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Check if given number is perfect square

  • Difficulty Level : Basic
  • Last Updated : 14 Jun, 2021

Given a number, check if it is a perfect square or not. 

Examples : 

Input : 2500
Output : Yes
Explanation:
2500 is a perfect square.
50 * 50 = 2500

Input  : 2555
Output : No

Approach:

  1. Take the square root of the number.
  2. Multiply the square root twice
  3. Use boolean equal operator to verify if the product of square root is equal to the number given.

C++




// CPP program to find if x is a
// perfect square.
#include <bits/stdc++.h>
using namespace std;
 
bool isPerfectSquare(long double x)
{
    // Find floating point value of
    // square root of x.
    if (x >= 0) {
 
        long long sr = sqrt(x);
         
        // if product of square root
        //is equal, then
        // return T/F
        return (sr * sr == x);
    }
    // else return false if n<0
    return false;
}
 
int main()
{
    long long x = 2502;
    if (isPerfectSquare(x))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

Java




// Java program to find if x is a
// perfect square.
class GFG {
 
    static boolean isPerfectSquare(int x)
    {
        if (x >= 0) {
           
            // Find floating point value of
            // square root of x.
            int sr = (int)Math.sqrt(x);
           
            // if product of square root
            // is equal, then
            // return T/F
 
            return ((sr * sr) == x);
        }
        return false;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int x = 2502;
 
        if (isPerfectSquare(x))
            System.out.print("Yes");
        else
            System.out.print("No");
    }
}
 
// This code is contributed by Anant Agarwal.

Python3




# Python program to find if x is a
# perfect square.
 
import math
 
 
def isPerfectSquare(x):
 
    #if x >= 0,
    if(x >= 0):
        sr = math.sqrt(x)
        # sqrt function returns floating value so we have to convert it into integer
        #return boolean T/F
        return ((sr*sr) == float(x))
    return false
 
# Driver code
 
 
x = 2502
if (isPerfectSquare(x)):
    print("Yes")
else:
    print("No")
 
# This code is contributed
# by Anant Agarwal.

C#




// C# program to find if x is a
// perfect square.
using System;
class GFG {
 
    static bool isPerfectSquare(double x)
    {
 
        // Find floating point value of
        // square root of x.
        if (x >= 0) {
 
            double sr = Math.Sqrt(x);
           
            // if product of square root
            // is equal, then
            // return T/F
            return (sr * sr == x);
        }
        // else return false if n<0
        return false;
    }
 
    // Driver code
    public static void Main()
    {
        double x = 2502;
 
        if (isPerfectSquare(x))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// PHP program to find if x is
// a perfect square.
 
function isPerfectSquare($x)
{
    // Find floating point value
    // of square root of x.
    $sr = sqrt($x);
     
    // If square root is an integer
    return (($sr - floor($sr)) == 0);
}
 
// Driver code
$x = 2502;
if (isPerfectSquare($x))
    echo("Yes");
else
    echo("No");
 
// This code is contributed by Ajit.
?>

Javascript




<script>
 
// JavaScript program to find if x is a
// perfect square.
 
function isPerfectSquare(x)
    {
        if (x >= 0) {
            
            // Find floating point value of
            // square root of x.
            let sr = Math.sqrt(x);
            
            // if product of square root
            // is equal, then
            // return T/F
  
            return ((sr * sr) == x);
        }
        return false;
    }
  
// Driver code
 
        let x = 2500;
  
        if (isPerfectSquare(x))
            document.write("Yes");
        else
            document.write("No");
 
// This code is contributed by souravghosh0416.
</script>
Output
No

To know more about the inbuilt sqrt function, refer this Stackoverflow and this StackExchange threads.



Another Approach :

  1. Use the floor and ceil function .
  2. If they are equal that implies the number is a perfect square.

C++




// C++ program for the above approach
#include <iostream>
#include <math.h>
using namespace std;
 
void checkperfectsquare(int n)
{
     
    // If ceil and floor are equal
    // the number is a perfect
    // square
    if (ceil((double)sqrt(n)) == floor((double)sqrt(n))) {
        cout << "perfect square";
    }
    else {
        cout << "not a perfect square";
    }
}
 
// Driver Code
int main()
{
 
    int n = 49;
    checkperfectsquare(n);
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
 
class GFG{
 
static void checkperfectsquare(int n)
{
     
    // If ceil and floor are equal
    // the number is a perfect
    // square
    if (Math.ceil((double)Math.sqrt(n)) ==
        Math.floor((double)Math.sqrt(n)))
    {
        System.out.print("perfect square");
    }
    else
    {
        System.out.print("not a perfect square");
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 49;
     
    checkperfectsquare(n);
}
}
 
// This code is contributed by subhammahato348

Python3




# Python3 program for the above approach
import math
 
def checkperfectsquare(x):
     
    # If ceil and floor are equal
    # the number is a perfect
    # square
    if (math.ceil(math.sqrt(n)) ==
       math.floor(math.sqrt(n))):
        print("perfect square")
    else:
        print("not a perfect square")
     
# Driver code
n = 49
  
checkperfectsquare(n)
 
# This code is contributed by jana_sayantan

C#




// C# program for the above approach
using System;
 
class GFG{
 
static void checkperfectsquare(int n)
{
     
    // If ceil and floor are equal
    // the number is a perfect
    // square
    if (Math.Ceiling((double)Math.Sqrt(n)) ==
        Math.Floor((double)Math.Sqrt(n)))
    {
        Console.Write("perfect square");
    }
    else
    {
        Console.Write("not a perfect square");
    }
}
 
// Driver Code
public static void Main()
{
    int n = 49;
 
    checkperfectsquare(n);
}
}
 
// This code is contributed by subhammahato348

Javascript




<script>
 
// Javascript program for the above approach
function checkperfectsquare(n)
{
     
    // If ceil and floor are equal
    // the number is a perfect
    // square
    if (Math.ceil(Math.sqrt(n)) ==
        Math.floor(Math.sqrt(n)))
    {
        document.write("perfect square");
    }
    else
    {
        document.write("not a perfect square");
    }
}
 
// Driver Code
let n = 49;
 
checkperfectsquare(n);
 
// This code is contributed by rishavmahato348
 
</script>
Output
perfect square

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