[LeetCode] Jewels and Stones

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".

예제

Example 1:

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

Example 2:

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

조건

• 1 <= jewels.length, stones.length <= 50
• jewels and stones consist of only English letters.
• All the characters of jewels are unique.

답

풀이 1

Counter를 활용한 내 풀이

class Solution:
    def numJewelsInStones(self, jewels: str, stones: str) -> int:
        ans = 0
        cnt = collections.Counter(stones)
        for i in jewels:
            ans += cnt[i]
        return ans

풀이 2

해시 테이블을 이용한 풀이

class Solution:
    def numJewelsInStones(self, jewels: str, stones: str) -> int:
        freqs = {}
        count = 0
        
        for char in stones:
            if char not in freqs:
                freqs[char] = 1
            else:
                freqs[char] += 1
        
        # 보석의 빈도 수 합산
        for char in jewels:
            if char in freqs:
                count += freqs[char]
        
        return count

풀이 3

defaultdict를 이용한 풀이

class Solution:
    def numJewelsInStones(self, jewels: str, stones: str) -> int:
        freqs = collections.defaultdict(int)
        count = 0
        
        # 비교 없이 돌(S) 빈도 수 계산
        for char in stones:
            freqs[char] += 1
        
        # 비교 없이 보석(J) 빈도 수 합산
        for char in jewels:
            count += freqs[char]
            
        return count

풀이 4

파이썬다운 방식

class Solution:
    def numJewelsInStones(self, jewels: str, stones: str) -> int:
        return sum(s in jewels for s in stones)