diff --git a/alien-dictionary/EGON.py b/alien-dictionary/EGON.py
new file mode 100644
index 000000000..6ca647020
--- /dev/null
+++ b/alien-dictionary/EGON.py
@@ -0,0 +1,106 @@
+from collections import deque
+from typing import List
+from unittest import TestCase, main
+
+
+class Solution:
+    def foreignDictionary(self, words: List[str]) -> str:
+        return self.solve_topological_sort(words)
+
+    """
+    LintCode 로그인이 안되어서 https://neetcode.io/problems/foreign-dictionary 에서 실행시키고 통과만 확인했습니다.
+
+    Runtime: ? ms (Beats ?%)
+    Time Complexity:
+        #0. 복잡도 변수 정의 
+        - words 배열의 길이를 n
+        - words 배열을 이루는 단어들의 평균 길이를 l
+        - words 배열을 이루는 단어를 이루는 문자들의 총 갯수를 c (= n * l)
+        - words 배열을 이루는 단어를 이루는 문자들의 중복 제거 집합의 크기를 s라 하자
+        
+        #1. 초기화
+        - words 배열을 이루는 단어를 이루는 문자들을 조회하며 char_set을 초기화하는데 O(c)
+        - 위상정렬에 사용할 graph 딕셔너리 초기화를 위해 char_set의 크기만큼 조회하므로 O(s)
+        - 마찬가지로 위상정렬에 사용할 rank 딕셔너리 초기화에 O(s)
+        > O(c) + O(s) + O(s) ~= O(c + s)
+        
+        #2. 위상정렬 간선 초기화
+        - words 배열을 조회하는데 O(n - 1)
+            - 단어 간 접두사 관계인 경우, 체크하는 startswith 메서드 사용에 * O(l)
+            - 단어 간 접두사 관계가 아닌 경우, first_char, second_char를 구하는데
+                - zip 생성에 O(l)
+                    - zip 조회에 * O(l)
+        > O(n - 1) * (O(l) + O(l) * O(l)) ~= O(n) * O(l ^ 2) ~= O(c * l) ~= O(c)
+        
+        #3. 위상정렬 실행
+        - dq 초기화에 rank 딕셔너리의 모든 키를 조회하는데 O(s)
+        - dq를 이용해서 graph 딕셔너리의 모든 values를 순회하는데, #2에서 각 first_char, second_char마다 1회 value가 추가되었으므로, 중복이 없는 경우 최대 O(n), upper bound
+        > O(s) + O(n) ~= O(s + n), upper bound
+        
+        #4. 최종 계산
+        > O(c + s) + O(c) + O(s + n) ~= O(c + s) + O(s + n) = O(n * l + s) + O(n + s) ~= O(n * l + s), upper bound
+
+    Memory: ? MB (Beats ?%)
+    Space Complexity: O(s + c)
+        - char_set의 크기에서 O(s)
+        - graph의 keys는 최대 s개이고 values는 최대 c개이므로 O(s + c), upper bound
+        - rank의 keys의 크기에서 O(s)
+        - dq의 최대 크기는 rank의 크기와 같으므로 O(s)
+        > O(s) + O(s + c) + O(s) + O(s) ~= O(s + c)
+    """
+    def solve_topological_sort(self, words: List[str]) -> str:
+        if not words:
+            return ""
+
+        char_set = set([char for word in words for char in word])
+        graph = {char: set() for char in char_set}
+        rank = {char: 0 for char in char_set}
+        for i in range(len(words) - 1):
+            first_word, second_word = words[i], words[i + 1]
+
+            if len(first_word) > len(second_word) and first_word.startswith(second_word):
+                return ""
+
+            first_char, second_char = next(((fc, sc) for fc, sc in zip(first_word, second_word) if fc != sc), ("", ""))
+            if (first_char and second_char) and second_char not in graph[first_char]:
+                graph[first_char].add(second_char)
+                rank[second_char] += 1
+
+        result = []
+        dq = deque([char for char in rank if rank[char] == 0])
+        while dq:
+            curr_char = dq.popleft()
+            result.append(curr_char)
+            for post_char in graph[curr_char]:
+                rank[post_char] -= 1
+                if rank[post_char] == 0:
+                    dq.append(post_char)
+
+        if len(result) != len(rank):
+            return ""
+        else:
+            return "".join(result)
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        words = ["z","o"]
+        output = "zo"
+        solution = Solution()
+        self.assertEqual(solution.foreignDictionary(words), output)
+
+    def test_2(self):
+        words = ["hrn","hrf","er","enn","rfnn"]
+        output = "hernf"
+        solution = Solution()
+        self.assertEqual(solution.foreignDictionary(words), output)
+
+    def test_3(self):
+        words = ["wrt","wrf","er","ett","rftt","te"]
+        output = "wertf"
+        solution = Solution()
