[LeetCode] 1544번 Make The String Great

문제

Given a string s of lower and upper case English letters.

A good string is a string which doesn’t have two adjacent characters s[i] and s[i + 1] where:

• 0 <= i <= s.length - 2
• s[i] is a lower-case letter and s[i + 1] is the same letter but in upper-case or vice-versa.

To make the string good, you can choose two adjacent characters that make the string bad and remove them. You can keep doing this until the string becomes good.

Return the string after making it good. The answer is guaranteed to be unique under the given constraints.

Notice that an empty string is also good.

예제

Example 1:

Input: s = "leEeetcode"
Output: "leetcode"
Explanation: In the first step, either you choose i = 1 or i = 2, both will result "leEeetcode" to be reduced to "leetcode".

Example 2:

Input: s = "abBAcC"
Output: ""
Explanation: We have many possible scenarios, and all lead to the same answer. For example:
"abBAcC" --> "aAcC" --> "cC" --> ""
"abBAcC" --> "abBA" --> "aA" --> ""

Example 3:

Input: s = "s"
Output: "s"

조건

Constraints:

• 0 <= s.length <= 100
• 0 <= t.length <= 10^4
• Both strings consists only of lowercase characters.

풀이과정

내 풀이

class Solution:
    def makeGood(self, s: str) -> str:
        li = list(s)
        count = 0
        while(count == 0):
            count2 = 0
            n = len(li)
            for i in range(len(li)-1):
                count2 += 1
                if (ord(li[i]) - 32 == ord(li[i+1])) or (ord(li[i]) + 32 == ord(li[i+1])):
                    del li[i:i+2]
                    break
            if count2 == n-1:
                return "".join(li)
            

s를 li리스트로 넣어준 뒤, 아스키 코드를 사용해서 소문자와 대문자가 인접한 경우 삭제해준다.

다른 풀이

class Solution:
    def makeGood(self, s: str) -> str:
        result = []
        for c in s:
            if not result:
                result.append(c)
            elif result[-1].isupper() and result[-1].lower() == c:
                result.pop()
            elif result[-1].islower() and result[-1].upper() == c:
                result.pop()
            else:
                result.append(c)
        return ''.join(result)

배워야할 점

1. 아스키코드보다 upper()나 lower()사용하면 편했을 것.
2. 굳이 리스트에 넣지 않고도 for c in s로 전개.

가볍게 풀어보려고 했는데 생각보다 오래걸림. 열심히 해야겠다. 생각보다 문자열에서 배울게 많았던 문제.