# Check whether the number can be made perfect square after adding 1

Given an integer N, the task is to check whether N the given number can be made a perfect square after adding 1 to it.

Examples:

Input: 3
Output: Yes
3 + 1 = 4 which is a perfect square i.e. 22

Input: 5
Output: No
5 + 1 = 6 which is not a perfect square.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Check whether n + 1 is a perfect square or not by taking the square root of n + 1 and checking whether it is an integer. If it is then n + 1 is a perfect square and n is a sunny number.

Below is the implementation of the above approach:

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function that returns true ` `// if x is a perfect square ` `bool` `isPerfectSquare(``long` `double` `x) ` `{ ` ` `  `    ``// Find floating point value of ` `    ``// square root of x ` `    ``long` `double` `sr = ``sqrt``(x); ` ` `  `    ``// If square root is an integer ` `    ``return` `((sr - ``floor``(sr)) == 0); ` `} ` ` `  `// Function that returns true ` `// if n is a sunny number ` `bool` `isSunnyNum(``int` `n) ` `{ ` ` `  `    ``// If (n + 1) is a perfect square ` `    ``if` `(isPerfectSquare(n + 1)) ` `        ``return` `true``; ` `    ``return` `false``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 3; ` ` `  `    ``if` `(isSunnyNum(n)) ` `        ``cout << ``"Yes"``; ` `    ``else` `        ``cout << ``"No"``; ` ` `  `    ``return` `0; ` `} `

Output:

```Yes
```

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