added 0058_length_of_last_word.py

Author: nyierr

problems/easy/0058_length_of_last_word.py

@@ -0,0 +1,57 @@
+"""
+Given a string s consisting of some words separated by some
+number of 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 <= 10^4
+  * s consists of only English letters and spaces ' '
+  * There will be at least one word in s
+"""
+
+class Solution:   
+    # O(n) approach, we are starting from the beginning
+    #def lengthOfLastWord(self, s: str) -> int:
+    #    word = ""
+    #    for i in range(len(s)):
+    #        if word != "" and s[i-1] == " " and s[i] != " ":
+    #            word = ""
+    #            
+    #        if s[i] != " ":
+    #            word += s[i]
+    #            
+    #    return len(word)
+    
+    # O(n) approach, start from the end
+    # it will be O(n) if the string consist of a single word
+    # (with or without spaces at the end)
+    # otherwise it will be O(k) where k = length of last word
+    # + the number of space after the last word
+    def lengthOfLastWord(self, s: str) -> int:
+        c = 0
+        for i in range(len(s)-1, -1, -1):
+            if s[i] != " ":
+                c += 1
+            
+            if c > 0 and s[i] == " ":
+                break
+                
+        return c