Merge pull request #883 from HodaeSsi/main

[HodaeSsi] Week 5

Author: TonyKim9401

best-time-to-buy-and-sell-stock/HodaeSsi.py

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+# 시간 복잡도 : O(n)
+# 공간 복잡도 : O(1)
+class Solution:
+	def maxProfit(self, prices: List[int]) -> int:
+		min_price = float('inf')
+		max_profit = 0
+
+		for price in prices:
+			min_price = min(min_price, price)
+			max_profit = max(max_profit, price - min_price)
+
+		return max_profit
+	

encode-and-decode-strings/HodaeSsi.py

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+# 시간복잡도: O(n)
+# 공간복잡도: O(1)
+
+class Solution:
+	"""
+	@param: strs: a list of strings
+	@return: encodes a list of strings to a single string.
+	"""
+	def encode(self, strs):
+		if not strs:
+			return ""
+		return chr(257).join(strs)
+
+	"""
+	@param: str: A string
+	@return: decodes a single string to a list of strings
+	"""
+	def decode(self, str):
+		if not str:
+			return []
+		return str.split(chr(257))
+	

group-anagrams/HodaeSsi.py

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+# 시간복잡도 : O(n*mlogm) (n은 strs의 길이, m은 strs의 원소의 길이)
+# 공간복잡도 : O(n)
+
+class Solution:
+	def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+		anagrams = collections.defaultdict(list)
+		for word in strs:
+			anagrams[''.join(sorted(word))].append(word)
+		return list(anagrams.values())
+	

implement-trie-prefix-tree/HodaeSsi.py

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+# 시간복잡도: O(n) (n은 문자열의 길이)
+# 공간복잡도: O(n) (n은 문자열의 길이)
+class Trie:
+	def __init__(self):
+		self.root = {}
+		self.end = False
+
+	def insert(self, word: str) -> None:
+		node = self.root
+		for char in word:
+			if char not in node:
+				node[char] = {}
+			node = node[char]
+		node[self.end] = True
+	
+	def search(self, word: str) -> bool:
+		node = self.root
+		for char in word:
+			if char not in node:
+				return False
+			node = node[char]
+		return self.end in node
+	
+	def startsWith(self, prefix: str) -> bool:
+		node = self.root
+		for char in prefix:
+			if char not in node:
+				return False
+			node = node[char]
+		return True
+

word-break/HodaeSsi.py

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+# 시간복잡도 : O(n^2)
+# 공간복잡도 : O(n)
+class Solution:
+	def wordBreak(self, s: str, wordDict: List[str]) -> bool:
+		dp = [False] * (len(s) + 1)
+		dp[0] = True
+
+		for i in range(1, len(s) + 1):
+			for j in range(i):
+				if dp[j] and s[j:i] in wordDict:
+					dp[i] = True
+					break
+
+		return dp[-1]
+