Approach: Scan from the end

def lengthOfLastWord(s):
    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

Complexity: O(n) time, O(1) space.

Full Solution — all languages

class Solution:
    def lengthOfLastWord(self, s):
        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

import java.util.*;
class Solution {
    public int lengthOfLastWord(String s) {
        int i = s.length() - 1;
        while (i >= 0 && s.charAt(i) == ' ') i--;
        int length = 0;
        while (i >= 0 && s.charAt(i) != ' ') {
            length++;
            i--;
        }
        return length;
    }
}

#include <bits/stdc++.h> using namespace std;
class Solution {
    public: int lengthOfLastWord(string s) {
        int i = (int)s.size() - 1;
        while (i >= 0 && s[i] == ' ') i--;
        int length = 0;
        while (i >= 0 && s[i] != ' ') {
            length++;
            i--;
        }
        return length;
    }
};

/**
 * @param {string} s
 * @return {number}
 */ var lengthOfLastWord = function(s) {
    let i = s.length - 1;
    while (i >= 0 && s[i] === ' ') i--;
    let length = 0;
    while (i >= 0 && s[i] !== ' ') {
        length++;
        i--;
    }
    return length;
};

Length of Last Word

Example 1

Example 2

Example 3

Constraints