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

Recommended Practice

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 ` `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

 ``

PHP

 ``

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 ` `#include ` `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

 ``

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 ` `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)

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