Merge pull request DaleStudy#356 from HC-kang/main

[강희찬] WEEK 2 Solution

Author: HC-kang

construct-binary-tree-from-preorder-and-inorder-traversal/HC-kang.ts

@@ -0,0 +1,53 @@
+// Definition for a binary tree node.
+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;
+  }
+}
+
+// T.C: O(N)
+// S.C: O(N^2) - Slice makes n-1, n-2, ..., 1 for n times. So, it's O(N^2).
+function buildTree(preorder: number[], inorder: number[]): TreeNode | null {
+  if (preorder.length === 0 || inorder.length === 0) {
+    return null;
+  }
+  const root = new TreeNode(preorder[0]);
+  const idx = inorder.indexOf(preorder[0]);
+  root.left = buildTree(preorder.slice(1, idx + 1), inorder.slice(0, idx));
+  root.right = buildTree(preorder.slice(idx + 1), inorder.slice(idx + 1));
+
+  return root;
+}
+
+// Not using slice. but I think it's not necessary... first solution is more readable. and that's not so bad.
+// T.C: O(N)
+// S.C: O(N)
+function buildTree(preorder: number[], inorder: number[]): TreeNode | null {
+  // this tree is consist of unique values
+  const inorderMap = new Map<number, number>();
+  for (const [i, val] of inorder.entries()) {
+    inorderMap.set(val, i);
+  }
+
+  function helper(preLeft: number, preRight: number, inLeft: number, inRight: number): TreeNode | null {
+    if (preLeft > preRight) return null;
+
+    const rootValue = preorder[preLeft];
+    const root = new TreeNode(rootValue);
+    const inRootIdx = inorderMap.get(rootValue)!;
+
+    const leftSize = inRootIdx - inLeft;
+
+    root.left = helper(preLeft + 1, preLeft + leftSize, inLeft, inRootIdx - 1);
+    root.right = helper(preLeft + leftSize + 1, preRight, inRootIdx + 1, inRight);
+
+    return root;
+  }
+
+  return helper(0, preorder.length - 1, 0, inorder.length - 1);
+}

counting-bits/HC-kang.ts

@@ -0,0 +1,26 @@
+// T.C: O(n)
+// S.C: O(n)
+function countBits(n: number): number[] {
+  // T.C: O(1)
+  // S.C: O(1)
+  function countBit(num: number): number {
+    num = num - ((num >>> 1) & 0x55555555);
+    num = (num & 0x33333333) + ((num >>> 2) & 0x33333333);
+    num = (num + (num >>> 4)) & 0x0f0f0f0f;
+    num = num + (num >>> 8);
+    num = num + (num >>> 16);
+    return num & 0x3f;
+  }
+
+  return new Array(n + 1).fill(0).map((_, i) => countBit(i));
+}
+
+// T.C: O(n)
+// S.C: O(n)
+function countBits(n: number): number[] {
+  const dp = new Array(n + 1).fill(0);
+  for (let i = 1; i <= n; i++) {
+    dp[i] = dp[i >> 1] + (i & 1);
+  }
+  return dp;
+}

decode-ways/HC-kang.ts

@@ -0,0 +1,31 @@
+// T.C: O(n)
+// S.C: O(n)
+function numDecodings(s: string): number {
+  const NUM_OF_ALPHA = 26;
+  const memo = new Map<number, number>();
+
+  function dfs(idx: number): number {
+    if (idx === s.length) {
+      return 1;
+    }
+    if (s[idx] === '0') {
+      return 0;
+    }
+    if (memo.has(idx)) {
+      return memo.get(idx)!;
+    }
+
+    let count = dfs(idx + 1);
+    if (
+      idx + 2 <= s.length && // check if idx + 2 is in the range
+      parseInt(s.slice(idx, idx + 2), 10) <= NUM_OF_ALPHA
+    ) {
+      count += dfs(idx + 2);
+    }
+
+    memo.set(idx, count);
+    return count;
+  }
+
+  return dfs(0);
+}

encode-and-decode-strings/HC-kang.ts

@@ -0,0 +1,29 @@
+/**
+ * The most simple way
+ */
+function encode(strs: string[]): string {
+	return strs.join('🏖️');
+};
+function decode(s: string): string[] {
+	return s.split('🏖️');
+};
+
+// T.C: O(n)
+// S.C: O(n)
+function encode(strs: string[]): string {
+  return strs.map((s) => s.length + '#' + s).join('');
+}
+
+// T.C: O(n)
+// S.C: O(n)
+function decode(s: string): string[] {
+  const res: string[] = [];
+  let curIdx = 0;
+  while (curIdx < s.length) {
+    const sepIdx = s.indexOf('#', curIdx);
+    const len = parseInt(s.slice(curIdx, sepIdx), 10);
+    res.push(s.slice(sepIdx + 1, sepIdx + 1 + len));
+    curIdx = sepIdx + 1 + len;
+  }
+  return res;
+}

valid-anagram/HC-kang.ts

@@ -0,0 +1,16 @@
+// T.C: O(n)
+// S.C: O(1)
+function isAnagram(s: string, t: string): boolean {
+  if (s.length !== t.length) return false;
+
+  const NUM_OF_ALPHA = 26;
+  const A_CODE = 'a'.charCodeAt(0);
+  const bucket = new Array(NUM_OF_ALPHA).fill(0); // S.C: O(1)
+
+  for (let i = 0; i < s.length; i++) { // T.C: O(n)
+    bucket[s.charCodeAt(i) - A_CODE]++;
+    bucket[t.charCodeAt(i) - A_CODE]--;
+  }
+
+  return bucket.every(count => count === 0);
+}