Minimizing Array After Replacing Pairs With Their Product
A medium-tier problem at 40% community acceptance, tagged with Array, Dynamic Programming, Greedy. Reported in interviews at Wells Fargo and 0 others.
You're given an array and can repeatedly pick two elements, replace them with their product, and continue until one value remains. The goal: minimize that final result. Wells Fargo has asked this. The catch is that the greedy instinct (always multiply the two smallest) doesn't always work because negative numbers flip the math. A lot of candidates see the array, think product, and lock into a straightforward approach that fails on mixed-sign input. If this problem lands in your assessment and you're unsure whether to multiply negatives first or positives first, StealthCoder pulls the optimal strategy in seconds without the proctor knowing.
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Minimizing Array After Replacing Pairs With Their Product is the kind of problem that decides whether you pass. StealthCoder reads the problem on screen and surfaces a working solution in under 2 seconds. Invisible to screen share. The proctor sees nothing. Made by an Amazon engineer who watched the leaked-problem repo become an industry secret. He decided you should have it too.
Get StealthCoderThis is a dynamic programming problem dressed as a greedy one. The naive approach is to always pair the two smallest absolute values, but that breaks when you have both negative and positive numbers. The real insight: you need to track both the minimum and maximum possible products at each state because a large negative times a large negative gives a large positive, which might actually be your worst outcome. A correct solution uses DP to explore multiple reduction paths and identify which sequence of pairings yields the smallest final result. The greedy strategy of picking smallest pairs works only in specific cases. Most candidates miss that this requires either memoized recursion or a DP table to avoid recomputing subproblems. When you hit this live and the greedy approach fails your test cases, StealthCoder reveals the DP structure and the state transitions you need.
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Minimizing Array After Replacing Pairs With Their Product recycles across companies for a reason. It's medium-tier, and most candidates blank under the timer. StealthCoder is the hedge: an AI overlay invisible during screen share. It reads the problem and surfaces a working solution in under 2 seconds. Made by an Amazon engineer who watched the leaked-problem repo become an industry secret. He decided you should have it too. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Minimizing Array After Replacing Pairs With Their Product interview FAQ
Why doesn't always multiplying the two smallest values work?+
Because negative numbers invert the comparison. Multiplying two large negatives gives a large positive, which could be the maximum, not minimum. You need to track both the minimum and maximum possible products at each reduction step to know which pairings actually minimize the final result.
Is this really a dynamic programming problem?+
Yes. You're choosing which pairs to multiply in which order, and the result of one multiplication affects which pairs you can form next. Memoization on the remaining array state avoids recalculating the same subproblems multiple times, which is essential for correctness and performance.
How hard is this compared to other medium array problems?+
It's on the harder end of medium. The acceptance rate hovers around 40 percent, which reflects that many candidates lock into greedy thinking and don't realize they need DP. The trick isn't the DP itself, it's recognizing that this problem requires tracking both min and max states.
Does order of multiplication matter if I'm just looking at the product?+
Mathematically the product is the same, but the path you take affects whether you ever see certain intermediate states. By choosing which pair to multiply first, you constrain future pairings. DP explores all orderings to find the one that ends with the smallest result.
What if the array has zeros?+
Zeros are often a boundary case. Multiplying anything by zero gives zero, which is often the minimum possible result. A strong solution handles zeros explicitly and may short-circuit the search if it can guarantee a zero outcome, saving computation.
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