58. Length of Last Word

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.

Approach 01:

#include <bits/stdc++.h>

using namespace std;

class Solution {
public:
    int lengthOfLastWord(string s) {
        int n = s.size();
        int result =0, flag = 1;
        for(int i=n-1; i>=0; i--){
            if(s[i]!=' '){
                result++;
                flag = 0;
            }
            if(s[i]==' ' and flag == 0){
                return result;
            }
        }
        return result;
    }
};

from typing import *

class Solution:
    def lengthOfLastWord(self, s: str) -> int:
        n = len(s)
        result = 0
        flag = True

        for i in range(n - 1, -1, -1):
            if s[i] != ' ':
                result += 1
                flag = False
            if s[i] == ' ' and not flag:
                return result

        return result

Time Complexity

• The time complexity of the lengthOfLastWord function is \( O(n) \), where \( n \) is the length of the input string.
• The algorithm processes each character of the string exactly once, resulting in a linear time complexity.

Space Complexity

• The space complexity of the lengthOfLastWord function is \( O(1) \).
• This is because the function uses a constant amount of extra space regardless of the size of the input string.

Approach 02:

#include <string>
#include <vector>
#include <sstream>
#include <algorithm>

using namespace std;

class Solution {
public:
    int lengthOfLastWord(string s) {
        stringstream ss(s);
        string word;
        string result;

        // Split the string into words using a stringstream
        while (ss >> word) {
            result = word;
        }

        // Return the length of the last word
        return result.length();
    }
};

from typing import *

class Solution:
    def lengthOfLastWord(self, s: str) -> int:
        data = s.split(" ")
        result = ''
        for word in data:
            if word != "":
                result = word
        return len(result)

Time Complexity

• The time complexity of the lengthOfLastWord function is \( O(n) \), where \( n \) is the length of the input string.
• The algorithm processes each character of the string exactly once, resulting in a linear time complexity.

Space Complexity

• The space complexity of the lengthOfLastWord function is \( O(n) \).
• This is because the function uses additional space to store the words and the result string.