# Leetcode Happy-Number

#### 2022-07-26 09:12

> ##### Algorithms:
>
> #algorithm #Floyd_s_cycle_finding_algorithm
>
> ##### Difficulty:
>
> #coding_problem #difficulty-easy
>
> ##### Additional tags:
>
> #leetcode
>
> ##### Revisions:
>
> N/A

##### Related topics:

##### Links:

- [Link to problem](https://leetcode.com/problems/happy-number/)

---

### Problem

Write an algorithm to determine if a number `n` is happy.

A **happy number** is a number defined by the following process:

- Starting with any positive integer, replace the number by the sum of the squares of its digits.
- Repeat the process until the number equals 1 (where it will stay), or it **loops endlessly in a cycle** which does not include 1.
- Those numbers for which this process **ends in 1** are happy.

Return `true` _if_ `n` _is a happy number, and_ `false` _if not_.

#### Examples

**Example 1:**

**Input:** n = 19
**Output:** true
**Explanation:**
12 + 92 = 82
82 + 22 = 68
62 + 82 = 100
12 + 02 + 02 = 1

**Example 2:**

**Input:** n = 2
**Output:** false

#### Constraints

- `1 <= n <= 231 - 1`

### Thoughts

> [!summary]
> This is a #Floyd_s_cycle_finding_algorithm

This works, because as the problem mentioned, this will result in a endless loop.

So, by using fast ans slow, we can determine whether there is a loop.

And, when fast hit 1, we know slow will eventually reach the
answer, so we return early. (but in the cost of time of checking).

### Solution

```cpp
class Solution {
  int getDigitSqrt(int i) {
    int sum = 0;

    while (i) {
      sum += (i % 10) * (i % 10);
      i /= 10;
    }

    return sum;
  }

public:
  bool isHappy(int n) {
    // Floyd cycle finding algorighm.
    int slow, fast;
    slow = fast = n;

    do {
      slow = getDigitSqrt(slow);
      fast = getDigitSqrt(fast);
      fast = getDigitSqrt(fast);
      if (fast == 1) {
        return true;
      }
    } while (fast != slow);

    return false;
  }
};
```
