Check if given number is perfect square

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)