58. Length of Last Word LeetCode Solution
In this guide, you will get 58. Length of Last Word LeetCode Solution with the best time and space complexity. The solution to Length of Last Word problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Table of Contents
- Problem Statement
- Complexity Analysis
- Length of Last Word solution in C++
- Length of Last Word solution in Java
- Length of Last Word solution in Python
- Additional Resources
Problem Statement of 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.
Complexity Analysis
- Time Complexity: O(n)
- Space Complexity: O(1)
58. Length of Last Word LeetCode Solution in C++
class Solution {
public:
int lengthOfLastWord(string s) {
int i = s.length() - 1;
while (i >= 0 && s[i] == ' ')
--i;
const int lastIndex = i;
while (i >= 0 && s[i] != ' ')
--i;
return lastIndex - i;
}
};
/* code provided by PROGIEZ */
58. Length of Last Word LeetCode Solution in Java
class Solution {
public int lengthOfLastWord(String s) {
int i = s.length() - 1;
while (i >= 0 && s.charAt(i) == ' ')
--i;
final int lastIndex = i;
while (i >= 0 && s.charAt(i) != ' ')
--i;
return lastIndex - i;
}
}
// code provided by PROGIEZ
58. Length of Last Word LeetCode Solution in Python
class Solution:
def lengthOfLastWord(self, s: str) -> int:
i = len(s) - 1
while i >= 0 and s[i] == ' ':
i -= 1
lastIndex = i
while i >= 0 and s[i] != ' ':
i -= 1
return lastIndex - i
# code by PROGIEZ
Additional Resources
- Explore all LeetCode problem solutions at Progiez here
- Explore all problems on LeetCode website here
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