Check if given number is perfect square

Last Updated : 17 Sep, 2024

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

Examples :

Input : n = 36
Output : Yes

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

Input : n = 8
Output : No

Table of Content

Using sqrt()
Using ceil, floor and sqrt() functions
Using Binary search
Using Mathematical Properties

Using sqrt()

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.

Code Implementation:

C++

// CPP program to find if x is a
// perfect square.
#include <bits/stdc++.h>
using namespace std;

bool isPerfectSquare(long long 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 = 49;
    if (isPerfectSquare(x))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

// C program to find if x is a
// perfect square.
#include <stdio.h>
#include <math.h>

int isPerfectSquare(long long 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 0;
}

int main()
{
    long long x = 49;
    if (isPerfectSquare(x))
        printf("Yes");
    else
        printf("No");
    return 0;
}

Java

// Java program to find if x is a perfect square.
public class GfG {

    public static boolean isPerfectSquare(long x)
    {
        // Find floating point value of
        // square root of x.
        if (x >= 0) {
            long sr = (long)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;
    }

    public static void main(String[] args)
    {
        long x = 49;
        if (isPerfectSquare(x))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

Python

# Python program to find if x is a
# perfect square.
def is_perfect_square(x):
    # Find floating point value of
    # square root of x.
    if x >= 0:
        sr = int(x ** 0.5)

        # if product of square root
        # is equal, then
        # return T/F
        return (sr * sr == x)
    # else return false if n<0
    return False

x = 49
if is_perfect_square(x):
    print("Yes")
else:
    print("No")

// C# program to find if x is a
// perfect square.
using System;

class GfG {
    static bool IsPerfectSquare(long x) {
        // Find floating point value of
        // square root of x.
        if (x >= 0) {
            long sr = (long) 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;
    }

    static void Main() {
        long x = 49;
        if (IsPerfectSquare(x))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

JavaScript

// JavaScript program to find if x is a
// perfect square.
function isPerfectSquare(x) {
    // Find floating point value of
    // square root of x.
    if (x >= 0) {
        let sr = Math.floor(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;
}

const x = 49;
if (isPerfectSquare(x))
    console.log("Yes");
else
    console.log("No");

Output

Yes

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

Using ceil, floor and sqrt() functions

Use the floor and ceil and sqrt() function.
If they are equal that implies the number is a perfect square.

Code Implementation:

C++

#include <bits/stdc++.h>
using namespace std;

bool isPerfectSquare(long long n)
{
    // If ceil and floor are equal
    // the number is a perfect
    // square
    if (ceil((double)sqrt(n)) == floor((double)sqrt(n))){
        return true;
    }
    else{
        return false;
    }
}

int main()
{
    long long x = 49;
    if (isPerfectSquare(x))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

#include <stdio.h>
#include <math.h>

// If ceil and floor are equal
// the number is a perfect
// square
int isPerfectSquare(long long n) {
    if (ceil((double)sqrt(n)) == floor((double)sqrt(n))) {
        return 1;  // true
    }
    else {
        return 0;  // false
    }
}

int main() {
    long long x = 49;
    if (isPerfectSquare(x))
        printf("Yes");
    else
        printf("No");
    return 0;
}

Java

public class GfG {
    public static boolean isPerfectSquare(long n) {
        // If ceil and floor are equal
        // the number is a perfect
        // square
        if (Math.ceil(Math.sqrt(n)) == Math.floor(Math.sqrt(n))) {
            return true;
        } else {
            return false;
        }
    }

    public static void main(String[] args) {
        long x = 49;
        if (isPerfectSquare(x))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

Python

import math

def is_perfect_square(n):
    # If ceil and floor are equal
    # the number is a perfect
    # square
    if math.ceil(math.sqrt(n)) == math.floor(math.sqrt(n)):
        return True
    else:
        return False

x = 49
if is_perfect_square(x):
    print("Yes")
else:
    print("No")

using System;

class GfG {
    public static bool IsPerfectSquare(long n) {
        // If ceil and floor are equal
        // the number is a perfect
        // square
        if (Math.Ceiling(Math.Sqrt(n)) == Math.Floor(Math.Sqrt(n))) {
            return true;
        } else {
            return false;
        }
    }

    static void Main() {
        long x = 49;
        if (IsPerfectSquare(x))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

JavaScript

function isPerfectSquare(n) {
    // If ceil and floor are equal
    // the number is a perfect
    // square
    if (Math.ceil(Math.sqrt(n)) === Math.floor(Math.sqrt(n))) {
        return true;
    } else {
        return false;
    }
}

const x = 49;
if (isPerfectSquare(x)) {
    console.log("Yes");
} else {
    console.log("No");
}

Output

Yes

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

Using Binary search:

