- Choose the invariant that makes a window valid or useful.
- Advance the right boundary and add the new element.
- Move the left boundary only as needed while maintaining the invariant and updating the answer.
Code notes
- 39 lines of Python from the credited upstream file 1932.py.
- The implementation visibly relies on sequence storage, hash lookup.
- No explicit loop blocks detected.
Complexity
Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.
Check the problem constraints before deciding whether this complexity will pass.
Use this to learn the idea, then write your own version.
1class Solution:2 def canMerge(self, trees: list[TreeNode]) -> TreeNode | None:3 valToNode = {} 4 count = collections.Counter() 5 6 for tree in trees:7 valToNode[tree.val] = tree8 count[tree.val] += 19 if tree.left:10 count[tree.left.val] += 111 if tree.right:12 count[tree.right.val] += 113 14 def isValidBST(tree: TreeNode | None, minNode: TreeNode | None,15 maxNode: TreeNode | None) -> bool:16 if not tree:17 return True18 if minNode and tree.val <= minNode.val:19 return False20 if maxNode and tree.val >= maxNode.val:21 return False22 if not tree.left and not tree.right and tree.val in valToNode:23 val = tree.val24 tree.left = valToNode[val].left25 tree.right = valToNode[val].right26 del valToNode[val]27 28 return isValidBST(29 tree.left, minNode, tree) and isValidBST(30 tree.right, tree, maxNode)31 32 for tree in trees:33 if count[tree.val] == 1:34 if isValidBST(tree, None, None) and len(valToNode) <= 1:35 return tree36 return None37 38 return None39