LeetCode #58 — EASY

Length of Last Word

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

The Problem

Problem Statement

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.

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 <= 10⁴
s consists of only English letters and spaces ' '.
There will be at least one word in s.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

"Hello World"

Example 2

"   fly me   to   the moon  "

Example 3

"luffy is still joyboy"

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

class Solution {
    public int lengthOfLastWord(String s) {
        int i = s.length() - 1;

        // Skip trailing spaces.
        while (i >= 0 && s.charAt(i) == ' ') i--;

        int len = 0;
        while (i >= 0 && s.charAt(i) != ' ') {
            len++;
            i--;
        }
        return len;
    }
}

func lengthOfLastWord(s string) int {
    i := len(s) - 1
    for i >= 0 && s[i] == ' ' {
        i--
    }

    length := 0
    for i >= 0 && s[i] != ' ' {
        length++
        i--
    }

    return length
}

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

        length = 0
        while i >= 0 and s[i] != ' ':
            length += 1
            i -= 1

        return length

impl Solution {
    pub fn length_of_last_word(s: String) -> i32 {
        let bytes = s.as_bytes();
        let mut i = bytes.len() as i32 - 1;

        while i >= 0 && bytes[i as usize] == b' ' {
            i -= 1;
        }

        let mut len = 0i32;
        while i >= 0 && bytes[i as usize] != b' ' {
            len += 1;
            i -= 1;
        }

        len
    }
}

function lengthOfLastWord(s: string): number {
  let i = s.length - 1;
  while (i >= 0 && s[i] === ' ') i--;

  let len = 0;
  while (i >= 0 && s[i] !== ' ') {
    len++;
    i--;
  }
  return len;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.