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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 <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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