My first instinct was to do pairwise comparisons between all numbers and BFS from there.
Model the problem as a graph where each number is a node and edges connect numbers sharing a digit, then find the largest connected component using Union-Find (DSU) or BFS/DFS. Since numbers are two-digit, you can optimize by grouping via digit buckets (0-9) and unioning numbers that share a digit.
Pro tip: Mention that with only 10 possible digits, you can use a Union-Find structure with 10 digit nodes and union each number's two digits, then count the size of each digit group. This reduces the problem to O(n α(10)) time and avoids building a full graph.
Confirm the input format (list of two-digit integers) and that grouping is transitive. Restate the goal: return the size of the largest group of numbers connected by shared digits.
Decide between Union-Find (DSU) or graph traversal (BFS/DFS). Union-Find is efficient for dynamic connectivity and easy to implement with path compression and union by rank.
For each number, extract its two digits and union them. Alternatively, create a mapping from each digit to a list of numbers and union numbers that share a digit.
After processing all numbers, count the size of each group by finding the root of each number's digits and tallying. Track the maximum size.
State time complexity: O(n α(10)) ≈ O(n) with Union-Find, and space O(10) for digit nodes plus O(n) for mapping numbers to digits if needed.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.