Went recursive, which was the right call under time pressure.
Use a recursive DFS that leverages the BST property to prune branches: if the current node's value is less than low, only recurse right; if greater than high, only recurse left; otherwise include the value and recurse both sides. This yields O(n) worst-case but often much faster in practice.
Pro tip: Mention that the pruning makes the algorithm O(k) where k is the number of nodes visited, which can be much smaller than n. Also, clarify that the sum can be computed without modifying the tree, and discuss handling large sums with appropriate data types.
Ask about edge cases: empty tree, low > high, negative values, and whether the tree is guaranteed to be a BST. Confirm the expected return type (e.g., integer, long).
Describe how the BST property allows skipping entire subtrees: if node.val < low, all left descendants are too small; if node.val > high, all right descendants are too large.
Write pseudocode: if node is null return 0; if node.val < low return rangeSum(node.right); if node.val > high return rangeSum(node.left); else return node.val + rangeSum(node.left) + rangeSum(node.right).
State that time is O(n) worst-case but O(k) where k is the number of nodes in range, and space is O(h) for recursion stack, where h is tree height.
Mention that an iterative stack-based approach can avoid recursion depth issues, and walk through an example to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This one tripped me up more than I'd like to admit.
Design a streaming interface that maintains a sliding window of the last K values and efficiently computes their product. Handle zeros by tracking the count of zeros in the window and the product of non-zero elements, so the product is zero if any zero is present, otherwise the product of non-zeros. Use a queue to manage the window and update the product and zero count on each insertion and removal.
Pro tip: Discuss the trade-off between maintaining a running product and recomputing from scratch; the zero-handling approach avoids division by zero and keeps operations O(1) amortized. Also, clarify assumptions about data types (e.g., integers vs floats) and potential overflow.
Ask about the data type, whether K is fixed or dynamic, and if the product can overflow. Confirm that the interface should support adding new values and querying the product of the last K values.
Use a queue (or circular buffer) to store the last K values. Maintain a running product of non-zero values and a count of zeros in the current window.
When a new value arrives, add it to the queue. If it's zero, increment zero count; otherwise, multiply it into the running product. If the queue size exceeds K, remove the oldest value: if it's zero, decrement zero count; otherwise, divide the running product by it (or recompute if division is problematic).
If zero count > 0, return 0; otherwise, return the running product. Ensure that the product is only valid when the window has exactly K elements (or handle initial partial windows as per requirements).
State that each operation is O(1) amortized. Discuss edge cases: K=0, K=1, all zeros, no zeros, and overflow handling (e.g., using modular arithmetic or big integers).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.