+        self.assertEqual(solution.foreignDictionary(words), output)
+
+
+if __name__ == '__main__':
+    main()
diff --git a/longest-palindromic-substring/EGON.py b/longest-palindromic-substring/EGON.py
new file mode 100644
index 000000000..a288e3966
--- /dev/null
+++ b/longest-palindromic-substring/EGON.py
@@ -0,0 +1,126 @@
+from unittest import TestCase, main
+
+
+class Solution:
+    def longestPalindrome(self, s: str) -> str:
+        return self.solve_manacher_algorithm(s)
+
+    """
+    Runtime: 47 ms (Beats 96.97%)
+    Time Complexity: O(n ^ 3)
+        - s의 길이를 n이라 하면, s의 길이 - 1 만큼 조회하는데 O(n - 1)
+            - 각 문자마다 sliding_window를 2회 호출하는데, 각 호출마다 최대 s의 길이만큼 반복하므로, * 2 * O(n), upper bound
+                - 반복 후 s를 slicing하는데 최대 * O(n), upper bound
+        > O(n - 1) * (2 * O(n)) * O(n) ~= O(n ^ 3)
+
+    Memory: 16.54 MB (Beats 88.85%)
+    Space Complexity: O(n)
+        - sliding_window의 결과로 생성되는 문자열의 최대 길이는 n이고, 조회마다 2회 생성되므로 2 * O(n), upper bound
+        > 2 * O(n) ~= O(n)
+    """
+    def solve_sliding_window(self, s: str) -> str:
+
+        def sliding_window(left: int, right: int) -> str:
+            while 0 <= left and right < len(s) and s[left] == s[right - 1]:
+                left -= 1
+                right += 1
+
+            return s[left + 1: right - 1]
+
+        if len(s) < 2 or s == s[::-1]:
+            return s
+
+        result = ''
+        for i in range(len(s) - 1):
+            result = max(result, sliding_window(i, i + 1), sliding_window(i, i + 2), key=len)
+
+        return result
+
+    """
+    Runtime: 36 ms (Beats 98.09%)
+    Time Complexity: O(n ^ 2)
+        - s의 길이를 n이라 하면, s의 길이 - 1 만큼 조회하는데 O(n - 1)
+        - 각 문자마다 two_pointer 2회 호출하는데, 각 호출마다 최대 s의 길이만큼 반복하므로, * 2 * O(n), upper bound
+        > O(n - 1) * (2 * O(n)) ~= O(n ^ 2)
+
+    Memory: 16.85 MB (Beats 24.42%)
+    Space Complexity: O(1)
+        > 모든 변수는 result를 제외하고 인덱스를 위한 정수 변수만 사용하므로 O(1)
+    """
+    def solve_two_pointer(self, s: str) -> str:
+
+        if len(s) < 2 or s == s[::-1]:
+            return s
+
+        def two_pointer(left: int, right: int) -> (int, int):
+            while left >= 0 and right < len(s) and s[left] == s[right]:
+                left -= 1
+                right += 1
+
+            return left + 1, right - 1
+
+        start, end = 0, 0
+        for i in range(len(s) - 1):
+            first_left, first_right = two_pointer(i, i)
+            second_left, second_right = two_pointer(i, i + 1)
+
+            if first_right - first_left > end - start:
+                start, end = first_left, first_right
+            if second_right - second_left > end - start:
+                start, end = second_left, second_right
+
+        return s[start: end + 1]
+
+    """
+    Time Complexity: O(n)
+    Space Complexity: O(n)
+    """
+    def solve_manacher_algorithm(self, s: str) -> str:
+        SEPARATOR = '@'
+        # Step 1: Transform the string
+        t = SEPARATOR + SEPARATOR.join(s) + SEPARATOR
+        n = len(t)
+        p = [0] * n
+        center = right = 0  # Center and right boundary
+        max_length = 0
+        max_center = 0
+
+        # Step 2: Calculate palindrome radius for each character
+        for c in range(n):
+            # Use previously calculated information (symmetry)
+            if c < right:
+                p[c] = min(p[2 * center - c], right - c)
+
+            # Try to expand around i
+            while (0 <= c - p[c] - 1 and c + p[c] + 1 < n) and (t[c - p[c] - 1] == t[c + p[c] + 1]):
+                p[c] += 1
+
+            # Update center and right boundary if expanded beyond current right
+            if c + p[c] > right:
+                center = c
+                right = c + p[c]
+
+            # Update max palindrome length and center
+            if p[c] > max_length:
+                max_length = p[c]
+                max_center = c
