771 보석과 돌

• 771 보석과 돌
  • 문제
  • 조건
  • 예제
    • 예제 1
    • 예제 2
  • 해결
    • 1st
      • 1.1 해결해야 할 사항
      • 1.2 생각
      • 1.3 구현
        • 1.3.1 dictionary 사용
        • 1.3.2 Couter 사용
        • 1.3.3 문자열 그대로 사용
        • 1.3.4 문자열 그대로 사용2

771 보석과 돌

문제

You’re given strings jewels representing the types of stones that are jewels, and stones representing the stones you have. Each character in stones is a type of stone you have. You want to know how many of the stones you have are also jewels. Letters are case sensitive, so “a” is considered a different type of stone from “A”.

• jewels: 보석인 stones의 타입을 나타낸다
• stones: 내가 가진 돌을 나타내며, stones의 각 문자는 내가 가진 돌의 타입을 나타낸다
• 알고 싶은 것? 내가 가진 돌 중 얼마나 보석이 있는지
• 문자들은 대소문자를 구별한다

조건

• $1 \le jewels.length, stones.length \le 50$
• jewels and stones consist of only English letters.
• All the characters of jewels are unique.

예제

예제 1

Input: jewels = "aA", stones = "aAAbbbb"
Output: 3

예제 2

Input: jewels = "z", stones = "ZZ"
Output: 0

해결

1st

1.1 해결해야 할 사항

• jewels에 해당하는 문자가 stones에 얼마나 있는지 카운트

1.2 생각

• jewels의 각 문자는 유일하다고 하므로 중복은 고려하지 않아도 될 듯
• stones를 잘라서 jewels에 해당하는 문자는 딕셔너리에 카운트해서 담아서 카운트?
• Counter를 쓰면?

1.3 구현

1.3.1 dictionary 사용

def first(self, jewels: str, stones: str) -> int:
    # https://docs.python.org/ko/3/library/stdtypes.html?highlight=dict#dict
    jewels_dict = {}
    ans = 0
    for jewel in jewels:
        jewels_dict[jewel] = True

    for stone in stones:
        if stone in jewels_dict:
            ans += 1
    
    return ans

Runtime: 32 ms, faster than 63.11% of Python3 online submissions for Jewels and Stones. Memory Usage: 14.2 MB, less than 76.02% of Python3 online submissions for Jewels and Stones.

1.3.2 Couter 사용

def second(self, jewels: str, stones: str) -> int:
    # https://docs.python.org/ko/3/library/collections.html?highlight=counter#collections.Counter
    c = Counter(stones)
    ans = 0
    for jewel in jewels:
        if jewel in c:
            ans += c[jewel]

    return ans

Runtime: 36 ms, faster than 21.95% of Python3 online submissions for Jewels and Stones. Memory Usage: 14.3 MB, less than 48.17% of Python3 online submissions for Jewels and Stones.

1.3.3 문자열 그대로 사용

def third(self, jewels: str, stones: str) -> int:
    ans = 0
    for stone in stones:
        if stone in jewels:
            ans += 1
    
    return ans

Runtime: 32 ms, faster than 63.11% of Python3 online submissions for Jewels and Stones. Memory Usage: 14.1 MB, less than 91.67% of Python3 online submissions for Jewels and Stones.

1.3.4 문자열 그대로 사용2

• 리트 코드에 답으로 올라가 있는 코드 중 가장 빠른 코드로 기록된 걸로 테스트

def fourth(self, jewels: str, stones: str) -> int:
    count = 0
    for c in jewels:
        if c in stones:
            count += stones.count(c) 
    return count

Runtime: 28 ms, faster than 85.95% of Python3 online submissions for Jewels and Stones. Memory Usage: 14.4 MB, less than 13.91% of Python3 online submissions for Jewels and Stones.