Length of Last Word

Spiral Matrix

Find the Index of the First Occurrence in a String

Measure the final whitespace-delimited word in a string.

Return the number of characters in the final word of `s`. A word is a maximal run of non-space characters. Ignore trailing spaces and return `0` when no word exists.

**Input / output**
- Input: `s: string`
- Output: integer length

**Examples**
1. `"Hello World"` returns `5`.
2. `"   fly me   to   the moon  "` returns `4`.

**Constraints**
- `0 <= s.length <= 100,000`
- The separator is the ASCII space character.

**Follow-up**
Can you solve it with one reverse scan and no substring allocation?

- A word is any maximal run of non-space characters.
- Leading, trailing, and repeated spaces should not count as part of a word.
- If the string contains only spaces, the answer is `0`.

Understand What Counts as a Word

- One straightforward approach is to split the string on whitespace.
- The last token is the last word, and its length is the answer.
- This is easy to read, but it creates a list of all words.

Simple Split-Based Idea

- The last word is closest to the end of the string.
- First skip trailing spaces.
- Then count characters until you reach another space or the beginning of the string.

Scan from the End

- After trailing spaces are skipped, `i` points at the last character of the last word.
- Move left while characters are not spaces, increasing a counter.
- If there was no word, the counter stays at `0`.

Count the Last Word Characters

- The final solution uses the end scan to avoid storing extra words.
- Time complexity is O(n) in the worst case.
- Space complexity is O(1).