diff --git a/alien-dictionary/HoonDongKang.ts b/alien-dictionary/HoonDongKang.ts
new file mode 100644
index 000000000..b7178ebdf
--- /dev/null
+++ b/alien-dictionary/HoonDongKang.ts
@@ -0,0 +1,75 @@
+/**
+ * [Problem]: [892] Alien Dictionary
+ * (https://www.lintcode.com/problem/892/)
+ */
+
+//시간복잡도 O(n+e)
+//공간복잡도 O(n+e)
+export class Solution {
+    /**
+     * @param words: a list of words
+     * @return: a string which is correct order
+     */
+    alienOrder(words: string[]): string {
+        // Write your code here
+        const graph: Map<string, Set<string>> = new Map();
+
+        // 모든 문자를 그래프 노드로 초기화
+        for (const word of words) {
+            for (const char of word) {
+                if (!graph.has(char)) {
+                    graph.set(char, new Set());
+                }
+            }
+        }
+
+        // 문자 간 선후관계(edge) 설정
+        for (let i = 1; i < words.length; i++) {
+            const prev = words[i - 1];
+            const curr = words[i];
+            let isDiff = false;
+
+            for (let j = 0; j < Math.min(prev.length, curr.length); j++) {
+                const p = prev[j];
+                const c = curr[j];
+
+                if (p !== c) {
+                    graph.get(p)!.add(c);
+                    isDiff = true;
+                    break;
+                }
+            }
+
+            if (!isDiff && prev.length > curr.length) {
+                return "";
+            }
+        }
+
+        const visited: Set<string> = new Set();
+        const visiting: Set<string> = new Set();
+        const output: string[] = [];
+
+        function dfs(node: string): boolean {
+            if (visited.has(node)) return true;
+            if (visiting.has(node)) return false;
+
+            visiting.add(node);
+
+            for (const neighbor of graph.get(node)!) {
+                if (!dfs(neighbor)) return false;
+            }
+
+            visiting.delete(node);
+            visited.add(node);
+            output.push(node);
+
+            return true;
+        }
+
+        for (const node of graph.keys()) {
+            if (!dfs(node)) return "";
+        }
+
+        return output.reverse().join("");
+    }
+}
diff --git a/construct-binary-tree-from-preorder-and-inorder-traversal/HoonDongKang.ts b/construct-binary-tree-from-preorder-and-inorder-traversal/HoonDongKang.ts
new file mode 100644
index 000000000..572c965e2
--- /dev/null
+++ b/construct-binary-tree-from-preorder-and-inorder-traversal/HoonDongKang.ts
@@ -0,0 +1,41 @@
+/**
+ * [Problem]: [105] Construct Binary Tree From Preorder And Inorder Traversal
+ * (https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/description/)
+ */
+
+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;
+    }
+}
+
+//시간복잡도 O(n)
+//공간복잡도 O(n)
+function buildTree(preorder: number[], inorder: number[]): TreeNode | null {
+    if (!preorder.length || !inorder.length) return null;
+
+    const map = new Map<number, number>();
+    let idx = 0;
+
+    inorder.forEach((v, i) => map.set(v, i));
+
+    function dfs(left: number, right: number): TreeNode | null {
+        if (left > right) return null;
+
+        const value = preorder[idx++];
+        const root = new TreeNode(value);
+        const index = map.get(value)!;
+
+        root.left = dfs(left, index - 1);
+        root.right = dfs(index + 1, right);
+
+        return root;
+    }
+
+    return dfs(0, inorder.length - 1);
+}
diff --git a/longest-palindromic-substring/HoonDongKang.ts b/longest-palindromic-substring/HoonDongKang.ts
new file mode 100644
index 000000000..98ecf666e
--- /dev/null
+++ b/longest-palindromic-substring/HoonDongKang.ts
@@ -0,0 +1,31 @@
+/**
+ * [Problem]: [5] Longest Palindromic Substring
+ * (https://leetcode.com/problems/longest-palindromic-substring/)
+ */
+// 시간복잡도 O(n^2)
+// 공간복잡도 O(1)
+function longestPalindrome(s: string): string {
+    if (s.length < 2) return s;
+
+    let start = 0;
+    let end = 0;
+
+    function findPalindrome(left: number, right: number): void {
+        while (0 <= left && right < s.length && s[left] === s[right]) {
+            left--;
+            right++;
+        }
+
+        if (right - left - 1 > end - start) {
+            start = left + 1;
+            end = right - 1;
+        }
+    }
+
+    for (let i = 0; i < s.length; i++) {
+        findPalindrome(i, i);
+        findPalindrome(i, i + 1);
+    }
+
+    return s.substring(start, end + 1);
+}
diff --git a/rotate-image/HoonDongKang.ts b/rotate-image/HoonDongKang.ts
new file mode 100644
index 000000000..ee0113941
--- /dev/null
+++ b/rotate-image/HoonDongKang.ts
@@ -0,0 +1,22 @@
+/**
+ * [Problem]: [48] Rotate Image
+ * (https://leetcode.com/problems/rotate-image/)
+ */
+/**
+ Do not return anything, modify matrix in-place instead.
+ */
+// 시간복잡도 O(n^2)
+// 공간복잡도 O(1)
+function rotate(matrix: number[][]): void {
+    const length = matrix.length;
+
+    for (let i = 0; i < length; i++) {
+        for (let j = i + 1; j < length; j++) {
+            [matrix[i][j], matrix[j][i]] = [matrix[j][i], matrix[i][j]];
+        }
+    }
+
+    for (let i = 0; i < length; i++) {
+        matrix[i].reverse();
+    }
+}
diff --git a/subtree-of-another-tree/HoonDongKang.ts b/subtree-of-another-tree/HoonDongKang.ts
new file mode 100644
index 000000000..6adb2b194
--- /dev/null
+++ b/subtree-of-another-tree/HoonDongKang.ts
@@ -0,0 +1,26 @@
+/**
+ * [Problem]: [572] Subtree of Another Tree
+ * (https://leetcode.com/problems/subtree-of-another-tree/description/)
+ */
+
+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;
+    }
+}
+
+//시간복잡도 O(n*m)
+//공간복잡도 O(n+m)
+function isSubtree(root: TreeNode | null, subRoot: TreeNode | null): boolean {
+    const pre = (node: TreeNode | null) => {
+        if (!node) return "N";
+        return `(${node.val},${pre(node.left)},${pre(node.right)})`;
+    };
+
+    return pre(root).includes(pre(subRoot));
+}