My first instinct was to overthink it and reach for some graph-based approach.
Recognize that a valid group is defined by a common digit, so for each digit 0-9, count how many numbers contain that digit. The maximum count across all digits is the answer. This reduces the problem to a simple frequency count per digit.
Pro tip: Clarify that numbers are two-digit, so each number contains exactly two digits (possibly the same). This means we can efficiently check membership for each digit without iterating over all digits of the number.
A group is valid if there exists at least one digit that appears in every number of the group. We need the maximum size of such a group.
The common digit must be one of the digits present in the numbers. Since numbers are between 10 and 99, each number has exactly two digits. Thus, we can consider each digit 0-9 separately.
Initialize an array of size 10 to zero. For each number, extract its tens and ones digits, and increment the count for each digit. Then, the maximum count among all digits is the answer.
The algorithm runs in O(N) time and O(1) space, which is optimal. Mention that this is efficient for large N.
Discuss cases where no digit is common (e.g., numbers with disjoint digits) or when the array is empty. Also, note that digits can repeat within a number (e.g., 11), but that doesn't affect the count.
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