58. Length of Last Word - Explanation

Problem Link

Description

You are given a string s consisting of words and spaces, return the length of the last word in the string.

A word is a maximal substring consisting of non-space characters only.

Note: A substring is a contiguous non-empty sequence of characters within a string.

Example 1:

Input: s = "Hello World"

Output: 5

Explanation: The last word is "World" with length 5.

Example 2:

Input: s = "   fly me   to   the moon  "

Output: 4

Explanation: The last word is "moon" with length 4.

Example 3:

Input: s = "luffy is still joyboy"

Output: 6

Explanation: The last word is "joyboy" with length 6.

Constraints:

• 1 <= s.length <= 10,000
• s consists of only English letters and spaces ' '.
• There will be at least one word in s.

Company Tags

Please upgrade to NeetCode Pro to view company tags.

1. Iteration - I

class Solution:
    def lengthOfLastWord(self, s: str) -> int:
        length = i = 0
        while i < len(s):
            if s[i] == ' ':
                while i < len(s) and s[i] == ' ':
                    i += 1
                if i == len(s):
                    return length
                length = 0
            else:
                length += 1
                i += 1
        return length

Time & Space Complexity

• Time complexity: $O(n)$
• Space complexity: $O(1)$

---

2. Iteration - II

class Solution:
    def lengthOfLastWord(self, s: str) -> int:
        i, length = len(s) - 1, 0
        while s[i] == ' ':
            i -= 1
        while i >= 0 and s[i] != ' ':
            i -= 1
            length += 1
        return length

Time & Space Complexity

• Time complexity: $O(n)$
• Space complexity: $O(1)$

---

3. Built-In Function

class Solution:
    def lengthOfLastWord(self, s: str) -> int:
        return len(s.split().pop())

Time & Space Complexity

• Time complexity: $O(n)$
• Space complexity: $O(n)$