Merge pull request #1401 from hi-rachel/main

[hi-rachel] Week 05 Solutions

Author: hi-rachel

best-time-to-buy-and-sell-stock/hi-rachel.py

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+# TC: O(N), SC: O(1)
+
+class Solution:
+    def maxProfit(self, prices: List[int]) -> int:
+        max_profit = 0
+        min_price = prices[0]
+
+        for price in prices:
+            max_profit = max(price - min_price, max_profit)
+            min_price = min(price, min_price)
+        return max_profit
+
+# TS 풀이
+# 배열 요소(숫자값)을 직접 순회하려면 for ... of 사용 혹은 forEach
+# for ... in -> 인덱스를 가져옴
+
+# function maxProfit(prices: number[]): number {
+#     let max_profit: number = 0;
+#     let min_price: number = prices[0];
+
+#     for (let price of prices) {
+#         max_profit = Math.max(max_profit, price - min_price);
+#         min_price = Math.min(min_price, price);
+#     }
+#     return max_profit;
+# };

encode-and-decode-strings/hi-rachel.py

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+# 문제: https://neetcode.io/problems/string-encode-and-decode
+# TC: O(N), SC: O(1)
+# ASCII 문자열이 아닌 이모티콘으로 구분자 선택
+
+class Solution:
+    def encode(self, strs: List[str]) -> str:
+        return '🤍'.join(strs)
+
+    def decode(self, s: str) -> List[str]:
+        return s.split('🤍')
+
+# ASCII 문자열에 포함된 기호로 구분자를 써야할 때
+# -> 글자 수 표시
+class Solution:
+    def encode(self, strs: List[str]) -> str:
+        text = ""
+        for str in strs:
+            text += f"{len(str)}:{str}"
+        return text
+
+    def decode(self, s: str) -> List[str]:
+        ls, start = [], 0
+        while start < len(s):
+            mid = s.find(":", start)
+            length = int(s[start : mid])
+            word = s[mid + 1 : mid + 1 + length]            
+            ls.append(word)
+            start = mid + 1 + length
+        return ls
+    

group-anagrams/hi-rachel.py

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+# TC O(N * K log K), SC O(N * K)
+
+from collections import defaultdict
+
+class Solution:
+    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+        anagram_map = defaultdict(list)
+
+        for word in strs:
+            key = ''.join(sorted(word))
+            anagram_map[key].append(word)
+
+        return list(anagram_map.values())
+    

group-anagrams/hi-rachel.ts

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+// TC O(N * K log K), N: 단어 수, K: 평균 단어 길이, 각 단어를 정렬
+// SC O(N * K)
+
+/**
+ * Array.from() => 이터러블 -> 일반 배열로 변환
+ */
+function groupAnagrams(strs: string[]): string[][] {
+  const anagramMap: Map<string, string[]> = new Map();
+
+  for (const word of strs) {
+    const key = word.split("").sort().join("");
+    const group = anagramMap.get(key) || [];
+    group.push(word);
+    anagramMap.set(key, group);
+  }
+  return Array.from(anagramMap.values());
+}
+
+/**
+ * reduce 풀이
+ */
+// function groupAnagrams(strs: string[]): string[][] {
+//   const anagramMap = strs.reduce((map, word) => {
+//     const key = word.split("").sort().join("");
+//     if (!map.has(key)) {
+//       map.set(key, []);
+//     }
+//     map.get(key).push(word);
+//     return map;
+//   }, new Map<string, string[]>());
+
+//   return Array.from(anagramMap.values());
+// }
+
+/**  간결한 버전
+ * x ||= y (ES2021)
+ * x가 falsy 값이면 y를 할당
+ * JS에서 falsy 값 = false, 0, '', null, undefined, NaN
+ *
+ * Object.values() => 객체의 값들만 배열로 추출할때 사용
+ */
+// function groupAnagrams(strs: string[]): string[][] {
+//     const anagramMap: {[key: string]: string[]} = {};
+
+//     strs.forEach((word) => {
+//         const key = word.split('').sort().join('');
+//         (anagramMap[key] ||= []).push(word);
+//     });
+
+//     return Object.values(anagramMap);
+// }

