The null case is what trips people up here.
Use a post-order DFS that returns a status for each subtree: whether each target was found and the LCA if both are found in that subtree. At each node, combine results from children; if both targets are found and LCA not yet determined, the current node is the LCA. Handle missing nodes by returning null if either target is not found in the entire tree.
Pro tip: Clarify upfront whether the two target nodes are guaranteed to be distinct and whether the tree can be empty; this shows attention to edge cases and avoids incorrect assumptions. Also, mention that if the targets are the same node, the LCA is that node itself.
Ask if the tree can be empty, if the two targets can be the same node, and if nodes have parent pointers (which would allow a different approach). Confirm that if either node is missing, return null.
Decide on a recursive post-order DFS that processes children first and returns information about found targets and LCA. This naturally handles N-ary trees by iterating over all children.
The function returns a pair: (foundA, foundB, lca). If the current node is one of the targets, mark it as found. Recursively process each child and merge results. If both targets are found in the current subtree and lca is not set, set lca to the current node.
After the DFS, if either target was not found in the entire tree, return null. Otherwise, return the LCA found.
State that time complexity is O(N) since each node is visited once, and space complexity is O(H) for recursion stack, where H is tree height. Walk through a small example to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Classic two-pointer merge but with the overlap-collapsing logic on top.
Use a two-pointer technique to traverse both sorted interval lists simultaneously, merging intervals on the fly by comparing start times and handling overlaps. Maintain a result list and a 'current' interval that is extended when overlaps occur.
Pro tip: Clarify upfront whether the input lists are truly non-overlapping and sorted, and confirm the expected output format (e.g., list of intervals). This shows attention to detail and avoids incorrect assumptions.
Confirm that each list is sorted and internally non-overlapping, and ask about edge cases like empty lists or single intervals. This ensures you understand the problem constraints.
Set two pointers i and j to 0 for the two lists, and create an empty result list. Also initialize a 'current' interval to None to track the merged interval being built.
While both pointers are within bounds, pick the interval with the smaller start time. If it overlaps with 'current', merge them by updating the end; otherwise, add 'current' to result and set 'current' to the new interval. Advance the corresponding pointer.
After one list is exhausted, continue processing the remaining intervals from the other list in the same manner, merging with 'current' as needed.
After all intervals are processed, add the last 'current' interval to the result if it exists, and return the merged list.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This one is a bitmask DP problem and I did not see it immediately.
Model each word as a 26-bit mask of its letters, discarding any word with duplicate letters. Then use backtracking with pruning to explore subsets of valid words, tracking the combined mask and total character count to find the maximum. Alternatively, use DP over masks if the number of valid words is small.
Pro tip: Precompute valid words and their masks first; this reduces the search space and avoids repeated duplicate checks. Also, sort words by length descending to find a good initial solution early, which improves pruning.
For each word, compute a 26-bit integer where each bit represents a letter. If a word has duplicate letters, discard it since it can never be part of a valid subset.
Decide between backtracking with pruning or dynamic programming over masks. Backtracking is simpler and works well for typical constraints; DP is better if the number of valid words is small.
Recursively consider each valid word: include it only if its mask doesn't overlap with the current combined mask. Track the maximum total length. Prune branches where the remaining possible length cannot exceed the current best.
Sort words by length descending to find a strong initial solution early. Optionally, memoize states (index, combined mask) if the same state can be reached multiple times.
After exploring all valid subsets, return the maximum total character count found.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.