Happy Number

Excel Sheet Column Number

Sqrt(x)

Decide whether repeatedly summing the squares of a number's digits eventually reaches one.

A number is **happy** if the following process ends at `1`: replace the number with the sum of the squares of its digits, then repeat. A number is **not** happy if the process loops endlessly without ever reaching `1`.

Return `true` if `n` is happy and `false` otherwise.

**Input / output**
- Input: `n: integer` (positive)
- Output: `true` if `n` is happy, else `false`

**Examples**
1. `n = 19` returns `true` (`1^2+9^2=82`, `8^2+2^2=68`, `6^2+8^2=100`, `1^2+0^2+0^2=1`).
2. `n = 2` returns `false` (it enters the cycle `4, 16, 37, 58, 89, 145, 42, 20, 4, ...`).
3. `n = 1` returns `true`.

**Constraints**
- `1 <= n <= 2,147,483,647`

**Follow-up**
The digit-square process eventually cycles. Can you detect the cycle with Floyd's slow/fast pointers in `O(1)` extra space instead of a `seen` set?

- Each step replaces `n` with the sum of the squares of its digits.
- Because the values stay bounded, the sequence must eventually repeat.
- The number is happy only if that sequence reaches `1`.

Understand the Pattern

- Peel digits off with `% 10` and `// 10`.
- Square each digit and accumulate the total.
- This helper drives every step of the process.

Compute the Digit-Square Sum

- Keep a `seen` set of numbers already visited.
- Stop with `False` if a value repeats, or `True` if it reaches `1`.
- This is `O(1)`-ish space because the reachable values are bounded.

Detect the Cycle with a Set

- Floyd's slow/fast pointers detect the cycle without a set.
- Time is `O(log n)` per step across a bounded number of steps.
- Space is `O(1)`.