[byol-han] WEEK 09 solutions

clone-graph/byol-han.js

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+/**
+ * https://leetcode.com/problems/clone-graph/
+ * // Definition for a _Node.
+ * function _Node(val, neighbors) {
+ *    this.val = val === undefined ? 0 : val;
+ *    this.neighbors = neighbors === undefined ? [] : neighbors;
+ * };
+ * 시간 복잡도: O(N) — 노드 수만큼 순회
+ * 공간 복잡도: O(N) — visited 맵과 재귀 호출 스택
+ */
+
+/**
+ * @param {_Node} node
+ * @return {_Node}
+ */
+var cloneGraph = function (node) {
+  if (!node) return null;
+
+  const visited = new Map();
+
+  const dfs = (n) => {
+    if (visited.has(n)) {
+      return visited.get(n);
+    }
+
+    const clone = new Node(n.val);
+    visited.set(n, clone);
+
+    for (let neighbor of n.neighbors) {
+      clone.neighbors.push(dfs(neighbor));
+    }
+
+    return clone;
+  };
+
+  return dfs(node);
+};

longest-common-subsequence/byol-han.js

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+/**
+ * https://leetcode.com/problems/longest-common-subsequence/submissions/1644426037/
+ * @param {string} text1
+ * @param {string} text2
+ * @return {number}
+ */
+var longestCommonSubsequence = function (text1, text2) {
+  const m = text1.length;
+  const n = text2.length;
+
+  // Create 2D array initialized with 0
+  const dp = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
+
+  // Fill the dp table
+  for (let i = 1; i <= m; i++) {
+    for (let j = 1; j <= n; j++) {
+      if (text1[i - 1] === text2[j - 1]) {
+        // Characters match
+        dp[i][j] = dp[i - 1][j - 1] + 1;
+      } else {
+        // No match, take the max from left or top cell
+        dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
+      }
+    }
+  }
+
+  // The length of the longest common subsequence
+  return dp[m][n];
+};

palindromic-substrings/byol-han.js

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+/**
+ * https://leetcode.com/problems/palindromic-substrings/submissions/1644425061/
+ * @param {string} s
+ * @return {number}
+ */
+var countSubstrings = function (s) {
+  let count = 0;
+
+  // Helper function to expand around the center
+  function expandAroundCenter(left, right) {
+    while (left >= 0 && right < s.length && s[left] === s[right]) {
+      count++; // Found a palindrome
+      left--;
+      right++;
+    }
+  }
+
+  for (let i = 0; i < s.length; i++) {
+    expandAroundCenter(i, i); // Odd-length palindromes
+    expandAroundCenter(i, i + 1); // Even-length palindromes
+  }
+
+  return count;
+};