Check if given number is perfect square

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 floor()ed 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 = 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.

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

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

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

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

Another Approach : without using sqrtx() function

we will use binary search to do this

Below is the implementation of the above approach:

C++

#include <iostream>
using namespace std;
bool isPerfectSquare(int num)
{
    int64_t n = num;
    int64_t lo = 1;
    int64_t hi = num;
    while (hi - lo > 1) {
        int64_t mid = (hi + lo) / 2;
        if (num == 1) {
            return true;
        }
        else if (mid * mid == num) {
            return true;
        }
        else if (mid * mid > num) {
            hi = mid - 1;
        }
        else if (mid * mid < num) {
            lo = mid + 1;
        }
    }
    if (lo * lo == num || hi * hi == num) {
        return true;
    }
    return false;
}
int main()
{
    int n = 676;
    cout << isPerfectSquare(n);
    return 0;
}

Java

public class PerfectSquare {
  static boolean isPerfectSquare(int num)
  {
    long n = num;
    long lo = 1;
    long hi = num;
    while (hi - lo > 1) {
      long mid = (hi + lo) / 2;
      if (num == 1) {
        return true;
      }
      else if (mid * mid == num) {
        return true;
      }
      else if (mid * mid > num) {
        hi = mid - 1;
      }
      else if (mid * mid < num) {
        lo = mid + 1;
      }
    }
    if (lo * lo == num || hi * hi == num) {
      return true;
    }
    return false;
  }

  public static void main(String[] args)
  {
    int n = 676;
    System.out.println(isPerfectSquare(n));
  }
}

// This code is contributed by factworx412

Python3

def isPerfectSquare( num):

    n = num;
    lo = 1;
    hi = num;
    while (hi - lo > 1) :
        mid = (int)(hi + lo) / 2;
        if (num == 1) :
            return 1;

        elif (mid * mid == num) :
            return 1;

        elif (mid * mid > num) :
            hi = mid - 1;

        elif (mid * mid < num) :
            lo = mid + 1;


    if (lo * lo == num or hi * hi == num) :
        return 1;

    return 0;

n = 676;
print(isPerfectSquare(n));

# This code is contributed by ratiagrawal.

C#

using System;

class Program {
  static bool IsPerfectSquare(int num)
  {
    long n = num;
    long lo = 1;
    long hi = num;

    while (hi - lo > 1) {
      long mid = (hi + lo) / 2;
      if (num == 1) {
        return true;
      }
      else if (mid * mid == num) {
        return true;
      }
      else if (mid * mid > num) {
        hi = mid - 1;
      }
      else if (mid * mid < num) {
        lo = mid + 1;
      }
    }

    if (lo * lo == num || hi * hi == num) {
      return true;
    }
    return false;
  }
  static void Main(string[] args)
  {
    int n = 676;
    Console.WriteLine(IsPerfectSquare(n));
  }
}

// This code is contributed by lokeshpotta20.

Javascript

function isPerfectSquare(num)
{
    let n = num;
    let lo = 1;
    let hi = num;
    while (hi - lo > 1) {
        let mid = Math.floor(hi + lo) / 2;
        if (num == 1) {
            return 1;
        }
        else if (mid * mid == num) {
            return 1;
        }
        else if (mid * mid > num) {
            hi = mid - 1;
        }
        else if (mid * mid < num) {
            lo = mid + 1;
        }
    }
    if (lo * lo == num || hi * hi == num) {
        return 1;
    }
    return 0;
}

let n = 676;
document.write(isPerfectSquare(n));

Output

1

Time Complexity: O(log n)

The time complexity of this algorithm is O(log n) since the algorithm will run for log n iterations at most.

Space Complexity: O(1)

The algorithm uses constant space, as it only stores a few variables which don’t increase with the size of the input.

METHOD 4: Using ** operator

APPROACH:

 This program checks whether a given number is a perfect square or not using a brute-force approach.

ALGORITHM:

1.Take the input number, n.
2.Loop through all numbers from 0 to n.
3.Check whether the square of the current number is equal to n.
4.If it is, print “Yes” and break out of the loop.
5.If the loop completes without finding a perfect square, print “No”.

Python3

n = 2500

for i in range(n+1):
    if i**2 == n:
        print("Yes")
        break
else:
    print("No")

Output

Yes

Time complexity: O(n)
Space complexity: O(1)