# Solving Leetcode Interviews in Seconds with AI: Length of Last Word


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "58" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# 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.  

	# Explanation
	Here's an efficient solution to find the length of the last word in a string, along with explanations:

*   **Traverse from Right to Left:** The most efficient way is to start from the end of the string and move backward.
*   **Skip Trailing Spaces:** First, skip any trailing spaces.
*   **Count Characters:** Once a non-space character is encountered, start counting until another space is found or the beginning of the string is reached.

*   **Time Complexity:** O(n), where n is the length of the string.
*   **Space Complexity:** O(1)

	
	# Code
	```python
	def lengthOfLastWord(s: str) -> int:
    """
    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.
    """
    s = s.rstrip()  # Remove trailing spaces
    length = 0
    for i in range(len(s) - 1, -1, -1):
        if s[i] == ' ':
            break
        length += 1
    return length
	```