[Mock] Random 9

애플이 문제~

단순하게 푸는 것도 괜찮다는 것을 느낌

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1002. Find Common Characters

Given an array words of strings made only from lowercase letters, return a list of all characters that show up in all strings within the list (including duplicates). For example, if a character occurs 3 times in all strings but not 4 times, you need to include that character three times in the final answer.

You may return the answer in any order.

My Answer 1: Accepted (Runtime: 144 ms - 5.04% / Memory Usage: 14.3 MB - 62.34%)

class Solution:
    def commonChars(self, words: List[str]) -> List[str]:
        arr = list(words[0])
        for i in range(1, len(words)):
            tmp = list(words[i])
            for j in range(len(arr)):
                if arr[j] not in tmp:
                    arr[j] = "0"
                else:
                    tmp[tmp.index(arr[j])] = "1"
        
        answer = []
        
        for i in range(len(arr)):
            if arr[i] != "0":
                answer.append(arr[i])
        
        return answer

words[0] 을 기준으로 words[i] 에 words[0] 문자들이 있는지 확인

있으면 words[0] 문자 값은 0 으로, words[i] 문자 값은 1 로 바꿔주고

마지막에 0 이 아닌 값들만 answer 에 넣어줌

My Answer 2: Accepted (Runtime: 40 ms - 92.75% / Memory Usage: 14.4 MB - 31.41%)

class Solution:
    def commonChars(self, A: List[str]) -> List[str]:
        result = []
        for i in range(len(A[0])):
            result.append(A[0][i])
            for j in range(1, len(A)):
                if A[0][i] not in A[j]:
                    result.pop()
                    break
                else:
                    # 한번 찾은 부분은 또 사용하면 안되니까 없애줌
                    tmp = A[j].index(A[0][i])
                    A[j] = A[j][:tmp] + A[j][tmp+1:]
        
        return result

A[0] 을 기준으로 각 문자들이 모든 문자열에 있는지 확인
한번 찾은 부분은 또 사용하면 안되니까 잘라내줌

저번에 푼 방식인데 훨씬 나은듯

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1221. Split a String in Balanced Strings

Balanced strings are those that have an equal quantity of 'L' and 'R' characters.

Given a balanced string s, split it in the maximum amount of balanced strings.

Return the maximum amount of split balanced strings.

My Answer 1: Accepted (Runtime: 200 ms - 6.03% / Memory Usage: 14.4 MB - 11.23%)

class Solution:
    def balancedStringSplit(self, s: str) -> int:
        stack = ""
        cnt = 0
        
        for i in range(len(s)):
            stack += s[i]
            
            if len(stack) != 0:
                tmp = collections.Counter(stack)
                if tmp["L"] == tmp["R"]:
                    cnt += 1
        
        return cnt

stack 변수에 L, R 넣어주고 개수가 같으면 cnt += 1

My Answer 2: Accepted (Runtime: 32 ms - 56.85% / Memory Usage: 13.9 MB - 98.16%)

class Solution:
    def balancedStringSplit(self, s: str) -> int:
        left = 0
        right = 0
        result = 0
        
        for i in range(len(s)):
            if s[i] == "L":
                left += 1
            if s[i] == "R":
                right += 1
                
            if left == right:
                result += 1
        
        return result

L, R 개수 세줘서 같을때마다 result + 1

이건 저번에 풀었던 건데 훨씬 나은듯2