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The naive approach of just tracking a running max will burn you the second you hit a negative number.
Use dynamic programming to track both the maximum and minimum product ending at each position, because a negative number can turn a minimum into a maximum. Iterate through the array, updating these values and the global maximum product. This yields an O(n) time, O(1) space solution.
Pro tip: Clarify edge cases upfront, such as empty array, zeros, and negative numbers, and mention that the algorithm handles them naturally. Also, briefly discuss why a brute-force approach is inefficient and how your solution optimizes it.
Confirm the definition of subarray (contiguous) and discuss edge cases like empty array, single element, zeros, and negative numbers.
Explain that because multiplying by a negative flips sign, we must keep track of both the maximum and minimum product ending at the current position.
Define max_prod and min_prod as the max/min product of subarrays ending at the current index. Update them using the current number and the previous max/min.
Traverse the array, update max_prod and min_prod, and keep a running global maximum of max_prod.
State O(n) time and O(1) space. Walk through a small example to verify correctness.
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