My first instinct was some kind of greedy with frequency counting, which was actually right but I took forever to convince myself.
Recognize that the problem reduces to finding the digit (0-9) that appears in the most numbers, since any set of numbers sharing a common digit must all contain that digit. Count the frequency of each digit across all numbers, then return the maximum frequency. This yields an O(n) solution with O(1) extra space.
Pro tip: Clarify that numbers like 11 contain the same digit twice but should be counted once per digit; also mention that if no numbers are given, the answer is 0. This shows attention to edge cases and precision.
Restate the problem: select the maximum subset of two-digit numbers such that all share at least one common digit. Note that the common digit must be the same across all selected numbers.
Realize that the problem reduces to finding the digit that appears in the most numbers. Because if a set of numbers shares a digit d, then every number in the set contains d, so the size of the set is at most the frequency of d.
Initialize an array of size 10 to count occurrences of each digit. For each number, extract its tens and units digits, and increment the count for each unique digit in that number (avoid double-counting if both digits are the same).
After processing all numbers, the answer is the maximum value in the digit count array. If the list is empty, return 0.
Time complexity is O(n) since we process each number once. Space complexity is O(1) for the fixed-size count array. Discuss edge cases: empty list, numbers with repeated digits (e.g., 11), and ties (any digit with max count works).
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