Data Modelling & AI Data Structure & Algorithm

Check if given number is perfect square

20 June 2025

2

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

Examples :

Input : n = 2500
Output : Yes
Explanation: 2500 is a perfect square of 50

Input : n = 2555
Output : No

Approach:

Take the floor()ed square root of the number.
Multiply the square root twice.
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 = int(math.sqrt(x))
        # sqrt function returns floating value so we have to convert it into integer
        #return boolean T/F
        return ((sr*sr) == 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.

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>

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.
?>

Output

No

Time Complexity: O(log(x))
Auxiliary Space: O(1)

Check if given number is perfect square using ceil, floor and sqrt() function.

Use the floor and ceil and sqrt() function.
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

Time Complexity : O(sqrt(n))
Auxiliary space: O(1)

Check if given number is perfect square using Binary search:

Below is the implementation of the above approach:

C++14

#include <bits/stdc++.h>
using namespace std;
 
// Function to check if a number is a perfect square using
// binary search
bool isPerfectSquare(int n)
{
    // Base case: 0 and 1 are perfect squares
    if (n <= 1) {
        return true;
    }
 
    // Initialize boundaries for binary search
    long long left = 1, right = n;
 
    while (left <= right) {
 
        // Calculate middle value
        long long mid = left + (right - left) / 2;
 
        // Calculate square of the middle value
        long long square = mid * mid;
 
        // If the square matches n, n is a perfect square
        if (square == n) {
            return true;
        }
 
        // If the square is smaller than n, search the right
        // half
        else if (square < n) {
 
            left = mid + 1;
        }
 
        // If the square is larger than n, search the left
        // half
        else {
 
            right = mid - 1;
        }
    }
 
    // If the loop completes without finding a perfect
    // square, n is not a perfect square
    return false;
}
 
int main()
{
    int n = 2500;
 
    if (isPerfectSquare(n)) {
        cout << n << " is a perfect square." << endl;
    }
    else {
        cout << n << " is not a perfect square."
             << std::endl;
    }
 
    return 0;
}

Java

public class PerfectSquareCheck {
    // Function to check if a number is a perfect square
    // using binary search
    static boolean isPerfectSquare(int n)
    {
        // Base case: 0 and 1 are perfect squares
        if (n <= 1) {
            return true;
        }
 
        // Initialize boundaries for binary search
        long left = 1, right = n;
 
        while (left <= right) {
            // Calculate middle value
            long mid = left + (right - left) / 2;
 
            // Calculate square of the middle value
            long square = mid * mid;
 
            // If the square matches n, n is a perfect
            // square
            if (square == n) {
                return true;
            }
            // If the square is smaller than n, search the
            // right half
            else if (square < n) {
                left = mid + 1;
            }
            // If the square is larger than n, search the
            // left half
            else {
                right = mid - 1;
            }
        }
 
        // If the loop completes without finding a perfect
        // square, n is not a perfect square
        return false;
    }
 
    public static void main(String[] args)
    {
        int n = 2500;
 
        if (isPerfectSquare(n)) {
            System.out.println(n + " is a perfect square.");
        }
        else {
            System.out.println(
                n + " is not a perfect square.");
        }
    }
}

Python

# Function to check if a number is a perfect square using binary search
def isPerfectSquare(n):
    # Base case: 0 and 1 are perfect squares
    if n <= 1:
        return True
 
    # Initialize boundaries for binary search
    left, right = 1, n
 
    while left <= right:
        # Calculate middle value
        mid = left + (right - left) // 2
 
        # Calculate square of the middle value
        square = mid * mid
 
        # If the square matches n, n is a perfect square
        if square == n:
            return True
        # If the square is smaller than n, search the right half
        elif square < n:
            left = mid + 1
        # If the square is larger than n, search the left half
        else:
            right = mid - 1
 
    # If the loop completes without finding a perfect square, n is not a perfect square
    return False
 
 
n = 2500
if isPerfectSquare(n):
    print(n, "is a perfect square.")
else:
    print(n, "is not a perfect square.")

C#

using System;
 
class Program
{
    // Function to check if a number is a perfect square using binary search
    static bool IsPerfectSquare(int n)
    {
        // Base case: 0 and 1 are perfect squares
        if (n <= 1)
        {
            return true;
        }
 
        // Initialize boundaries for binary search
        long left = 1, right = n;
 
        while (left <= right)
        {
            // Calculate middle value
            long mid = left + (right - left) / 2;
 
            // Calculate square of the middle value
            long square = mid * mid;
 
            // If the square matches n, n is a perfect square
            if (square == n)
            {
                return true;
            }
            // If the square is smaller than n, search the right half
            else if (square < n)
            {
                left = mid + 1;
            }
            // If the square is larger than n, search the left half
            else
            {
                right = mid - 1;
            }
        }
 
        // If the loop completes without finding a perfect square, n is not a perfect square
        return false;
    }
 
    static void Main(string[] args)
    {
        int n = 2500;
 
        if (IsPerfectSquare(n))
        {
            Console.WriteLine(n + " is a perfect square.");
        }
        else
        {
            Console.WriteLine(n + " is not a perfect square.");
        }
    }
}

Javascript

// Function to check if a number is a perfect square using binary search
function isPerfectSquare(n) {
    // Base case: 0 and 1 are perfect squares
    if (n <= 1) {
        return true;
    }
 
    // Initialize boundaries for binary search
    let left = 1, right = n;
 
    while (left <= right) {
        // Calculate middle value
        let mid = Math.floor(left + (right - left) / 2);
 
        // Calculate square of the middle value
        let square = mid * mid;
 
        // If the square matches n, n is a perfect square
        if (square === n) {
            return true;
        }
        // If the square is smaller than n, search the right half
        else if (square < n) {
            left = mid + 1;
        }
        // If the square is larger than n, search the left half
        else {
            right = mid - 1;
        }
    }
 
    // If the loop completes without finding a perfect square, n is not a perfect square
    return false;
}
 
const n = 2500;
 
if (isPerfectSquare(n)) {
    console.log(n + " is a perfect square.");
} else {
    console.log(n + " is not a perfect square.");
}

Output

2500 is a perfect square.

Time Complexity: O(log n)
Auxiliary Space: O(1)

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Check if given number is perfect square

C++

Java

Python3

C#

Javascript

PHP

Check if given number is perfect square using ceil, floor and sqrt() function.

C++

Java

Python3

C#

Javascript

Check if given number is perfect square using Binary search:

C++14

Java

Python

C#

Javascript

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