How do you solve Length of Last Word on LeetCode?

Length of Last Word (LeetCode #58) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Trimming spaces is crucial to avoid counting them as part of the last word.

What is the brute force solution for Length of Last Word?

The brute force approach for Length of Last Word is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach involves splitting the string into words and then checking the last word's length. This is straightforward but can be inefficient for longer strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Length of Last Word?

Length of Last Word uses the following concepts: String. The recommended patterns to study are: Two Pointers, String Manipulation.

What are the interview tips for Length of Last Word?

Always clarify the problem requirements and constraints before diving into coding. Consider edge cases and test your solution against them. Explain your thought process clearly while coding; it helps the interviewer understand your approach.

What are common mistakes when solving Length of Last Word?

Forgetting to trim spaces before processing the string. Not handling edge cases like multiple spaces between words.

#58

Length of Last Word

Easy

StringTwo PointersString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal solution avoids unnecessary splitting by scanning the string from the end to find the last word directly. This is more efficient as it only requires a single pass.

⚙️

Algorithm

3 steps

1Step 1: Trim any trailing spaces from the string.
2Step 2: Initialize a counter for the length of the last word.
3Step 3: Traverse the string from the end until a space is encountered or the beginning of the string is reached, counting characters.

solution.py10 lines

1def length_of_last_word(s):
2    s = s.rstrip()
3    length = 0
4    for char in reversed(s):
5        if char == ' ':
6            if length > 0:
7                break
8        else:
9            length += 1
10    return length

ℹ

Complexity note: The time complexity is O(n) because we traverse the string once. The space complexity is O(1) since we use a constant amount of extra space.

1Trimming spaces is crucial to avoid counting them as part of the last word.
2Iterating from the end of the string is more efficient than splitting the entire string.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.