[hu6r1s] WEEK 05 Solutions

best-time-to-buy-and-sell-stock/hu6r1s.py

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+class Solution:
+    """
+    1. 브루트포스
+    2중 for문이라서 O(n^2)임.
+    prices의 길이가 10^5이므로 10^10이 되면서 시간초과가 발생
+
+    2. 이분탐색으로 가능할까 했지만 DP를 사용해야 하는 문제 같음
+    시간복잡도는 O(n)이 나옴
+    """
+    """
+    def maxProfit(self, prices: List[int]) -> int:
+        max_profit = 0
+        for i in range(len(prices)-1):
+            for j in range(i, len(prices)):
+                profit = prices[j] - prices[i]
+                max_profit = max(max_profit, profit)
+        return max_profit
+    """
+    def maxProfit(self, prices: List[int]) -> int:
+        max_profit = 0
+        min_price = prices[0]
+        for price in prices:
+            max_profit = max(max_profit, price - min_price)
+            min_price = min(price, min_price)
+        return max_profit

encode-and-decode-strings/hu6r1s.py

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+class Solution:
+    """
+    @param: strs: a list of strings
+    @return: encodes a list of strings to a single string.
+    """
+    def encode(self, strs):
+        return "secretKey!@#".join(strs)
+
+    """
+    @param: str: A string
+    @return: decodes a single string to a list of strings
+    """
+    def decode(self, str):
+        return str.split("secretKey!@#")

group-anagrams/hu6r1s.py

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+from collections import defaultdict
+
+class Solution:
+    """
+    1. 정렬된 값이 딕셔너리에 있으면 리스트 형식으로 삽입
+    """
+    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+        dict_strs = defaultdict(list)
+
+        for word in strs:
+            dict_strs[str(sorted(word))].append(word)
+        return list(dict_strs.values())

implement-trie-prefix-tree/hu6r1s.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
+
+
+# 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)

word-break/hu6r1s.java

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+import java.util.*;
+
+class Solution {
+    public boolean wordBreak(String s, List<String> wordDict) {
+        Set<String> wordSet = new HashSet<>(wordDict);
+        boolean[] dp = new boolean[s.length() + 1];
+        dp[0] = true;
+
+        for (int i = 1; i < s.length() + 1; i++) {
+            for (int j = 0; j < i; j++) {
+                if (dp[j] && (wordSet.contains(s.substring(j, i)))) {
+                    dp[i] = true;
+                    break;
+                }
+            }
+        }
+        return dp[s.length()];
+    }
+}

word-break/hu6r1s.py

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+class Solution:
+    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
+        word_set = set(wordDict)   # O(1) 조회를 위해 set으로 변환
+        n = len(s)
+        dp = [False] * (n + 1)
+        dp[0] = True   # 공집합은 항상 가능
+
+        for i in range(1, n + 1):
+            for j in range(i):
+                if dp[j] and s[j:i] in word_set:
+                    dp[i] = True
+                    break   # i번째까지 나눌 수 있으면 더 확인할 필요 없음
+
+        return dp[-1]