+
+        # Step 3: Extract the original string's palindrome
+        start = (max_center - max_length) // 2
+        return s[start:start + max_length]
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        s = "babad"
+        output = "bab"
+        self.assertEqual(Solution().longestPalindrome(s), output)
+
+    def test_2(self):
+        s = "cbbd"
+        output = "bb"
+        self.assertEqual(Solution().longestPalindrome(s), output)
+
+
+if __name__ == '__main__':
+    main()
diff --git a/rotate-image/EGON.py b/rotate-image/EGON.py
new file mode 100644
index 000000000..a81f6faed
--- /dev/null
+++ b/rotate-image/EGON.py
@@ -0,0 +1,47 @@
+from typing import List
+from unittest import TestCase, main
+
+
+class Solution:
+    def rotate(self, matrix: List[List[int]]) -> None:
+        return self.solve(matrix)
+
+    """
+    Runtime: 0 ms (Beats 100.00%)
+    Time Complexity: O(n ^ 2)
+        - 행렬의 행과 열을 교환하기 위해 이중 for문 사용에 O(n ^ 2)
+        - 행렬의 각 행을 뒤집기 위해, 행을 조회하는데 O(n)
+            - 각 행을 뒤집는데 * O(n)
+        > O(n ^ 2) + O(n) * O(n) ~= O(n ^ 2) + O(n ^ 2) ~= O(n ^ 2) 
+
+    Memory: 16.76 MB (Beats 14.84%)
+    Space Complexity: O(1)
+        > in-place 풀이이므로 상수 변수 할당을 제외한 메모리 사용 없음, O(1)
+    """
+    def solve(self, matrix: List[List[int]]) -> None:
+        N = len(matrix)
+
+        for i in range(N):
+            for j in range(i, N):
+                matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]
+
+        for row in matrix:
+            row.reverse()
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        matrix = [[1,2,3],[4,5,6],[7,8,9]]
+        output = [[7,4,1],[8,5,2],[9,6,3]]
+        Solution().rotate(matrix)
+        self.assertEqual(matrix, output)
+
+    def test_2(self):
+        matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]]
+        output = [[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]
+        Solution().rotate(matrix)
+        self.assertEqual(matrix, output)
+
+
+if __name__ == '__main__':
+    main()
diff --git a/subtree-of-another-tree/EGON.py b/subtree-of-another-tree/EGON.py
new file mode 100644
index 000000000..bce47da53
--- /dev/null
+++ b/subtree-of-another-tree/EGON.py
@@ -0,0 +1,116 @@
+from typing import Optional
+from unittest import TestCase, main
+
+
+# Definition for a binary tree node.
+class TreeNode:
+    def __init__(self, val=0, left=None, right=None):
+        self.val = val
+        self.left = left
+        self.right = right
+
+
+class Solution:
+    def isSubtree(self, root: Optional[TreeNode], subRoot: Optional[TreeNode]) -> bool:
+        return self.solve_dfs(root, subRoot)
+
+    """
+    Runtime: 35 ms (Beats 90.24%)
+    Time Complexity: O(n)
+        - root 트리의 크기를 n이라 하면, root 트리의 모든 노드를 조회하는데 O(n)
+            - 각 노드마다 is_same_tree 실행하는데, subRoot 트리의 크기를 m이라 하면, 최대 subRoot의 노드의 크기만큼 조회하므로 * O(m)
+        > O(n) * O(m) ~= O(n * m)
+
+    Memory: 17.09 (Beats 9.93%)
+    Space Complexity: O(n + m)
+        - stack의 최대 크기는 root 트리가 편향된 경우이며, 이는 root 트리의 노드의 총 갯수와 같으므로 O(n), upper bound
+        - is_same_tree 함수의 재귀 스택의 최대 깊이는 subRoot 트리가 편향된 경우이며, 이는 subRoot 트리의 노드의 총 갯수와 같으므로 O(m), upper bound
+        > O(n) + O(m) ~= O(n + m)
+    """
+    def solve_dfs(self, root: Optional[TreeNode], subRoot: Optional[TreeNode]) -> bool:
+
+        def is_same_tree(p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
+            if p is None and q is None:
+                return True
+            elif (p is not None and q is not None) and (p.val == q.val):
+                return is_same_tree(p.left, q.left) and is_same_tree(p.right, q.right)
+            else:
+                return False
+
+        result = False
+        stack = [root]
+        while stack:
+            curr = stack.pop()
+            if (curr and subRoot) and curr.val == subRoot.val:
+                result = result or is_same_tree(curr, subRoot)
