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[thispath98] Week 6 #895
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[thispath98] Week 6 #895
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class Solution: | ||
def maxProfit(self, prices: List[int]) -> int: | ||
minimum = prices[0] | ||
answer = 0 | ||
for i in range(1, len(prices)): | ||
if minimum > prices[i]: | ||
minimum = prices[i] | ||
else: | ||
diff = prices[i] - minimum | ||
if answer < diff: | ||
answer = diff | ||
return answer |
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class Codec: | ||
def encode(self, strs: List[str]) -> str: | ||
return "\n".join(strs) | ||
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def decode(self, s: str) -> List[str]: | ||
return s.split("\n") |
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from collections import defaultdict | ||
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class Solution: | ||
def groupAnagrams(self, strs: List[str]) -> List[List[str]]: | ||
anagram_dict = defaultdict(list) | ||
for string in strs: | ||
anagram_dict[tuple(sorted(string))].append(string) | ||
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answer = list(anagram_dict.values()) | ||
return answer | ||
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 리뷰가 늦어져서 죄송합니다. 저는 if-else문을 활용해서 키가 존재하는지의 여부를 체크했는데, defaultdict(list)를 사용하면 예외처리를 더 간단하게 할 수 있다는 걸 알게 됐습니다! 이번 주도 고생하셨습니다! |
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class Node: | ||
def __init__(self, is_end=False): | ||
self.child = {} | ||
self.is_end = is_end | ||
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class Trie: | ||
def __init__(self): | ||
self.root = Node() | ||
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def insert(self, word: str) -> None: | ||
node = self.root | ||
for ch in word: | ||
if ch not in node.child: | ||
node.child[ch] = Node() | ||
node = node.child[ch] | ||
node.is_end = True | ||
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def search(self, word: str) -> bool: | ||
node = self.root | ||
for ch in word: | ||
if ch not in node.child: | ||
return False | ||
node = node.child[ch] | ||
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return node.is_end | ||
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def startsWith(self, prefix: str) -> bool: | ||
node = self.root | ||
for ch in prefix: | ||
if ch not in node.child: | ||
return False | ||
node = node.child[ch] | ||
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return True | ||
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# Your Trie object will be instantiated and called as such: | ||
# obj = Trie() | ||
# obj.insert(word) | ||
# param_2 = obj.search(word) | ||
# param_3 = obj.startsWith(prefix) |
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class Solution: | ||
def isValid(self, s: str) -> bool: | ||
""" | ||
Intuition: | ||
stack 자료구조를 사용해서 닫히는 괄호가 올 경우 | ||
stack의 마지막과 일치하는지 확인한다. | ||
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Time Complexity: | ||
O(N): | ||
문자열을 한번 스캔하면서 조건문을 확인하므로 | ||
O(N)의 시간복잡도가 소요된다. | ||
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Space Complexity: | ||
O(N): | ||
최악의 경우 문자열 개수만큼 stack에 저장한다. | ||
""" | ||
stack = [] | ||
for ch in s: | ||
if ch in ["(", "{", "["]: | ||
stack.append(ch) | ||
elif ch in [")", "}", "]"]: | ||
if stack and ( | ||
(ch == ")" and stack[-1] == "(") | ||
or (ch == "}" and stack[-1] == "{") | ||
or (ch == "]" and stack[-1] == "[") | ||
): | ||
stack.pop() | ||
else: | ||
return False | ||
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if stack: | ||
return False | ||
else: | ||
return True | ||
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. return 시 stack의 사이즈를 비교하는 걸로 하면 한줄로 줄일 수 있을것 같아요!
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 코드가 직관적이라 이해하기 쉬웠습니다! 딕셔너리를 활용하여 코드를 조금 더 간결하게 개선할 수 있을 것 같습니다. |
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class Solution: | ||
def wordBreak(self, s: str, wordDict: List[str]) -> bool: | ||
checked = set() | ||
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def dfs(idx): | ||
if idx in checked: | ||
return | ||
checked.add(idx) | ||
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if idx == len(s): | ||
return True | ||
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for word in wordDict: | ||
word_len = len(word) | ||
if s[idx: idx + word_len] == word: | ||
if dfs(idx + word_len): | ||
return True | ||
return False | ||
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return dfs(0) |
There was a problem hiding this comment.
Choose a reason for hiding this comment
The reason will be displayed to describe this comment to others. Learn more.
아하핳... 저도 처음에 이렇게 풀었었는데, 문제에 숨은 의도를 좀 더 파악해 보는것도 괜찮을것 같아요!