Related Articles

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

• Last Updated : 10 May, 2021

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 = 36
Output : 5
Numbers are 4 = 12 * 22,
9 = 12 * 32,
16 = 12 * 42,
25 = 12 * 52,
36 = 12 * 62

Input : N = 1000000
Output : 999```

Approach: Let us consider a number P = (a2 * b2) such that P <= N. So we have (a2 * b2) <= 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 (a2 * b2) <= 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 ``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

 ``

## Javascript

 ``
Output:
`5`

Time Complexity: O(log(N))

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up