Week 06 solution commit

Author: juhui-jeong

container-with-most-water/juhui-jeong.ts

@@ -0,0 +1,24 @@
+/*
+시간복잡도: O(n)
+공간복잡도: O(1)
+포인터까지는 생각했으나, 그이후로는 해결법이 떠오르지 않아, 알고달레 풀이를 보고 해결.
+
+*/
+function maxArea(height: number[]): number {
+  let maxArea = 0;
+  let left = 0;
+  let right = height.length - 1;
+
+  while (left < right) {
+    let area = (right - left) * Math.min(height[left], height[right]);
+
+    if (maxArea < area) maxArea = area;
+
+    if (height[left] > height[right]) {
+      right -= 1;
+    } else {
+      left += 1;
+    }
+  }
+  return maxArea;
+}

design-add-and-search-words-data-structure/juhui-jeong.ts

@@ -0,0 +1,61 @@
+/*
+시간복잡도: O(L)
+공간복잡도: O(L)
+Trie는 검색을 빠르게 해주지만,
+. 와일드카드는 DFS 분기를 유발해 최악의 경우 시간 복잡도가 커질 수 있다.
+
+DFS가 필요한 검색 문제에 대한 공부 필요.
+*/
+class TrieNode {
+  children: Map<string, TrieNode>;
+  isEnd: boolean;
+
+  constructor() {
+    this.children = new Map();
+    this.isEnd = false;
+  }
+}
+
+class WordDictionary {
+  private root: TrieNode;
+
+  constructor() {
+    this.root = new TrieNode();
+  }
+
+  addWord(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)!;
+    }
+    node.isEnd = true;
+  }
+
+  search(word: string): boolean {
+    const dfs = (index: number, node: TrieNode): boolean => {
+      if (index === word.length) {
+        return node.isEnd;
+      }
+
+      const ch = word[index];
+
+      if (ch === '.') {
+        for (const child of node.children.values()) {
+          if (dfs(index + 1, child)) return true;
+        }
+        return false;
+      }
+
+      const next = node.children.get(ch);
+      if (!next) return false;
+
+      return dfs(index + 1, next);
+    };
+
+    return dfs(0, this.root);
+  }
+}

valid-parentheses/juhui-jeong.ts

@@ -0,0 +1,35 @@
+/*
+시간복잡도: O(n)
+공간복잡도: O(n)
+*/
+function isValid(s: string): boolean {
+  if (s.length % 2 !== 0) {
+    return false;
+  }
+
+  const stack: string[] = [];
+  const pair: Map<string, string> = new Map<string, string>([
+    [')', '('],
+    ['}', '{'],
+    [']', '['],
+  ]);
+
+  for (const char of s) {
+    // close brackts인지 확인
+    if (pair.has(char)) {
+      let topItem = stack.pop();
+      // pair 확인
+      if (topItem !== pair.get(char)) {
+        return false;
+      }
+    } else {
+      stack.push(char);
+    }
+  }
+
+  if (stack.length === 0) {
+    return true;
+  } else {
+    return false;
+  }
+}