Check if given number is perfect square (original) (raw)

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

Try It Yourself

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

// 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#

// 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");

**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; }

C

#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")

C#

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"); }

**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; }

C

#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")

C#

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"); }

**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; }

C

#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")

C#

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"); }

**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
= n2'

**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