+
+            if curr.left:
+                stack.append(curr.left)
+
+            if curr.right:
+                stack.append(curr.right)
+
+        return result
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        root = TreeNode(3)
+        root_1 = TreeNode(4)
+        root_2 = TreeNode(5)
+        root.left = root_1
+        root.right = root_2
+        root_3 = TreeNode(1)
+        root_4 = TreeNode(2)
+        root.left.left = root_3
+        root.left.right = root_4
+
+        subRoot = TreeNode(4)
+        sub_1 = TreeNode(1)
+        sub_2 = TreeNode(2)
+        subRoot.left = sub_1
+        subRoot.right = sub_2
+
+        output = True
+        self.assertEqual(Solution.isSubtree(Solution(), root, subRoot), output)
+
+    def test_2(self):
+        root = TreeNode(3)
+        root_1 = TreeNode(4)
+        root_2 = TreeNode(5)
+        root.left = root_1
+        root.right = root_2
+        root_3 = TreeNode(1)
+        root_4 = TreeNode(2)
+        root.left.left = root_3
+        root.left.right = root_4
+        root_5 = TreeNode(0)
+        root_4.left = root_5
+
+        subRoot = TreeNode(4)
+        sub_1 = TreeNode(1)
+        sub_2 = TreeNode(2)
+        subRoot.left = sub_1
+        subRoot.right = sub_2
+
+        output = False
+        self.assertEqual(Solution.isSubtree(Solution(), root, subRoot), output)
+
+    def test_3(self):
+        root = TreeNode(1)
+        root.right = TreeNode(1)
+        root.right.right = TreeNode(1)
+        root.right.right.right = TreeNode(1)
+        root.right.right.right.right = TreeNode(1)
+        root.right.right.right.right.right = TreeNode(2)
+
+        subRoot = TreeNode(1)
+        subRoot.right = TreeNode(1)
+        subRoot.right.right = TreeNode(2)
+
+        output = True
+        self.assertEqual(Solution.isSubtree(Solution(), root, subRoot), output)
+
+
+if __name__ == '__main__':
+    main()
diff --git a/validate-binary-search-tree/EGON.py b/validate-binary-search-tree/EGON.py
new file mode 100644
index 000000000..1d646a47b
--- /dev/null
+++ b/validate-binary-search-tree/EGON.py
@@ -0,0 +1,63 @@
+from typing import Optional
+from unittest import TestCase, main
+
+
+# Definition for a binary tree node.
+class TreeNode:
+    def __init__(self, val=0, left=None, right=None):
+        self.val = val
+        self.left = left
+        self.right = right
+
+
+class Solution:
+    def isValidBST(self, root: Optional[TreeNode]) -> bool:
+        return self.solve_dfs(root)
+
+    """
+    Runtime: 0 ms (Beats 100.00%)
+    Time Complexity: O(n)
+        > root 트리의 노드의 갯수를 n이라 하면, 모든 node를 조회하므로 O(n)
+
+    Memory: 18.91 MB (Beats 5.70%)
+    Space Complexity: O(n)
+        > root 트리의 높이만큼 is_valid_bst 함수가 재귀 호출 스택에 쌓이나, 문제의 제약조건에서 이진트리라고만 했으므로 편향될 수 있으므로, O(n), upper bound
+    """
+    def solve_dfs(self, root: Optional[TreeNode]) -> bool:
+
+        def is_valid_bst(node: Optional[TreeNode], lower_bound=float('-inf'), upper_bound=float('inf')):
+            if not node:
+                return True
+
+            if lower_bound < node.val < upper_bound:
+                return is_valid_bst(node.left, lower_bound, node.val) and is_valid_bst(node.right, node.val, upper_bound)
+            else:
+                return False
+
+        return is_valid_bst(root)
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        root = TreeNode(2)
+        root.left = TreeNode(1)
+        root.right = TreeNode(3)
+
+        output = True
+
+        self.assertEqual(Solution().isValidBST(root), output)
+
+    def test_2(self):
+        root = TreeNode(5)
+        root.left = TreeNode(4)
+        root.right = TreeNode(6)
+        root.right.left = TreeNode(3)
+        root.right.right = TreeNode(7)
+
+        output = False
+
+        self.assertEqual(Solution().isValidBST(root), output)
+
+
+if __name__ == '__main__':
+    main()