Count square and non-square numbers before n

Given a number n, we need to count square numbers smaller than or equal to n.
Examples :

Input : n = 5
Output : Square Number : 2
         Non-square numbers : 3
Explanation : Square numbers are 1 and 4.
Non square numbers are 2, 3 and 5.

Input : n = 10
Output : Square Number : 3
         Non-square numbers : 7
Explanation : Square numbers are 1, 4 and 9.
Non square numbers are 2, 3, 5, 6, 7, 8 and 10.

A simple solution is to traverse through all numbers from 1 to n and for every number check if n is perfect square or not. 
An efficient solution is based on below formula.
Count of square numbers that are greater than 0 and smaller than or equal to n are floor(sqrt(n)) or ??(n)?
Count of non-square numbers = n – ??(n)?

C++

// CPP program to count squares and
// non-squares before a number.
#include <bits/stdc++.h>
using namespace std;

void countSquaresNonSquares(int n)
{
    int sc = floor(sqrt(n));
    cout << "Count of squares "
         << sc << endl;
    cout << "Count of non-squares "
         << n - sc << endl;
}

// Driver Code
int main()
{
    int n = 10;
    countSquaresNonSquares(n);
    return 0;
}

Java

// Java program to count squares and
// non-squares before a number.
import java.io.*;
import java.math.*;

class GFG
{
    static void countSquaresNonSquares(int n)
    {
        int sc = (int)(Math.floor(Math.sqrt(n)));
        System.out.println("Count of" +
                     " squares " + sc);
        System.out.println("Count of" +
                      " non-squares " +
                            (n - sc) );
    }

    // Driver code
    public static void main(String args[])
    {
        int n = 10;
        countSquaresNonSquares(n);
    }
}

// This code is contributed
// by Nikita Tiwari.

Python3

# Python 3 program to count
# squares and non-squares
# before a number.
import math

def countSquaresNonSquares(n) :
    sc = (math.floor(math.sqrt(n)))
    print("Count of squares ", sc)
    print("Count of non-squares ", (n - sc) )


# Driver code
n = 10
countSquaresNonSquares(n)

# This code is contributed
# by Nikita Tiwari.

C#

// C# program to count squares and
// non-squares before a number.
using System;

class GFG
{
static void countSquaresNonSquares(int n)
{
    int sc = (int)Math.Sqrt(n);
    Console.WriteLine( "Count of " +
                        "squares " +
                               sc) ;
    Console.WriteLine("Count of " +
                   "non-squares " +
                         (n - sc));
}

    // Driver Code
    static public void Main ()
    {
    int n = 10;
    countSquaresNonSquares(n);
    }
}

// This code is contributed by anuj_67.

PHP

<?php
// PHP program to count
// squares and non-squares
// before a number.

// function to count squares
// and non-squares before a
// number
function countSquaresNonSquares($n)
{
    $sc = floor(sqrt($n));
    echo("Count of squares " .
                  $sc . "\n");
    echo("Count of non-squares " .
                      ($n - $sc));
}

// Driver code
$n = 10;
countSquaresNonSquares($n);

// This code is contributed by Ajit.
?>

Javascript

<script>

// Javascript program to count squares and
// non-squares before a number.

function countSquaresNonSquares(n)
{
    let sc = Math.floor(Math.sqrt(n));
    document.write("Count of squares "
        + sc + "<br>");
    document.write("Count of non-squares "
        + (n - sc) + "<br>");
}

// Driver Code

    let n = 10;
    countSquaresNonSquares(n);

//This code is contributed by Mayank Tyagi

</script>

Output :

Count of squares 3
Count of non-squares 7

Time Complexity: O(logn) 

Auxiliary Space: O(1) as using only constant variables

Approach 2: Using Loops:

Another approach to count the number of squares and non-squares before a given number is to iterate over all the numbers from 1 to n and check if each number is a perfect square or not. If a number is a perfect square, we increment the count of squares, otherwise, we increment the count of non-squares.

Here’s the code implementing this approach:

C++

#include <iostream>
#include <cmath>

void countSquaresNonSquares(int n)
{
    int countSquares = 0;
    int countNonSquares = 0;

    for (int i = 1; i <= n; i++) {
        int sqrtI = std::sqrt(i);
        if (sqrtI * sqrtI == i) {
            countSquares++;
        } else {
            countNonSquares++;
        }
    }

    std::cout << "Count of squares " << countSquares << std::endl;
    std::cout << "Count of non-squares " << countNonSquares << std::endl;
}

int main()
{
    int n = 10;
    countSquaresNonSquares(n);
    return 0;
}

Java

import java.util.*;

public class Main {
    public static void countSquaresNonSquares(int n) {
        int countSquares = 0;
        int countNonSquares = 0;

        for (int i = 1; i <= n; i++) {
            int sqrtI = (int)Math.sqrt(i);
            if (sqrtI * sqrtI == i) {
                countSquares++;
            } else {
                countNonSquares++;
            }
        }

        System.out.println("Count of squares " + countSquares);
        System.out.println("Count of non-squares " + countNonSquares);
    }

    public static void main(String[] args) {
        int n = 10;
        countSquaresNonSquares(n);
    }
}

Python3

import math

def countSquaresNonSquares(n):
    countSquares = 0
    countNonSquares = 0

    for i in range(1, n+1):
        sqrtI = int(math.sqrt(i))
        if sqrtI * sqrtI == i:
            countSquares += 1
        else:
            countNonSquares += 1

    print("Count of squares", countSquares)
    print("Count of non-squares", countNonSquares)

n = 10
countSquaresNonSquares(n)

C#

using System;

class MainClass {
    public static void countSquaresNonSquares(int n) {
        int countSquares = 0;
        int countNonSquares = 0;

        for (int i = 1; i <= n; i++) {
            int sqrtI = (int)Math.Sqrt(i);
            if (sqrtI * sqrtI == i) {
                countSquares++;
            } else {
                countNonSquares++;
            }
        }

        Console.WriteLine("Count of squares: " + countSquares);
        Console.WriteLine("Count of non-squares: " + countNonSquares);
    }

    public static void Main() {
        int n = 10;
        countSquaresNonSquares(n);
    }
}

Javascript

function countSquaresNonSquares(n) {
  let countSquares = 0;
  let countNonSquares = 0;

  for (let i = 1; i <= n; i++) {
    let sqrtI = Math.sqrt(i);
    if (sqrtI * sqrtI === i) {
      countSquares++;
    } else {
        countNonSquares++;
    }
  }

  console.log("Count of squares " + countSquares);
  console.log("Count of non-squares " + countNonSquares);
}

let n = 10;

// function Call
countSquaresNonSquares(n);


// This code is contributed by shivhack999

Output : 

Count of squares 3
Count of non-squares 7

Time Complexity:  O(nsqrt(n)), where n is the input variable, as described in the problem statement

Auxiliary Space: O(1) as using only constant variables