Count square and non-square numbers before n
Last Updated :
22 Mar, 2023
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++
#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;
}
int main()
{
int n = 10;
countSquaresNonSquares(n);
return 0;
}
|
Java
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) );
}
public static void main(String args[])
{
int n = 10 ;
countSquaresNonSquares(n);
}
}
|
Python3
import math
def countSquaresNonSquares(n) :
sc = (math.floor(math.sqrt(n)))
print ( "Count of squares " , sc)
print ( "Count of non-squares " , (n - sc) )
n = 10
countSquaresNonSquares(n)
|
C#
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));
}
static public void Main ()
{
int n = 10;
countSquaresNonSquares(n);
}
}
|
PHP
<?php
function countSquaresNonSquares( $n )
{
$sc = floor (sqrt( $n ));
echo ( "Count of squares " .
$sc . "\n" );
echo ( "Count of non-squares " .
( $n - $sc ));
}
$n = 10;
countSquaresNonSquares( $n );
?>
|
Javascript
<script>
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>" );
}
let n = 10;
countSquaresNonSquares(n);
</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;
countSquaresNonSquares(n);
|
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
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