LeetCode Challenge: 58. Length of Last Word - JavaScript Solution🚀

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Top Interview 150

String manipulation is a common coding interview topic, and this problem is a great example of working with substrings efficiently. Let’s solve LeetCode 58: Length of Last Word using a straightforward approach.

🚀 Problem Description

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

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

💡 Examples

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.

🏆 JavaScript Solution

The problem can be solved by trimming unnecessary spaces and locating the last word efficiently.

Implementation

var lengthOfLastWord = function(s) {
    const words = s.trim().split(' ');

    return words[words.length - 1].length;
};

🔍 How It Works

Trim the String:
- Remove leading and trailing spaces using trim() to simplify the logic.
Split the String:
- Use split(' ') to divide the string into words based on spaces.
Get the Last Word:
- Access the last element of the array (words[words.length - 1]) and return its length.

🔑 Complexity Analysis

> Time Complexity: O(n) is the length of the string. Trimming and splitting the string both run in linear time.
> Space Complexity: O(n), as splitting creates an array of words.

Alternative Solution (No Split)

For improved space efficiency, traverse the string backward to find the last word:

var lengthOfLastWord = function(s) {
    let length = 0;
    let i = s.length - 1;

    // Skip trailing spaces
    while (i >= 0 && s[i] === ' ') {
        i--;
    }

    // Count characters of the last word
    while (i >= 0 && s[i] !== ' ') {
        length++;
        i--;
    }

    return length;
};

🔑 Complexity Analysis (Alternative Solution)

> Time Complexity: O(n), as we traverse the string once.
> Space Complexity: O(1), as no extra data structures are used.

📋 Dry Run

Input: s = "fly me to the moon"

Output: 4

✨ Pro Tips for Interviews

Clarify constraints:
- Can the string be empty?
- Are there multiple spaces between words?
Optimize for edge cases:
- Input with only spaces (" " → 0).
- Single-word input ("hello" → 5).