Merge pull request #1398 from sejineer/main

[sejineer] Week 05 solutions

Author: sejineer

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

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+"""
+시간 복잡도: O(N)
+공간 복잡도: O(1)
+"""
+class Solution:
+    def maxProfit(self, prices: List[int]) -> int:
+        buy = prices[0]
+        result = 0
+        
+        for price in prices:
+            profit = price - buy
+            buy = min(buy, price)
+            result = max(profit, result)
+        
+        return result

group-anagrams/sejineer.py

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+"""
+시간 복잡도: O(N * K) N=문자열 개수, 평균 문자열 길이 = K (최대 100)
+공간 복잡도: O(N * K) 
+"""
+from collections import Counter, defaultdict
+
+class Solution:
+    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+        strs_dict = defaultdict(list)
+        for s in strs:
+            s_counter = Counter(s)
+            strs_dict[frozenset(s_counter.items())].append(s)
+        
+        return list(strs_dict.values())

implement-trie-prefix-tree/sejineer.py

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+class Trie:
+
+    def __init__(self):
+        self.root = {"$": True}        
+
+    def insert(self, word: str) -> None:
+        node = self.root
+        for ch in word:
+            if ch not in node:
+                node[ch] = {"$": False}
+            node = node[ch]
+        node["$"] = True
+
+
+    def search(self, word: str) -> bool:
+        node = self.root
+        for ch in word:
+            if ch not in node:
+                return False
+            node = node[ch]
+        return node["$"]
+
+    def startsWith(self, prefix: str) -> bool:
+        node = self.root
+        for ch in prefix:
+            if ch not in node:
+                return False
+            node = node[ch]
+        return True

word-break/sejineer.py

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+"""
+시간 복잡도: O(n * D * L) n = 문자열 길이, D = 사전 크기, L = 단어 평균 길이
+공간 복잡도: O(n)
+"""
+class Solution:
+    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
+        
+        @cache
+        def dfs(k: int) -> bool:
+            if k == len(s):
+                return True
+            for word in wordDict:
+                if s[k : k + len(word)] == word:
+                    if dfs(k + len(word)):
+                        return True
+            return False
+        
+        return dfs(0)