[강희찬] WEEK 12 Solution

merge-intervals/HC-kang.ts

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+/**
+ * https://leetcode.com/problems/merge-intervals
+ * T.C. O(n logn)
+ * S.C. O(n)
+ */
+function merge(intervals: number[][]): number[][] {
+  intervals.sort((a, b) => a[0] - b[0]); // T.C. O(n logn)
+
+  const result = [intervals[0]]; // S.C. O(n)
+
+  // T.C. O(n)
+  for (let i = 1; i < intervals.length; i++) {
+    const last = result[result.length - 1];
+    const current = intervals[i];
+
+    if (last[1] >= current[0]) {
+      last[1] = Math.max(last[1], current[1]);
+    } else {
+      result.push(current);
+    }
+  }
+
+  return result;
+}

number-of-connected-components-in-an-undirected-graph/HC-kang.ts

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+/**
+ * https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph
+ * T.C. O(V + E)
+ * S.C. O(V + E)
+ */
+function countComponents(n: number, edges: number[][]): number {
+  const graph = new Map<number, number[]>();
+
+  for (let i = 0; i < n; i++) {
+    graph.set(i, []);
+  }
+
+  for (const [u, v] of edges) {
+    graph.get(u)!.push(v);
+    graph.get(v)!.push(u);
+  }
+
+  function dfs(node: number) {
+    visited.add(node);
+
+    for (const neighbor of graph.get(node)!) {
+      if (!visited.has(neighbor)) {
+        dfs(neighbor);
+      }
+    }
+  }
+
+  const visited = new Set<number>();
+
+  let components = 0;
+
+  for (let i = 0; i < n; i++) {
+    if (!visited.has(i)) {
+      components++;
+      dfs(i);
+    }
+  }
+
+  return components;
+}
+
+/**
+ * Using Union Find
+ * T.C. O(V + E)
+ *
+ */
+function countComponents(n: number, edges: number[][]): number {
+  const parent = Array.from({ length: n }, (_, i) => i);
+
+  // find and compress path
+  function find(x: number): number {
+    if (parent[x] !== x) {
+      parent[x] = find(parent[x]);
+    }
+    return parent[x];
+  }
+
+  // union two sets and check if they have the same root
+  function union(x: number, y: number): boolean {
+    const rootX = find(x);
+    const rootY = find(y);
+    parent[rootX] = rootY;
+    return true;
+  }
+
+  for (const [x, y] of edges) {
+    union(x, y);
+  }
+
+  return new Set(parent.map(find)).size;
+}

remove-nth-node-from-end-of-list/HC-kang.ts

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+class ListNode {
+  val: number;
+  next: ListNode | null;
+  constructor(val?: number, next?: ListNode | null) {
+    this.val = val === undefined ? 0 : val;
+    this.next = next === undefined ? null : next;
+  }
+}
+
+/**
+ * https://leetcode.com/problems/remove-nth-node-from-end-of-list
+ * T.C. O(N)
+ * S.C. O(1)
+ */
+function removeNthFromEnd(head: ListNode | null, n: number): ListNode | null {
+  let first = head;
+  let second = head;
+
+  for (let i = 0; i < n; i++)
+    first = first!.next;
+
+  if (!first)
+    return head!.next;
+
+  while (first.next) {
+    first = first.next;
+    second = second!.next;
+  }
+
+  second!.next = second!.next!.next;
+
+  return head;
+}

same-tree/HC-kang.ts

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+class TreeNode {
+  val: number;
+  left: TreeNode | null;
+  right: TreeNode | null;
+  constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
+    this.val = val === undefined ? 0 : val;
+    this.left = left === undefined ? null : left;
+    this.right = right === undefined ? null : right;
+  }
+}
+
+/**
+ * https://leetcode.com/problems/same-tree/description/
+ * T.C. O(n)
+ * S.C. O(n)
+ */
+function isSameTree(p: TreeNode | null, q: TreeNode | null): boolean {
+  if (p === null && q === null) return true;
+  if (p === null || q === null) return false;
+  if (p.val !== q.val) return false;
+  return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
+}
+
+/**
+ * T.C. O(n)
+ * S.C. O(n)
+ */
+function isSameTree(p: TreeNode | null, q: TreeNode | null): boolean {
+  const stack: [TreeNode | null, TreeNode | null][] = [[p, q]];
+
+  while (stack.length) {
+    const [node1, node2] = stack.pop()!;
+    if (node1 === null && node2 === null) continue;
+    if (node1 === null || node2 === null) return false;
+    if (node1.val !== node2.val) return false;
+    stack.push([node1.left, node2.left]);
+    stack.push([node1.right, node2.right]);
+  }
+
+  return true;
+}

serialize-and-deserialize-binary-tree/HC-kang.ts

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+class TreeNode {
+  val: number;
+  left: TreeNode | null;
+  right: TreeNode | null;
+  constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
+    this.val = val === undefined ? 0 : val;
+    this.left = left === undefined ? null : left;
+    this.right = right === undefined ? null : right;
+  }
+}
+
+/**
+ * https://leetcode.com/problems/serialize-and-deserialize-binary-tree
+ * T.C. O(n)
+ * S.C. O(n)
+ */
+function serialize(root: TreeNode | null): string {
+  return JSON.stringify(root);
+}
+
+function deserialize(data: string): TreeNode | null {
+  return JSON.parse(data);
+}
+
+/**
+ * Recursive
+ * T.C. O(n)
+ * S.C. O(n)
+ */
+function serialize(root: TreeNode | null): string {
+  return root
+    ? `${root.val},${serialize(root.left)},${serialize(root.right)}`
+    : 'null';
+}
+
+function deserialize(data: string): TreeNode | null {
+  const values = data.split(',');
+  let index = 0;
+
+  function dfs(): TreeNode | null {
+    const val = values[index++];
+    if (val === 'null') return null;
+
+    const node = new TreeNode(parseInt(val));
+    node.left = dfs();
+    node.right = dfs();
+    return node;
+  }
+
+  return dfs();
+}