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NeetCode #8LC-58EasyArrays & Hashing
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#8 · #58 · Length of Last Word(最后一个单词的长度)

📌 Problem Statement & Constraints

Given a string s consisting of words and spaces, return the length of the last word. A word is a maximal substring of non-space characters. Constraints: 1 <= s.length <= 10^4; s contains only English letters and spaces.

💡 Core Algorithmic Approaches

  1. Trailing spaces complicate a naive split, so strip them first.
  2. After stripping, the last word starts right after the final space. Use rfind(' ') to locate it and return the length of the suffix.
  3. If there is no space, the whole string is one word, so the answer is its length.
  4. Python one-liners: len(s.split()[-1]) or len(s.rstrip().rsplit(' ', 1)[-1]).

💻 Benchmark Python3 Implementation

class Solution:
    def lengthOfLastWord(self, s: str) -> int:
        i = len(s) - 1
        while i >= 0 and s[i] == " ":      # skip trailing spaces
            i -= 1
        length = 0
        while i >= 0 and s[i] != " ":      # count the last word
            length += 1
            i -= 1
        return length


# Concise equivalent
class Solution2:
    def lengthOfLastWord(self, s: str) -> int:
        return len(s.split()[-1])

⚡ Complexity Deep Dive

⏱️ Time Complexity
O(n) worst case (the string may be entirely spaces followed by one short word), but the scan starts from the end so it is usually much less.
💾 Space Complexity
O(1) for the pointer version; O(n) for split which materialises all words.

⚠️ Interview Pitfalls & Follow-ups

  • s.split(' ')[-1]: with explicit ' ' as the separator, trailing spaces produce empty strings, so the last element may be "". Use split() with no argument (which collapses whitespace) or strip first.
  • Forgetting the trailing-space skip: "a " should return 1, not 0.
  • Assuming exactly one space between words: the constraints allow multiple spaces.