# Count number less than N which are product of perfect squares

Given an Integer N. The task is count numbers P less than N such that P is a product of two distinct perfect squares.**Examples**:

Input: N = 36Output: 5 Numbers are 4 = 1^{2}* 2^{2}, 9 = 1^{2}* 3^{2}, 16 = 1^{2}* 4^{2}, 25 = 1^{2}* 5^{2}, 36 = 1^{2}* 6^{2}Input: N = 1000000Output: 999

**Approach:** Let us consider a number P = (a^{2} * b^{2}) such that P <= N. So we have (a^{2} * b^{2}) <= N. This can be written as (a * b) <= sqrt(N).

So we have to count pairs (a, b) such that (a * b) <= sqrt(N) and a <= b.

Let us take a number **Q = (a * b)** such that Q <= sqrt(N).

Taking a = 1, we have b = sqrt(N) – 1 numbers such that, ( a * b = Q <= sqrt(N)).

Thus we can have all sqrt(N) – 1 numbers such that (a^{2} * b^{2}) <= N.

Below is the implementation of the above approach:

## C++

`// C++ program to count number less` `// than N which are product of` `// any two perfect squares` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to count number less` `// than N which are product of` `// any two perfect squares` `int` `countNumbers(` `int` `N)` `{` ` ` `return` `int` `(` `sqrt` `(N)) - 1;` `}` `// Driver program` `int` `main()` `{` ` ` `int` `N = 36;` ` ` `cout << countNumbers(N);` ` ` `return` `0;` `}` |

## Java

`// Java program to count number less` `// than N which are product of` `// any two perfect squares` `import` `java.util.*;` `class` `solution` `{` `// Function to count number less` `// than N which are product of` `// any two perfect squares` `static` `int` `countNumbers(` `int` `N)` `{` ` ` `return` `(` `int` `)Math.sqrt(N) - ` `1` `;` `}` `// Driver program` `public` `static` `void` `main(String args[])` `{` ` ` `int` `N = ` `36` `;` ` ` `System.out.println(countNumbers(N));` ` ` `}` `}` `//This code is contributed by` `// Surendra_Gangwar` |

## Python 3

`# Python 3 program to count number` `# less than N which are product of` `# any two perfect squares` `import` `math` `# Function to count number less` `# than N which are product of` `# any two perfect squares` `def` `countNumbers(N):` ` ` `return` `int` `(math.sqrt(N)) ` `-` `1` `# Driver Code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `N ` `=` `36` ` ` `print` `(countNumbers(N))` `# This code is contributed` `# by ChitraNayal` |

## C#

`// C# program to count number less` `// than N which are product of` `// any two perfect squares` `using` `System;` `class` `GFG` `{` `// Function to count number less` `// than N which are product of` `// any two perfect squares` `static` `int` `countNumbers(` `int` `N)` `{` ` ` `return` `(` `int` `)(Math.Sqrt(N)) - 1;` `}` `// Driver Code` `public` `static` `void` `Main()` `{` ` ` `int` `N = 36;` ` ` `Console.Write(countNumbers(N));` `}` `}` `// This code is contributed` `// by Akanksha Rai` |

## PHP

`<?php` `// PHP program to count number less` `// than N which are product of` `// any two perfect squares` `// Function to count number less` `// than N which are product of` `// any two perfect squares` `function` `countNumbers(` `$N` `)` `{` ` ` `return` `(int)(sqrt(` `$N` `)) - 1;` `}` `// Driver Code` `$N` `= 36;` `echo` `countNumbers(` `$N` `);` `// This code is contributed by akt_mit` `?>` |

## Javascript

`<script>` ` ` `// Javascript program to count number less` ` ` `// than N which are product of` ` ` `// any two perfect squares` ` ` ` ` `// Function to count number less` ` ` `// than N which are product of` ` ` `// any two perfect squares` ` ` `function` `countNumbers(N)` ` ` `{` ` ` `return` `parseInt(Math.sqrt(N), 10) - 1;` ` ` `}` ` ` ` ` `let N = 36;` ` ` ` ` `document.write(countNumbers(N));` ` ` `</script>` |

**Output:**

5

**Time Complexity:** O(log(N))

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