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# 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```
Recommended Practice

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 ``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

 ``

## Javascript

 ``

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 ``#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)

Another Approach : without using sqrtx() function

we will use binary search to do this

Below is the implementation of the above approach:

## C++

 `#include ``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)

My Personal Notes arrow_drop_up