My first instinct was to just do pairwise comparisons and call it a day.
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. Alternatively, use Union-Find (Disjoint Set Union) to efficiently merge groups based on shared digits, tracking the size of each component.
Pro tip: Since numbers are two-digit, you can optimize by using the digits themselves (0-9) as intermediate nodes in a bipartite graph, reducing the number of nodes to at most 10 + N and simplifying the union operations.
Restate the problem to ensure understanding: group two-digit integers transitively if they share at least one digit, and return the largest group size. Ask about edge cases like empty input, single number, or numbers with repeated digits (e.g., 11).
Decide between graph traversal (BFS/DFS) or Union-Find. Union-Find is often more efficient for dynamic connectivity and easier to implement for transitive grouping.
For Union-Find: initialize each number as its own set. For each digit (0-9), keep a list of numbers containing that digit. Union all numbers in each digit's list. Then find the maximum set size. For graph: build adjacency list and run BFS/DFS to find largest component.
Discuss time and space complexity. Union-Find with path compression and union by rank: O(N α(N)) time, O(N) space. Graph approach: O(N + E) time, where E is number of edges (at most 10*N).
Walk through examples, including edge cases like [11, 22, 33] (all separate), [12, 23, 34] (all connected), and [10, 20, 30] (all connected via 0). Verify the algorithm returns correct sizes.
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