Given a positive integer n, check if it is perfect square or not using only addition/subtraction operations and in minimum time complexity.**We strongly recommend you to minimize your browser and try this yourself first.**

We can use the property of odd number for this purpose:

Addition of first n odd numbers is always perfect square 1 + 3 = 4, 1 + 3 + 5 = 9, 1 + 3 + 5 + 7 + 9 + 11 = 36 ...

Below is the implementation of above idea :

## C++

`// C++ program to check if n is perfect square` `// or not` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// This function returns true if n is` `// perfect square, else false` `bool` `isPerfectSquare(` `int` `n)` `{` ` ` `// sum is sum of all odd numbers. i is` ` ` `// used one by one hold odd numbers` ` ` `for` `(` `int` `sum = 0, i = 1; sum < n; i += 2) {` ` ` `sum += i;` ` ` `if` `(sum == n)` ` ` `return` `true` `;` ` ` `}` ` ` `return` `false` `;` `}` `// Driver code` `int` `main()` `{` ` ` `isPerfectSquare(35) ? cout << ` `"Yes\n"` `: cout << ` `"No\n"` `;` ` ` `isPerfectSquare(49) ? cout << ` `"Yes\n"` `: cout << ` `"No\n"` `;` ` ` `return` `0;` `}` |

## Java

`// Java program to check if n` `// is perfect square or not` `public` `class` `GFG {` ` ` `// This function returns true if n` ` ` `// is perfect square, else false` ` ` `static` `boolean` `isPerfectSquare(` `int` `n)` ` ` `{` ` ` `// sum is sum of all odd numbers. i is` ` ` `// used one by one hold odd numbers` ` ` `for` `(` `int` `sum = ` `0` `, i = ` `1` `; sum < n; i += ` `2` `) {` ` ` `sum += i;` ` ` `if` `(sum == n)` ` ` `return` `true` `;` ` ` `}` ` ` `return` `false` `;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String args[])` ` ` `{` ` ` `if` `(isPerfectSquare(` `35` `))` ` ` `System.out.println(` `"Yes"` `);` ` ` `else` ` ` `System.out.println(` `"NO"` `);` ` ` `if` `(isPerfectSquare(` `49` `))` ` ` `System.out.println(` `"Yes"` `);` ` ` `else` ` ` `System.out.println(` `"No"` `);` ` ` `}` `}` `// This code is contributed by Sam007` |

## Python3

`# This function returns true if n is` `# perfect square, else false` `def` `isPerfectSquare(n):` ` ` `# the_sum is sum of all odd numbers. i is` ` ` `# used one by one hold odd numbers` ` ` `i ` `=` `1` ` ` `the_sum ` `=` `0` ` ` `while` `the_sum < n:` ` ` `the_sum ` `+` `=` `i` ` ` `if` `the_sum ` `=` `=` `n:` ` ` `return` `True` ` ` `i ` `+` `=` `2` ` ` `return` `False` `# Driver code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `print` `(` `'Yes'` `) ` `if` `isPerfectSquare(` `35` `) ` `else` `print` `(` `'NO'` `)` ` ` `print` `(` `'Yes'` `) ` `if` `isPerfectSquare(` `49` `) ` `else` `print` `(` `'NO'` `)` `# This code works only in Python 3` |

## C#

`// C# program to check if n` `// is perfect square or not` `using` `System;` `public` `class` `GFG {` ` ` `// This function returns true if n` ` ` `// is perfect square, else false` ` ` `static` `bool` `isPerfectSquare(` `int` `n)` ` ` `{` ` ` `// sum is sum of all odd numbers. i is` ` ` `// used one by one hold odd numbers` ` ` `for` `(` `int` `sum = 0, i = 1; sum < n; i += 2) {` ` ` `sum += i;` ` ` `if` `(sum == n)` ` ` `return` `true` `;` ` ` `}` ` ` `return` `false` `;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `Main(String[] args)` ` ` `{` ` ` `if` `(isPerfectSquare(35))` ` ` `Console.WriteLine(` `"Yes"` `);` ` ` `else` ` ` `Console.WriteLine(` `"No"` `);` ` ` `if` `(isPerfectSquare(49))` ` ` `Console.WriteLine(` `"Yes"` `);` ` ` `else` ` ` `Console.WriteLine(` `"No"` `);` ` ` `}` `}` `// This code is contributed by Sam007.` |

## PHP

`<?php` `// PHP program to check if n is ` `// perfect square or not` `// This function returns true if n is` `// perfect square, else false` `function` `isPerfectSquare(` `$n` `)` `{` ` ` `// sum is sum of all odd numbers.` ` ` `// i is used one by one hold odd` ` ` `// numbers` ` ` `for` `( ` `$sum` `= 0, ` `$i` `= 1; ` `$sum` `< ` `$n` `;` ` ` `$i` `+= 2)` ` ` `{` ` ` `$sum` `+= ` `$i` `;` ` ` `if` `(` `$sum` `== ` `$n` `)` ` ` `return` `true;` ` ` `}` ` ` ` ` `return` `false;` `}` `// Driver code` `if` `(isPerfectSquare(35))` ` ` `echo` `"Yes\n"` `;` `else` ` ` `echo` `"No\n"` `;` ` ` `if` `(isPerfectSquare(49))` ` ` `echo` `"Yes\n"` `;` `else` ` ` `echo` `"No\n"` `;` `// This code is contributed by ajit.` `?>` |

Output :

No Yes

**Time Complexity: **O(n)

**Auxiliary Space: **O(1)**How does this work?**

Below is explanation of above approach.

1 + 3 + 5 + ... (2n-1) =∑(2*i - 1) where 1<=i<=n = 2*∑i -∑1 where 1<=i<=n = 2n(n+1)/2 - n = n(n+1) - n = n^{2}

**Reference:**

https://www.geeksforgeeks.org/sum-first-n-odd-numbers-o1-complexity/

http://blog.jgc.org/2008/02/sum-of-first-n-odd-numbers-is-always.html

This article is contributed by Utkarsh Trivedi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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