Day019 - 58. Length of Last Word / 14. Longest Common Prefix

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.

내 풀이

class Solution:
    def lengthOfLastWord(self, s: str) -> int:

        s = list(s)
        answer = 0

        while s:
            if s[-1] == " ": s.pop()
            else: break

        while s:
            if s[-1] != " ":
                answer += 1
                s.pop()
            else:
                break

        return answer

class Solution:
    def lengthOfLastWord(self, s: str) -> int:

        # print(s.strip())
        # print(s.strip().split())

        text = s.strip().split()
        target = text[-1]

        return len(target)

# Time Complexity : \(O(N)\)

14. Longest Common Prefix

Write a function to find the longest common prefix string amongst an array of strings.

If there is no common prefix, return an empty string "".

Example 1:

Input: strs = ["flower","flow","flight"]
Output: "fl"

Example 2:

Input: strs = ["dog","racecar","car"]
Output: ""
Explanation: There is no common prefix among the input strings.

Constraints:

• 1 <= strs.length <= 200
• 0 <= strs[i].length <= 200
• strs[i] consists of only lowercase English letters.

• 내 풀이

class Solution:
    def longestCommonPrefix(self, strs: List[str]) -> str:

        first_word = list(strs[0])
        first_word.reverse()
        answer = []

        strs = strs[1:]

        check_idx = 0

        while first_word:
            for i in strs:
                try:
                    if first_word[-1] == i[check_idx]:
                        pass
                    else:
                        if check_idx == 0:
                            return ""
                        else:
                            return ''.join(answer)
                except:
                    return ''.join(answer)
            check_idx += 1
            answer.append(first_word.pop())

        return ''.join(answer)

# Time Complexity : \(O(N)\)

• 친구의 풀이

class Solution:
    def longestCommonPrefix(self, strs: List[str]) -> str:
        if len(strs) == 0:
            return ""

        prefix = strs[0]
        for i in range(len(strs)):
            while strs[i].find(prefix) != 0:
                prefix = prefix[0:len(prefix)-1]
                if prefix == "":
                    return ""
        return prefix