From 095994abc46a0822ebce336d30ea513207d12600 Mon Sep 17 00:00:00 2001
From: Dusuna <94776135+dusunax@users.noreply.github.com>
Date: Tue, 18 Mar 2025 22:04:22 +0900
Subject: [PATCH 1/4] add solution: subtree-of-another-tree

---
 subtree-of-another-tree/dusunax.py | 62 ++++++++++++++++++++++++++++++
 1 file changed, 62 insertions(+)
 create mode 100644 subtree-of-another-tree/dusunax.py

diff --git a/subtree-of-another-tree/dusunax.py b/subtree-of-another-tree/dusunax.py
new file mode 100644
index 000000000..4f7850994
--- /dev/null
+++ b/subtree-of-another-tree/dusunax.py
@@ -0,0 +1,62 @@
+'''
+# 572. Subtree of Another Tree
+
+## 1. Recursive
+use a recursive function to check if the two trees are identical.
+- if they are identical, return True.
+- if they are not identical, recursively check the left subtree and right subtree.
+
+### base case
+- isSubtree
+  - if the current tree is None, return False.
+  - if the subRoot is None, return True. (empty tree is a subtree of any tree)
+- isIdentical
+  - if both trees are None, return True. (they are identical)
+  - if one of the trees is None, return False.
+  - if the values of the current nodes are different, return False.
+  - left subtree and right subtree's results are must be True.
+
+## 2. Snapshot
+> reference: https://www.algodale.com/problems/subtree-of-another-tree/
+
+use a snapshot string of the preorder traversal to check if the subRoot is a subtree of the current tree.
+- if the subRoot is a subtree of the current tree, return True.
+- if the subRoot is not a subtree of the current tree, return False.
+'''
+class Solution:
+    '''
+    1. Recursive
+    TC: O(n * m), m is the number of nodes in the subRoot.
+    SC: O(n) (recursive stack space)
+    '''
+    def isSubtree(self, root: Optional[TreeNode], subRoot: Optional[TreeNode]) -> bool:
+        if not root:
+            return False
+        if not subRoot:
+            return True
+
+        if self.isIdentical(root, subRoot): 
+            return True
+
+        return self.isSubtree(root.left, subRoot) or self.isSubtree(root.right, subRoot)
+
+    def isIdentical(self, s: Optional[TreeNode], t: Optional[TreeNode]) -> bool:
+        if not s and not t:
+            return True
+        if not s or not t:
+            return False
+            
+        return s.val == t.val and self.isIdentical(s.left, t.left) and self.isIdentical(s.right, t.right)
+
+    '''
+    2. Snapshot
+    TC: O(n + m), n is the number of nodes in the root, m is the number of nodes in the subRoot.
+    SC: O(n + m) (recursive stack space)
+    '''
+    def isSubtree(self, root: Optional[TreeNode], subRoot: Optional[TreeNode]) -> bool:
+        def preorder(node):
+            if not node:
+                return '#'
+            return f'({node.val},{preorder(node.left)},{preorder(node.right)})'
+
+        return preorder(subRoot) in preorder(root)

From 1574698c255b2a88bc4a80d336e46652ba716644 Mon Sep 17 00:00:00 2001
From: Dusuna <94776135+dusunax@users.noreply.github.com>
Date: Tue, 18 Mar 2025 22:38:16 +0900
Subject: [PATCH 2/4] add solution: validate-binary-search-tree

---
 validate-binary-search-tree/dusunax.py | 26 ++++++++++++++++++++++++++
 1 file changed, 26 insertions(+)
 create mode 100644 validate-binary-search-tree/dusunax.py

diff --git a/validate-binary-search-tree/dusunax.py b/validate-binary-search-tree/dusunax.py
new file mode 100644
index 000000000..cc39aedd9
--- /dev/null
+++ b/validate-binary-search-tree/dusunax.py
@@ -0,0 +1,26 @@
+'''
+# 98. Validate Binary Search Tree
+
+check if is a valid BST.
+- the left value is smaller than the current value
+- the right value is greater than the current value
+
+👉 all left and right subtrees must also valid continuously. 
+so, we need to check the low & high bound recursively. (not only check current node & immediate children)
+
+## base case
+- if the node is None, return True (empty tree is valid BST)
+- valid node value has range of left < node.val < right, if not, return False
+- recursively check the left and right subtree, update the low & high.
+'''
+class Solution:
+    def isValidBST(self, root: Optional[TreeNode]) -> bool:
+        def dfs(node, low = float('-inf'), high = float('inf')):
+            if not node:
+                return True
+            if not (low < node.val < high):
+                return False
+            
+            return dfs(node.left, low, node.val) and dfs(node.right, node.val, high)
+        
+        return dfs(root) 

