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I went straight to a hash-based bucket approach where you mask each character position and store the pattern.
Start by clarifying requirements: word length, character set, and whether the data structure should support dynamic updates. Then propose a solution using a hash map keyed by word patterns (e.g., replacing each character with a wildcard) to enable O(1) lookups for words differing by exactly one character, and discuss trade-offs with alternatives like trie-based approaches.
Pro tip: Mention that you would preprocess the dictionary to build the pattern map during the build method, and for search, generate all possible patterns for the query and check if any exists in the map, ensuring O(L) time per query where L is word length.
Ask about word length consistency, character set (e.g., lowercase letters), and whether the data structure needs to support updates or deletions. This determines the optimal approach.
Explain that for each word, you generate all patterns by replacing each character with a wildcard (e.g., '*'), and store these patterns in a hash map mapping to the original words. This allows O(1) average lookup for any query pattern.
For build, iterate through each word, generate all patterns, and add to the map. For search, generate all patterns of the query string and check if any pattern exists in the map; if yes, return true.
Discuss time and space complexity: build O(N*L), search O(L), space O(N*L). Compare with alternatives like trie with DFS (O(N*L) search) or brute force (O(N*L)), highlighting the efficiency of the hashing approach.
Consider cases like duplicate words, empty strings, and words of different lengths. Mention potential optimizations like grouping by length or using a trie if memory is a concern.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
DFS to collect (row, col, char) tuples then sort.
Perform a DFS traversal to assign coordinates to each node, storing them in a list. Then sort the list by column ascending, then row ascending, and finally by a left-to-right tiebreaker (e.g., original insertion order or a secondary key). Concatenate the node values in the sorted order.
Pro tip: Clarify the tie-breaking rule: 'left to right' likely means the order of nodes as they appear from left to right in the tree, which can be captured by a pre-order traversal index. Mention that you'll use a stable sort or include an index to ensure correct ordering.
Restate the problem to ensure understanding: assign coordinates via DFS, sort by column then row, and handle ties left to right. Ask if 'left to right' refers to the node's position in the tree's in-order traversal or its horizontal position.
Use DFS (pre-order) to traverse the tree, assigning (row, col) to each node. Store each node's value, row, col, and a traversal index (to break ties) in a list.
Sort the list by column ascending, then row ascending, and finally by the traversal index (or a left-to-right key) to resolve ties. Ensure the comparator is stable or includes all necessary keys.
Iterate through the sorted list and concatenate the node values into a string. Return the resulting string.
State that the time complexity is O(n log n) due to sorting, and space complexity is O(n) for storing the nodes. Mention that DFS is O(n).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.