[dolphinflow86] WEEK 15 Solutions

alien-dictionary/dolphinflow86.py

@@ -3,11 +3,12 @@
 # SC: O(1) - unique letters and adjacency list bounded by 26 characters
 
 from collections import deque
+from typing import List
 
 
 class Solution:
 
-    def alienOrder(self, words) -> str:
+    def alienOrder(self, words: List[str]) -> str:
         adj = {char: set() for word in words for char in word}
         indegree = {char: 0 for word in words for char in word}
 
@@ -41,3 +42,5 @@ def alienOrder(self, words) -> str:
             return ""
 
         return "".join(result)
+
+    alien_order = alienOrder

construct-binary-tree-from-preorder-and-inorder-traversal/dolphinflow86.py

@@ -0,0 +1,40 @@
+# N is the number of nodes in the binary tree.
+# TC: O(N) - building hashmap takes O(N), and each node is visited once in O(1) time
+# SC: O(N) - hashmap takes O(N) space, and recursion call stack takes O(H) up to O(N)
+
+from typing import List, Optional
+
+
+# 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 buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
+        inorder_map = {val: idx for idx, val in enumerate(inorder)}
+        preorder_idx = 0
+
+        def build(left: int, right: int) -> Optional[TreeNode]:
+            nonlocal preorder_idx
+
+            if left > right:
+                return None
+
+            root_val = preorder[preorder_idx]
+            root = TreeNode(root_val)
+            preorder_idx += 1
+
+            mid = inorder_map[root_val]
+
+            root.left = build(left, mid - 1)
+            root.right = build(mid + 1, right)
+
+            return root
+
+        return build(0, len(inorder) - 1)

longest-palindromic-substring/dolphinflow86.py

@@ -0,0 +1,31 @@
+# N is the length of the string s.
+# TC: O(N^2) - expanding around each of the 2N - 1 possible centers takes up to O(N)
+# SC: O(1) - constant auxiliary space tracking the start index and maximum length
+
+
+class Solution:
+
+    def longestPalindrome(self, s: str) -> str:
+        if len(s) <= 1:
+            return s
+
+        start, max_len = 0, 0
+
+        def expand(left: int, right: int) -> tuple[int, int]:
+            while left >= 0 and right < len(s) and s[left] == s[right]:
+                left -= 1
+                right += 1
+            return left + 1, right - left - 1
+
+        for i in range(len(s)):
+            # Odd length palindrome (center is i)
+            l1, len1 = expand(i, i)
+            if len1 > max_len:
+                start, max_len = l1, len1
+
+            # Even length palindrome (center is between i and i + 1)
+            l2, len2 = expand(i, i + 1)
+            if len2 > max_len:
+                start, max_len = l2, len2
+
+        return s[start:start + max_len]

rotate-image/dolphinflow86.py

@@ -0,0 +1,21 @@
+# N is the number of rows and columns in the matrix.
+# TC: O(N^2) - reversing rows takes O(N^2) and transposing across diagonal takes O(N^2)
+# SC: O(1) - in-place rotation with no extra memory allocation
+
+from typing import List
+
+
+class Solution:
+
+    def rotate(self, matrix: List[List[int]]) -> None:
+        """
+        Do not return anything, modify matrix in-place instead.
+        """
+        # 1. Flip vertically (reverse rows)
+        matrix.reverse()
+
+        # 2. Transpose across main diagonal
+        n = len(matrix)
+        for i in range(n):
+            for j in range(i + 1, n):
+                matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]

subtree-of-another-tree/dolphinflow86.py

@@ -0,0 +1,34 @@
+# M and N are the number of nodes in root and subRoot trees respectively.
+# TC: O(M * N) - in the worst case, is_same is checked for every node in root
+# SC: O(H) - recursion stack memory bounded by the height of root tree (up to O(M))
+
+from typing import Optional
+
+
+# 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:
+        if not subRoot:
+            return True
+        if not root:
+            return False
+
+        def is_same(s: Optional["TreeNode"], t: Optional["TreeNode"]) -> bool:
+            if not s and not t:
+                return True
+            if not s or not t or s.val != t.val:
+                return False
+            return is_same(s.left, t.left) and is_same(s.right, t.right)
+
+        if is_same(root, subRoot):
+            return True
+
+        return self.isSubtree(root.left, subRoot) or self.isSubtree(root.right, subRoot)