Checking every subarray is O(n squared). Prefix sums plus a remainder-frequency map does it in one O(n) pass: two prefixes with the same remainder mod k bracket a divisible subarray. The trap is negative remainders. Here is the clean answer.
Updated Aug 2026 · Grounded in real Applied AI Engineer interview loops and written to a senior-engineer editorial bar.
Checking every subarray is O(n squared). Prefix sums plus a remainder-frequency map does it in one O(n) pass: two prefixes with the same remainder mod k bracket a divisible subarray. The trap is negative remainders. Here is the clean answer.
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