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>
using namespace std;

// Function to check if a number is a perfect square using
// binary search
bool isPerfectSquare(long long 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()
{
    long long x = 49;
    if (isPerfectSquare(x))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

#include <stdio.h>

// Function to check if a number is a perfect square using
// binary search
int isPerfectSquare(long long n) {
    // Base case: 0 and 1 are perfect squares
    if (n <= 1) {
        return 1;
    }

    // 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 1;
        }
        // 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 0;
}

int main() {
    long long x = 49;
    if (isPerfectSquare(x))
        printf("Yes");
    else
        printf("No");
    return 0;
}

Java

public class GfG {
    // Function to check if a number is a perfect square using
    // binary search
    public static boolean isPerfectSquare(long 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) {
        long x = 49;
        if (isPerfectSquare(x))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

Python

def is_perfect_square(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

x = 49
if is_perfect_square(x):
    print("Yes")
else:
    print("No")

using System;

class GfG {
    // Function to check if a number is a perfect square using
    // binary search
    static bool IsPerfectSquare(long 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() {
        long x = 49;
        if (IsPerfectSquare(x))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

JavaScript

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 = left + Math.floor((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 x = 49;
if (isPerfectSquare(x)) {
    console.log("Yes");
} else {
    console.log("No");
}

Output

Yes

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

Using Mathematical Properties:

The idea is based on the fact that perfect squares are always some of first few odd numbers.

1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16
1 + 3 + 5 + 7 + 9 = 25
1 + 3 + 5 + 7 + 9 + 11 = 36
………………………………………………

Code Implementation:

C++

#include <bits/stdc++.h>
using namespace std;

bool isPerfectSquare(long long n) {
  
    // 0 is considered as a perfect
    // square
    if (n == 0) return true;
	
    long long odd = 1;
	while (n > 0) {
		n -= odd;
		odd += 2;
	}
	return n == 0;
}

int main()
{
    long long x = 49;
    if (isPerfectSquare(x))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

#include <stdio.h>

// 0 is considered as a perfect
// square
int isPerfectSquare(long long n) {
    if (n == 0) return 1; // true

    long long odd = 1;
    while (n > 0) {
        n -= odd;
        odd += 2;
    }
    return n == 0;
}

int main() {
    long long x = 49;
    if (isPerfectSquare(x))
        printf("Yes");
    else
        printf("No");
    return 0;
}

Java

public class GfG {
    public static boolean isPerfectSquare(long n) {
        // 0 is considered as a perfect
        // square
        if (n == 0) return true;

        long odd = 1;
        while (n > 0) {
            n -= odd;
            odd += 2;
        }
        return n == 0;
    }

    public static void main(String[] args) {
        long x = 49;
        if (isPerfectSquare(x))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

Python

def is_perfect_square(n):
    # 0 is considered as a perfect
    # square
    if n == 0:
        return True

    odd = 1
    while n > 0:
        n -= odd
        odd += 2
    return n == 0

x = 49
if is_perfect_square(x):
    print("Yes")
else:
    print("No")

using System;

class GfG {
    public static bool IsPerfectSquare(long n) {
        // 0 is considered as a perfect
        // square
        if (n == 0) return true;

        long odd = 1;
        while (n > 0) {
            n -= odd;
            odd += 2;
        }
        return n == 0;
    }

    static void Main() {
        long x = 49;
        if (IsPerfectSquare(x))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

JavaScript

function isPerfectSquare(n) {
    // 0 is considered as a perfect
    // square
    if (n === 0) return true;

    let odd = 1;
    while (n > 0) {
        n -= odd;
        odd += 2;
    }
    return n === 0;
}

const x = 49;
if (isPerfectSquare(x)) {
    console.log("Yes");
} else {
    console.log("No");
}

Output

Yes

How does this work?

1 + 3 + 5 + … (2n-1) = Sum(2*i – 1) where 1<=i<=n
= 2*Sum(i) – Sum(1) where 1<=i<=n
= 2n(n+1)/2 – n
= n(n+1) – n
= n^2′

Time Complexity Analysis

Let us assume that the above loop runs i times. The following series would have i terms.

1 + 3 + 5 …….. = n [Let there be i terms]
1 + (1 + 2) + (1 + 2 + 2) + …….. = n [Let there be i terms]
(1 + 1 + ….. i-times) + (2 + 4 + 6 …. (i-1)-times) = n
i + (2^(i-1) – 1) = n
2^(i-1) = n + 1 – i
Using this expression, we can say that i is upper bounded by Log n

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Aditya Darekar

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Article Tags :

Practice Tags :

Mathematical

Check if given number is perfect square

Using sqrt()

Using ceil, floor and sqrt() functions

Using Binary search:

Using Mathematical Properties:

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