implement-trie-prefix-tree/hi-rachel.py

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+# Trie Data Structure 구현
+# TC: O(n), SC: O(N)
+# autocomplete and spellchecker 등에 사용
+
+class Node:
+    def __init__(self, ending=False):
+        self.children = {}
+        self.ending = ending
+
+class Trie:
+    def __init__(self):
+        self.root = Node(ending=True)  # 노드 객체 생성
+
+    def insert(self, word: str) -> None:
+        node = self.root  #Trie의 최상위 노드부터 시작 
+        for ch in word:
+            if ch not in node.children:
+                node.children[ch] = Node()  # 새로운 노드 생성
+            node = node.children[ch]
+        node.ending = True
+
+    def search(self, word: str) -> bool:
+        node = self.root
+
+        for ch in word:
+            if ch not in node.children:
+                return False
+            node = node.children[ch]
+        return node.ending
+        
+    def startsWith(self, prefix: str) -> bool:
+        node = self.root
+        
+        for ch in prefix:
+            if ch not in node.children:
+                return False
+            node = node.children[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)

implement-trie-prefix-tree/hi-rachel.ts

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+class TrieNode {
+  children: Map<string, TrieNode>;
+  ending: boolean;
+
+  constructor(ending = false) {
+    this.children = new Map();
+    this.ending = ending;
+  }
+}
+
+class Trie {
+  private root: TrieNode;
+
+  constructor() {
+    this.root = new TrieNode(true);
+  }
+
+  insert(word: string): void {
+    let node = this.root;
+    for (const ch of word) {
+      if (!node.children.has(ch)) {
+        node.children.set(ch, new TrieNode());
+      }
+      node = node.children.get(ch)!; // 노드 포인터를 ch로 이동
+    }
+    node.ending = true;
+  }
+
+  search(word: string): boolean {
+    let node = this.root;
+    for (const ch of word) {
+      if (!node.children.has(ch)) return false;
+      node = node.children.get(ch)!;
+    }
+    return node.ending;
+  }
+
+  startsWith(prefix: string): boolean {
+    let node = this.root;
+    for (const ch of prefix) {
+      if (!node.children.has(ch)) return false;
+      node = node.children.get(ch)!;
+    }
+    return true;
+  }
+}
+
+/**
+ * Your Trie object will be instantiated and called as such:
+ * var obj = new Trie()
+ * obj.insert(word)
+ * var param_2 = obj.search(word)
+ * var param_3 = obj.startsWith(prefix)
+ */

word-break/hi-rachel.ts

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+// wordDict 조합으로 s를 만들 수 있는지 반환해라
+// TC: O(n^2) SC: O(n + m * k) => s의 길이 + (단어 수 * 평균 단어 길이)
+// .has(word) => O(1)
+
+function wordBreak(s: string, wordDict: string[]): boolean {
+  const wordSet = new Set(wordDict);
+  const dp: boolean[] = Array(s.length + 1).fill(false);
+  dp[0] = true;
+
+  for (let i = 1; i <= s.length; i++) {
+    for (let j = 0; j < i; j++) {
+      if (dp[j] && wordSet.has(s.slice(j, i))) {
+        dp[i] = true;
+        break;
+      }
+    }
+  }
+  return dp[s.length];
+}
+
+// PY 풀이
+// class Solution:
+//     def wordBreak(self, s: str, wordDict: List[str]) -> bool:
+//         wordSet = set(wordDict)
+//         dp = [False] * (len(s)+1)
+//         dp[0] = True
+
+//         for i in range(1, len(s)+1):
+//             for j in range(0, i):
+//                 if dp[j] and s[j:i] in wordSet:
+//                     dp[i] = True
+
+//         return dp[len(s)]