How do you approach Length of Last Word?

Splitting on spaces works but allocates the whole word list to use one element.

What is the time complexity of Length of Last Word?

Time complexity: O(n) worst case, typically far less.. Space complexity: O(1).

What are common mistakes on Length of Last Word?

Building the result with repeated concatenation in a loop - quadratic in most languages; collect into a list and join, or use a string builder. Assuming one character is one code unit. Non-ASCII input breaks index arithmetic. Comparing strings or boxed characters by reference rather than by value.

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Length of Last Word

EasyAlgorithms

Description

Given a string `s` of words separated by spaces, return the length of the **last** word. A word is a maximal run of non-space characters. The string may have trailing spaces.

Example 1

Input: s = "Hello World"

Output: 5

Example 2

Input: s = " fly me to the moon "

Output: 4

Example 3

Input: s = "luffy is still joyboy"

Output: 6

Constraints

- 1 <= s.length <= 10^4 - s consists of English letters and spaces - There is at least one word in s.

Length of Last Word: approach, complexity and pitfalls

An easy strings problem, graded against 6 test cases (3 of them hidden).

The pattern: Strings

Parsing, comparison, and character-frequency reasoning.

Reach for it when you see: String input where the work is scanning, comparing, or counting characters.

More Stringsproblems →

If you are stuck

Show the 3-step hint ladder

Splitting on spaces works but allocates the whole word list to use one element.
Scan from the END instead: skip trailing spaces, then count characters until the next space or the start of the string.
The trailing-space case is the one that breaks naive solutions - handle it before you start counting.

The optimal approach

Show the approach: Scan backwards and stop early

The obvious solution splits on whitespace, filters out empty strings, and takes the last element. It is correct and perfectly acceptable, but it allocates a list of every word to use exactly one of them.

Scanning from the end is O(1) in space and usually touches only a handful of characters. Two phases: skip any trailing spaces, then count non-space characters until you hit a space or run off the front.

The trailing-space case is the whole difficulty. A solution that starts counting immediately returns 0 for `"day "`, and one that splits naively on a single space produces empty strings that must be filtered - which is why `" a b "` breaks so many first attempts.

A language note worth knowing: Python's `s.split()` with no argument collapses runs of whitespace and drops leading and trailing empties, so `s.split()[-1]` is correct in one line. `s.split(" ")` with an explicit separator does not, and leaves the empty strings in. The difference catches people out well beyond this problem.

Time: O(n) worst case, typically far less.
Space: O(1)

The full reference solution in every supported language stays in the editor above - reveal it there once you have had a real attempt.

Common Strings mistakes

These apply to the pattern as a whole, not just this problem.

Building the result with repeated concatenation in a loop - quadratic in most languages; collect into a list and join, or use a string builder.
Assuming one character is one code unit. Non-ASCII input breaks index arithmetic.
Comparing strings or boxed characters by reference rather than by value.