My first instinct was to overthink it and reach for some kind of sliding window, but it's really just a linear scan.
Clarify that the problem reduces to finding the longest contiguous range of columns where every row has a 1. Then propose an efficient solution: for each column, compute the AND of all rows (or equivalently, check if any row has a 0), and find the longest run of columns where the AND is 1. This can be done in O(R*C) time by scanning columns and maintaining a running count, or in O(C) time after preprocessing each column with early termination.
Pro tip: Mention that you can optimize by stopping early when a 0 is found in a column, and that the problem is essentially finding the longest subarray of columns where the column-wise minimum is 1. Also, discuss how to handle large matrices with streaming or parallel processing if needed.
Confirm that the matrix has R rows (microservices) and C columns (days), and that a day is 'good' if all entries in that column are 1. The goal is the maximum number of consecutive good days.
For each column, determine if it is all 1s. This can be done by checking each row in that column, or by computing the column-wise AND. The longest contiguous sequence of such columns is the answer.
Scan columns from left to right. For each column, check all rows until a 0 is found (early exit). If all are 1, increment a current run length; otherwise reset it. Keep track of the maximum run length. Time complexity O(R*C) worst-case, but often faster with early exit.
Discuss potential optimizations like using bitwise operations if rows are represented as bitsets, or parallelizing column checks. Handle edge cases: empty matrix, single row/column, all 1s, all 0s.
State time and space complexity. Walk through a small example to verify correctness. Mention that the solution is optimal for the given constraints.
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