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()

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

Check if given number is perfect square

Aditya Darekar

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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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