Factorial

Move Zeroes

Fibonacci Number

Compute the product of all positive integers up to a non-negative input.

Return `n!`, the product of all positive integers from `1` through `n`. By definition, `0! = 1`.

**Input / output**
- Input: `n: integer`
- Output: integer factorial

**Examples**
1. `n = 0` returns `1`.
2. `n = 5` returns `120`.

**Constraints**
- `0 <= n <= 12` so the result fits a 32-bit signed integer

**Follow-up**
What overflow and recursion-depth issues appear when the input range grows?

- Factorial means multiplying all positive integers from `1` through `n`.
- By definition, `0!` equals `1`.
- The input is non-negative, so we do not need to handle negative numbers.

Understand the Definition

- Factorial has a direct recursive definition: `n! = n * (n - 1)!`.
- The base case is `0! = 1` or `1! = 1`.
- This works, but an iterative loop avoids recursion depth concerns.

Recursive Idea

- Start with `result = 1`.
- Multiply by each number from `2` through `n`.
- If `n` is `0` or `1`, the loop does not run and the answer remains `1`.

Iterative Accumulation

- Use `range(2, n + 1)` so the final multiplier `n` is included.
- Update the running product each iteration.
- Return the product after all factors have been applied.

Build the Loop

- The solution directly follows the mathematical definition.
- Time complexity is O(n).
- Space complexity is O(1) because it uses one accumulator.