[강희찬] Week1 문제 풀이

contains-duplicate/HC-kang.ts

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+/**
+217. Contains Duplicate
+
+Example 1:
+Input: nums = [1,2,3,1]
+Output: true
+
+Example 2:
+Input: nums = [1,2,3,4]
+Output: false
+
+Example 3:
+Input: nums = [1,1,1,3,3,4,3,2,4,2]
+Output: true
+ */
+
+// Time complexity: O(n)
+// Space complexity: O(n)
+function containsDuplicate(nums: number[]): boolean {
+  return nums.length !== new Set(nums).size;
+}

kth-smallest-element-in-a-bst/HC-kang.ts

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+/*
+230. Kth Smallest Element in a BST
+
+Example 1:
+    3
+   / \
+  1   4
+   \
+    2
+Input: root = [3,1,4,null,2], k = 1
+Output: 1
+
+Example 2:
+        5
+       / \
+      3   6
+     / \
+    2   4
+   /
+  1
+Input: root = [5,3,6,2,4,null,null,1], k = 3
+Output: 3
+ */
+
+// 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)
+//     }
+// }
+
+// Time complexity: O(n)
+// Space complexity: O(n)
+function kthSmallest(root: TreeNode | null, k: number): number {
+  function inorder(root: TreeNode | null, arr: number[]) {
+    if (!root) return;
+
+    inorder(root.left, arr);
+    arr.push(root.val);
+    inorder(root.right, arr);
+  }
+
+  const arr: number[] = [];
+  inorder(root, arr);
+  return arr[k - 1];
+}
+
+/**
+ * NOT MY SOLUTION - NO NEED TO REVIEW
+ * Awesome solution from leetcode
+ */
+function kthSmallest(root: TreeNode | null, k: number): number {
+  /*
+  define. return the kth smallest value in a BST.
+  assess. smallest value can be 0. At least k nodes, which can be at its smallest, 1.
+  approach. DFS with backtracking. Traverse down the left edges until we hit null. if k is 1, return that value. Else, backtrack, go right, then try to go left again. Use a stack.
+  */
+  const stack = [];
+
+  let currentRank = 0;
+
+  let currentNode = root;
+
+  while (currentNode || stack.length > 0) {
+
+      while (currentNode) {
+          stack.push(currentNode);
+
+          currentNode = currentNode.left;
+      }
+
+      currentNode = stack.pop();
+
+      currentRank++;
+
+      if (currentRank === k) return currentNode.val;
+
+      currentNode = currentNode.right;
+  }
+
+};

number-of-1-bits/HC-kang.ts

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+/**
+191. Number of 1 Bits
+
+Example 1:
+Input: n = 11
+Output: 3
+Explanation:
+The input binary string 1011 has a total of three set bits.
+
+Example 2:
+Input: n = 128
+Output: 1
+Explanation:
+The input binary string 10000000 has a total of one set bit.
+
+Example 3:
+Input: n = 2147483645
+Output: 30
+Explanation:
+The input binary string 1111111111111111111111111111101 has a total of thirty set bits.
+ */
+
+function hammingWeight(n: number): number {
+  // Time complexity: O(logn)
+  // Space complexity: O(logn)
+  // it has a better readability and not so bad in space complexity
+  return n.toString(2).split('1').length - 1;
+
+  // Time complexity: O(logn)
+  // Space complexity: O(1)
+  // it's better in space complexity, but sometimes the bitwise operation is not easy to understand
+  let count = 0;
+  while (n !== 0) {
+    count += n & 1;
+    n >>>= 1;
+  }
+  return count;
+}

palindromic-substrings/HC-kang.ts

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+/*
+647. Palindromic Substrings
+
+Example 1:
+Input: s = "abc"
+Output: 3
+Explanation: Three palindromic strings: "a", "b", "c".
+
+Example 2:
+Input: s = "aaa"
+Output: 6
+Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".
+ */
+
+// Time complexity: O(n^3)
+// Space complexity: O(n^3)
+function countSubstrings(s: string): number {
+  function isPalindrome(s: string): boolean {
+    if (s.length === 0) return false;
+    if (s.length === 1) return true;
+    let left = 0;
+    let right = s.length - 1;
+    while (left < right) {
+      if (s[left] !== s[right]) return false;
+      left++;
+      right--;
+    }
+    return true;
+  }
+
+  const candidates = new Map<string, number>();
+  for (let i = 0; i < s.length; i++) {
+    for (let j = i + 1; j <= s.length; j++) {
+      // this will cost both t.c. and s.c. O(n^3)
+      candidates.set(s.slice(i, j), (candidates.get(s.slice(i, j)) || 0) + 1);
+    }
+  }
+
+  let count = 0;
+  for (const [key, value] of candidates) {
+    if (isPalindrome(key)) count += value
+  }
+
+  return count;
+};
+
+// Time complexity: O(n^3)
+// Space complexity: O(1)
+function countSubstrings(s: string): number {
+  function isPalindrome(s: string): boolean {
+    if (s.length === 0) return false;
+    if (s.length === 1) return true;
+    let left = 0;
+    let right = s.length - 1;
+    while (left < right) {
+      if (s[left] !== s[right]) return false;
+      left++;
+      right--;
+    }
+    return true;
+  }
+
+  let count = 0;
+  for (let i = 0; i < s.length; i++) {
+    for (let j = i + 1; j <= s.length; j++) {
+      // this will cause t.c. O(n^3). need to optimize this part
+      if (isPalindrome(s.slice(i, j))) count++;
+    }
+  }
+
+  return count;
+}
+
+// Time complexity: O(n^2)
+// Space complexity: O(1)
+function countSubstrings(s: string): number {
+  function expandIsPalindrome(left: number, right: number): number {
+    let count = 0;
+    while (left >= 0 && right < s.length && s[left] === s[right]) {
+      count++;
+      left--;
+      right++;
+    }
+    return count;
+  }
+
+  let count = 0;
+  for (let i = 0; i < s.length; i++) {
+    count += expandIsPalindrome(i, i);
+    count += expandIsPalindrome(i, i + 1);
+  }
+  return count;
+}

top-k-frequent-elements/HC-kang.ts

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+/*
+347. Top K Frequent Elements
+
+Example 1:
+Input: nums = [1,1,1,2,2,3], k = 2
+Output: [1,2]
+
+Example 2:
+Input: nums = [1], k = 1
+Output: [1]
+ */
+
+// Time complexity: O(nlogn)
+// Space complexity: O(n)
+function topKFrequent(nums: number[], k: number): number[] {
+  const frequentMap = new Map<number, number>();
+  for (const num of nums) { // s.c. O(n)
+    frequentMap.set(num, (frequentMap.get(num) || 0) + 1);
+  }
+  return Array.from(frequentMap.entries())
+    .sort((a, b) => b[1] - a[1]) // this will cost t.c. O(nlogn)
+    .slice(0, k)
+    .map((v) => v[0]);
+}