[SunaDu] Week 11

graph-valid-tree/dusunax.py

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+'''
+# 261. Graph Valid Tree
+
+## What constitutes a 🌲
+1. it's a graph.
+2. Connected: edges == n - 1, visited node count == n
+3. Acyclic: there is no cycle.
+
+## Approach A. DFS
+use DFS to check if there is a cycle in the graph.
+- if there were no cycle & visited node count == n, return True.
+
+## Approach B. Disjoint Set Union (서로소 집합)
+use Disjoint Set Union to check if there is a cycle in the graph.
+- if you find a cycle, return False immediately.
+- if you find no cycle, return True.
+
+### Union Find Operation
+- Find: find the root of a node.
+  - if the root of two nodes is already the same, there is a cycle.
+- Union: connect two nodes. 
+
+## Approach Comparison
+- **A. DFS**: simple and easy to understand.
+- **B. Disjoint Set Union**: quicker to check if there is a cycle. if there were more edges, Union Find would be faster.
+'''
+class Solution:
+    def validTreeDFS(self, n: int, edges: List[List[int]]) -> bool:
+        if len(edges) != n - 1:
+            return False
+
+        graph = [[] for _ in range(n)]
+        for node, neighbor in edges:
+            graph[node].append(neighbor)
+            graph[neighbor].append(node)
+
+        visited = set()
+        def dfs(node):
+            visited.add(node)
+            for neighbor in graph[node]:
+                if neighbor not in visited:
+                    dfs(neighbor)
+
+        dfs(0)
+        return len(visited) == n
+
+    def validTreeUnionFind(self, n: int, edges: List[List[int]]) -> bool:
+        if len(edges) != n - 1:
+            return False
+
+        parent = [i for i in range(n)]
+
+        def find(x):
+            if x == parent[x]:
+                return x
+            parent[x] = find(parent[x])
+            return parent[x]
+
+        def union(x, y):
+            parent[find(x)] = find(y)
+
+        for node, neighbor in edges:
+            if find(node) == find(neighbor):
+                return False
+            union(node, neighbor)
+
+        return True

maximum-depth-of-binary-tree/dusunax.py

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+'''
+# 104. Maximum Depth of Binary Tree
+
+use DFS to find the maximum depth of the binary tree.
+
+## A. if we use a helper function (not a good idea)
+``` 
+helper function explanation:
+- store the max_depth in the nonlocal variable.(outside of the helper function)
+- base case: if the node is None, update the max_depth and return.
+- in the helper function, do recursive call for the left and right children of the node.
+  - update the count for the depth of the tree.
+- update the max_depth when the node is a leaf node's children.
+```
+
+## B. return the max_depth directly
+👉 why helper function is not necessary?
+recursion function can return the max_depth directly.  
+remove side effect & non-local variable.
+
+## TC is O(n)
+
+visit each node once for checking if it is a leaf node's children.
+
+## SC is O(h)
+
+h for height of the tree
+'''
+class Solution:
+    '''
+    A. first approach with side effect
+    '''
+    def maxDepthWithHelper(self, root: Optional[TreeNode]) -> int: 
+        max_depth = 0
+
+        def helper(node, count):
+            nonlocal max_depth 
+            if node is None:
+                max_depth = max(max_depth, count)
+                return
+
+            helper(node.left, count+1) 
+            helper(node.right, count + 1)
+
+        helper(root, max_depth)
+
+        return max_depth
+
+    '''
+    B. return the max_depth directly.
+    - more concise & readable
+    - no side effect & non-local variable
+    '''
+    def maxDepth(self, root: Optional[TreeNode]) -> int: 
+        if root is None:
+            return 0
+
+        left_count = self.maxDepth(root.left)
+        right_count = self.maxDepth(root.right)
+
+        return max(left_count, right_count) + 1

reorder-list/dusunax.py

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+'''
+# 143. Reorder list
+use two pointers for each steps.
+
+1. finding the middle
+    - two pointers: slow, fast
+    - move slow 1, fast 2 until fast reaches the end.
+2. reversing the second half
+    - two pointers: prev, curr
+    - start from slow to end, do common reverse linked list operation.
+    - need to break the links to halves beforehand.
+3. merging first & second
+    - two pointers: frist, second
+    - merge second between first, until second is None
+
+## TC is O(n)
+- find the middle: O(n)
+- reverse the second half: O(n)
+- merge the two halves: O(n)
+
+## SC is O(1)
+- no extra space is used.
+'''
+class Solution:
+    def reorderList(self, head: Optional[ListNode]) -> None:
+        # 1. finding the middle
+        slow, fast = head, head
+        while fast and fast.next:
+            slow = slow.next
+            fast = fast.next.next
+
+        # 2. reversing second half
+        second_not_reversed = slow.next
+        slow.next = None
+        prev, curr = None, second_not_reversed
+        while curr:
+            temp = curr.next
+            curr.next = prev
+            prev = curr
+            curr = temp
+
+        # 3. merging first & second
+        first, second = head, prev
+        while second:
+            temp1, temp2 = first.next, second.next
+            first.next = second
+            second.next = temp1
+            first = temp1
+            second = temp2