From e2661f121da3dfbe85cf8f7467f4b486b2a3ec15 Mon Sep 17 00:00:00 2001
From: Dusuna <94776135+dusunax@users.noreply.github.com>
Date: Fri, 21 Mar 2025 07:49:48 +0900
Subject: [PATCH 3/4] add solution: longest-palindromic-substring

---
 longest-palindromic-substring/dusunax.py | 135 +++++++++++++++++++++++
 1 file changed, 135 insertions(+)
 create mode 100644 longest-palindromic-substring/dusunax.py

diff --git a/longest-palindromic-substring/dusunax.py b/longest-palindromic-substring/dusunax.py
new file mode 100644
index 000000000..74cfb5d8a
--- /dev/null
+++ b/longest-palindromic-substring/dusunax.py
@@ -0,0 +1,135 @@
+'''
+# 5. Longest Palindromic Substring
+
+DP 테이블을 사용하여 팰린드롬 여부를 저장하고 최대 길이를 찾기.
+'''
+class Solution:
+    '''
+    TC: O(n^2)
+    SC: O(n^2)
+    '''
+    def longestPalindrome(self, s: str) -> str:
+        n = len(s)
+        if n == 1:
+            return s
+        
+        start, length = 0, 1
+        dp = [[False] * n for _ in range(n)] # SC: O(n^2)
+
+        # length 1, diagonal elements in 2d array
+        for i in range(n): # TC: O(n)
+            dp[i][i] = True
+
+        # length 2, are two elements same
+        for i in range(n - 1):
+            if s[i] == s[i + 1]:
+                dp[i][i + 1] = True
+                start, length = i, 2
+
+        # length 3+
+        for word_length in range(3, n + 1): # TC: O(n^2)
+            for i in range(n - word_length + 1):
+                j = i + word_length - 1
+
+                if s[i] == s[j] and dp[i + 1][j - 1]:
+                    dp[i][j] = True
+                    start, length = i, word_length
+                
+        return s[start:start + length] 
+    
+'''
+## Check Leetcode Hints
+- How can we reuse a previously computed palindrome to compute a larger palindrome?
+  - use dp table that stores isPalindrome computation.
+- If “aba” is a palindrome, is “xabax” a palindrome? Similarly is “xabay” a palindrome?
+  - if it is palindrome, we only need to check the outermost chars, that wrapping our palindrome.
+- Palindromic checks can be O(1) by reusing prev computations.
+  - DP!
+
+## 동적 프로그래밍 풀이
+- s[i:j]의 팰린드롬 여부를 저장하기 위해서는 2차원 배열을 사용하는 것이 간편하다.
+- 길이가 1인 경우, 팰린드롬
+- 길이가 2인 경우, 두 문자가 같다면 팰린드롬
+- 3 이상은 dp 테이블을 활용하며 확인
+  - ex) 길이가 5인 문자열을 다음과 같이 탐색
+  ```
+  len: 3, i: 0, j: 2
+  len: 3, i: 1, j: 3
+  len: 3, i: 2, j: 4
+  len: 4, i: 0, j: 3
+  len: 4, i: 1, j: 4
+  len: 5, i: 0, j: 4
+  ```
+
+## 탐구
+- can we use sliding window to optimize?
+슬라이딩 윈도우는 연속적인 구간을 유지하며 최적해를 찾을 때 유용하지만, 팰린드롬은 중앙 확장과 이전 계산 재사용이 핵심.
+
+- can we solve this by counting char?
+문자 개수를 세면 "어떤 문자가 몇 번 나왔는지"만 알 수 있지만, 팰린드롬 여부는 문자 순서가 중요하다.
+
+- 그렇다면 개선 방법은? 
+> Manacher's Algorithm: https://www.geeksforgeeks.org/manachers-algorithm-linear-time-longest-palindromic-substring-part-1/  
+>
+>팰린드롬의 대칭성을 활용해 중앙을 기준으로 확장하는 Manacher's Algorithm을 적용할 수 있다. 모든 문자 사이에 #을 넣어 짝수, 홀수를 동일하게 다루는 기법인데 구현이 다소 복잡하다. 시간 복잡도는 O(n)이다.
+'''
+class Solution:
+    '''
+    TC: O(n)
+    SC: O(n)
+    '''
+    def longestPalindromeManacher(self, s: str) -> str:
+        '''
+        1. 문자열 변환하기 (가운데에 # 추가)
+        2. 저장공간 초기화
+        3. 변환된 문자열 순회하기
+          (a) 미러 인덱스 계산
+          (b) 팰린드롬 반지름 확장
+            - i에서 양쪽 문자들을 비교해서 확장
+          (c) 새로운 팰린드롬 찾기
+            - 새로운 팰린드롬을 i로 설정
+            - 새로운 팰린드롬의 오른쪽 끝을 i + p[i]로 설정
+          (d) 가장 긴 팰린드롬 업데이트
+        '''
+        transformed = '#' + '#'.join(s) + '#'
+        n = len(transformed)
+        p = [0] * n # i 중심의 팰린드롬 크기 
+        c = 0  # 현재 팰린드롬의 중심
+        r = 0  # 현재 팰린드롬의 오른쪽 끝
+        max_len = 0  # 가장 긴 팰린드롬의 길이
+        center = 0  # 가장 긴 팰린드롬의 중심
+
+        for i in range(n):
+            # 현재 위치 i의 미러 인덱스 (중앙을 기준으로 대칭되는 인덱스)
+            # ex) ababbaa
+            #     0123456
+            # i가 3이고 center가 2일 때, 2*2 - 3 = 1, 미러 인덱스는 1
+            # i가 5이고 center가 4일 때, 2*4 - 5 = 3, 미러 인덱스는 3
+            mirror = 2 * c - i
+
+            if i < r:
+                # r - i: 얼마나 더 확장 될 수 있는가 => 현재 팰린드롬의 오른쪽 끝에서 현재 인덱스까지의 거리
+                # p[mirror]: 미러 인덱스에서의 팰린드롬 반지름
+                p[i] = min(r - i, p[mirror])
+                # 작은 값만큼만 확장이 가능하다. 예를 들어, r - i가 더 작은 값이라면 팰린드롬을 그만큼만 확장할 수 있고, p[mirror]가 더 작은 값이라면 이미 그만큼 확장이 된 상태
+                # r - i가 3이고 p[mirror]가 2라면 팰린드롬을 2만큼 확장할 수 있고, r - i가 2이고 p[mirror]가 3이라면 팰린드롬을 2만큼 확장할 수 있다.
+
+            # 현재 중심에서 팰린드롬을 확장
+            # 양 끝이 같다면 팰린드롬 반지름 p[i]를 1씩 증가
+            while i + p[i] + 1 < n and i - p[i] - 1 >= 0 and transformed[i + p[i] + 1] == transformed[i - p[i] - 1]:
+                p[i] += 1
+
+            # 기존에 찾은 팰린드롬보다 더 큰 팰린드롬을 찾은 경우 
+            # 현재 중심과 오른쪽 끝을 설정
+            if i + p[i] > r:
+                c = i
+                r = i + p[i]
+
+            # 가장 긴 팰린드롬 업데이트
+            if p[i] > max_len:
+                max_len = p[i]
+                center = i
+
+        # 변환된 문자열에서 가장 긴 팰린드롬을 원래 문자열에서 추출
+        start = (center - max_len) // 2
+        return s[start:start + max_len]

