[jinah92] Week 5

best-time-to-buy-and-sell-stock/jinah92.py

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+# O(N) times, O(1) spaces
+# price[i] 가격으로 팔아서 가장 수익을 많이 내려면, i-1번째까지 중 가장 값이 싼 날짜에서 주식을 사면 된다
+class Solution:
+    def maxProfit(self, prices: List[int]) -> int:
+        max_profit = 0
+        min_price = prices[0]
+
+        for price in prices:
+            profit = price - min_price
+            max_profit = max(max_profit, profit)
+            min_price = min(min_price, price)
+
+        return max_profit

encode-and-decode-strings/jinah92.py

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+# O(N) times, O(1) spaces
+class Solution:
+    """
+    @param: strs: a list of strings
+    @return: encodes a list of strings to a single string.
+    """
+    def encode(self, strs):
+        encoded_str = ""
+        for str in strs:
+            encoded_str += f"{len(str)}:{str}"
+        return encoded_str
+
+    """
+    @param: str: A string
+    @return: decodes a single string to a list of strings
+    """
+    def decode(self, str):
+        list_strs, start = [], 0
+        while start < len(str):
+            offset = str.find(":", start)
+            s_length = int(str[start:offset])
+            list_strs.append(str[offset+1:offset+1+s_length])
+            start = offset+1+s_length
+        return list_strs

group-anagrams/jinah92.py

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+# O(N * NlogN) times, O(N) spaces
+class Solution:
+    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+        anagrams_dict = {}
+        result = []
+
+        for str in strs: # N 시간 소요
+            str_count = {}
+            for s in str:
+                if not s in str_count:
+                    str_count[s] = 1
+                else:
+                    str_count[s] += 1
+            anagrams_keys = []
+            for key, val in sorted(str_count.items()): # 최대 N개의 dict items를 배열하므로 NlogN 시간 소요
+                anagrams_keys.append(tuple([key, val]))
+
+            anagrams_key = tuple(anagrams_keys)
+            if tuple(anagrams_keys) not in anagrams_dict:
+                anagrams_dict[tuple(anagrams_keys)] = []
+            anagrams_dict[tuple(anagrams_keys)].append(str)
+
+
+        return list(anagrams_dict.values())

implement-trie-prefix-tree/jinah92.py

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+# 시간복잡도
+## insert, search는 해시 기반 조회로 인해 O(1), startsWith은 O(n * k) 시간이 소요 (n: 저장된 단어수, k: prefix 길이)
+# 공간복잡도
+## O(n * m) 공간이 필요 (n: 저장된 단어 수, m : 각 단어의 평균 길이)
+class Trie:
+    def __init__(self):
+        self.trie = set()
+
+    def insert(self, word: str) -> None:
+        self.trie.add(word)
+
+    def search(self, word: str) -> bool:
+        return word in self.trie
+
+    def startsWith(self, prefix: str) -> bool:
+        return any(item.startswith(prefix) for item in self.trie)

word-break/jinah92.py

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+# O(2^s*w) times, O(s) space
+class Solution:
+    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
+        @cache
+        def dfs(start):
+            if start == len(s):
+                return True
+            for word in wordDict:
+                if s[start:start+len(word)] == word:
+                    if dfs(start+len(word)):
+                        return True
+            return False
+
+        return dfs(0)