approach` `#include <iostream>` `using` `namespace` `std;` `// Function to find the count of valid` `// triplets (a, b, c) such that the value` `// of a * b * c <= N and a <= b <= c` `long` `long` `validTriplets(` `int` `N)` `{` ` ` `// Stores count of triplets` ` ` `long` `long` `ans = 0;` ` ` `// Loop to iterate in the` ` ` `// range [1, N^(1/3)]` ` ` `for` `(` `long` `long` `a = 1; a * a * a <= N; a++) {` ` ` `// Loop to iterate in the` ` ` `// range [a, (N/a)^(1/2)]` ` ` `for` `(` `long` `long` `b = a; a * b * b <= N; b++) {` ` ` `// Add the count of valid` ` ` `// values of c for a fixed` ` ` `// value of a and b` ` ` `ans += N / a / b - b + 1;` ` ` `}` ` ` `}` ` ` `// Return Answer` ` ` `return` `ans;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `N = 5;` ` ` `cout << validTriplets(N);` ` ` `return` `0;` `}` |

## Java

`// Java implementation of the above approach` `import` `java.util.*;` `class` `GFG{` `// Function to find the count of valid` `// triplets (a, b, c) such that the value` `// of a * b * c <= N and a <= b <= c` `static` `long` `validTriplets(` `int` `N)` `{` ` ` ` ` `// Stores count of triplets` ` ` `long` `ans = ` `0` `;` ` ` `// Loop to iterate in the` ` ` `// range [1, N^(1/3)]` ` ` `for` `(` `long` `a = ` `1` `; a * a * a <= N; a++) {` ` ` `// Loop to iterate in the` ` ` `// range [a, (N/a)^(1/2)]` ` ` `for` `(` `long` `b = a; a * b * b <= N; b++) {` ` ` `// Add the count of valid` ` ` `// values of c for a fixed` ` ` `// value of a and b` ` ` `ans += N / a / b - b + ` `1` `;` ` ` `}` ` ` `}` ` ` `// Return Answer` ` ` `return` `ans;` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `N = ` `5` `;` ` ` `System.out.print(validTriplets(N));` `}` `}` `// This code is contributed by 29AjayKumar` |

## Python3

`# Python implementation of the above approach` `# Function to find the count of valid` `# triplets (a, b, c) such that the value` `# of a * b * c <= N and a <= b <= c` `def` `validTriplets(N):` ` ` ` ` `# Stores count of triplets` ` ` `ans ` `=` `0` `;` ` ` `# Loop to iterate in the` ` ` `# range [1, N^(1/3)]` ` ` `for` `a ` `in` `range` `(` `1` `, ` `int` `(N ` `*` `*` `(` `1` `/` `3` `)) ` `+` `1` `):` ` ` ` ` `# Loop to iterate in the` ` ` `# range [a, (N/a)^(1/2)]` ` ` `b ` `=` `a;` ` ` `while` `(a ` `*` `b ` `*` `b <` `=` `N):` ` ` `# Add the count of valid` ` ` `# values of c for a fixed` ` ` `# value of a and b` ` ` `ans ` `+` `=` `N ` `/` `a ` `/` `b ` `-` `b ` `+` `1` `;` ` ` `b ` `+` `=` `1` ` ` ` ` `# Return Answer` ` ` `return` `int` `(ans);` `# Driver Code` `N ` `=` `5` `;` `print` `(validTriplets(N));` `# This code is contributed by gfgking` |

## C#

`// C# implementation of the above approach` `using` `System;` `class` `GFG {` ` ` `// Function to find the count of valid` ` ` `// triplets (a, b, c) such that the value` ` ` `// of a * b * c <= N and a <= b <= c` ` ` `static` `long` `validTriplets(` `int` `N)` ` ` `{` ` ` `// Stores count of triplets` ` ` `long` `ans = 0;` ` ` `// Loop to iterate in the` ` ` `// range [1, N^(1/3)]` ` ` `for` `(` `long` `a = 1; a * a * a <= N; a++) {` ` ` `// Loop to iterate in the` ` ` `// range [a, (N/a)^(1/2)]` ` ` `for` `(` `long` `b = a; a * b * b <= N; b++) {` ` ` `// Add the count of valid` ` ` `// values of c for a fixed` ` ` `// value of a and b` ` ` `ans += N / a / b - b + 1;` ` ` `}` ` ` `}` ` ` `// Return Answer` ` ` `return` `ans;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `N = 5;` ` ` `Console.WriteLine(validTriplets(N));` ` ` `}` `}` `// This code is contributed by ukasp.` |

## Javascript

`<script>` `// JavaScript implementation of the above approach` `// Function to find the count of valid` `// triplets (a, b, c) such that the value` `// of a * b * c <= N and a <= b <= c` `function` `validTriplets(N)` `{` ` ` ` ` `// Stores count of triplets` ` ` `let ans = 0;` ` ` `// Loop to iterate in the` ` ` `// range [1, N^(1/3)]` ` ` `for` `(let a = 1; a * a * a <= N; a++)` ` ` `{` ` ` ` ` `// Loop to iterate in the` ` ` `// range [a, (N/a)^(1/2)]` ` ` `for` `(let b = a; a * b * b <= N; b++)` ` ` `{` ` ` ` ` `// Add the count of valid` ` ` `// values of c for a fixed` ` ` `// value of a and b` ` ` `ans += N / a / b - b + 1;` ` ` `}` ` ` `}` ` ` ` ` `// Return Answer` ` ` `return` `Math.floor(ans);` `}` `// Driver Code` `let N = 5;` `document.write(validTriplets(N));` `// This code is contributed by Potta Lokesh` `</script>` |

**Output**

6

**Time Complexity:** O(N^{2/3})**Auxiliary Space:** O(1)