From e3b5bde24e7d8b32d5e09fa7887e6ca91c868762 Mon Sep 17 00:00:00 2001
From: Dusuna <94776135+dusunax@users.noreply.github.com>
Date: Fri, 21 Mar 2025 08:26:14 +0900
Subject: [PATCH 4/4] add solution: rotate-image

---
 rotate-image/dusunax.py | 77 +++++++++++++++++++++++++++++++++++++++++
 1 file changed, 77 insertions(+)
 create mode 100644 rotate-image/dusunax.py

diff --git a/rotate-image/dusunax.py b/rotate-image/dusunax.py
new file mode 100644
index 000000000..56fe2d32b
--- /dev/null
+++ b/rotate-image/dusunax.py
@@ -0,0 +1,77 @@
+'''
+# 48. Rotate Image
+
+rotate 2d matrix 90 degree in place.
+👉 transpose matrix and reverse each row.
+
+- original matrix
+1 2 3
+4 5 6
+7 8 9
+
+- transpose matrix (swap i, j) (flip diagonally)
+1 4 7
+2 5 8
+3 6 9
+
+- reverse each row (horizontal flip)
+7 4 1
+8 5 2
+9 6 3
+'''
+class Solution:
+    '''
+    TC: O(n^2)
+    SC: O(1)
+    '''
+    def rotate(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 i in range(n):
+            matrix[i].reverse()
+
+'''
+# 💡 other rotate in-place examples
+
+## rotate 180 degree
+
+### solution A. reverse each column (vertical flip) & reverse each row (horizontal flip)
+
+```python
+def rotate180(matrix: List[List[int]]) -> None:
+  n = len(matrix)
+  matrix.reverse()
+
+  for i in range(n):
+    matrix[i].reverse()
+```
+
+### solution B. (after transpose, reverse each row (horizontal flip)) * 2.
+
+90 degree * 2 = 180 degree
+
+```python
+def rotate180(matrix: List[List[int]]) -> None:
+  rotate90(matrix)
+  rotate90(matrix)
+```
+
+## rotate -90 degree
+
+after transpose, reverse each column (vertical flip)
+
+```python
+def rotate90CCW(matrix):
+    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]
+
+    matrix.reverse()
+```
+'''