Minimum Difference Across Cyclic Rotations
Reported by candidates from ZipRecruiter's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
The ZipRecruiter OA reported in October 2022 looks like a rotation puzzle, but the real question is which data structure you need. With n up to 2000, the answer is probably none. Plain arrays and a nested loop do the job. You try every cyclic left rotation of first, sum the absolute differences against second, and keep the minimum. Most people overthink it and reach for hashing or heaps. If you blank on the setup during the live assessment, StealthCoder runs invisibly on your desktop and gives you the working structure in real time. Know the shape before you open the invite.
The problem
You are given equal-length integer arrays first and second. For every cyclic left rotation of first, sum the absolute differences of aligned elements. Return the minimum sum across all rotations. Function minimumRotationDifference(first: int[], second: int[]) → long Examples Example 1 first = [1,2,3] second = [2,3,1] return = 0 One left rotation makes the arrays equal. Example 2 first = [1,4] second = [2,8] return = 5 The unrotated cost is 1 + 4 = 5, which is minimal. Constraints 1 <= first.length == second.length <= 2000 -1000000000 <= value <= 1000000000
Reported by candidates. Source: FastPrep
Pattern and pitfall
The trick is reading the constraint. Length is at most 2000, so n rotations times n elements is about 4 million operations. That's trivial. Loop shift from 0 to n-1, then loop i from 0 to n-1 and compare first[(i + shift) % n] against second[i]. Add Math.abs of the difference and track the minimum. The common pitfall is overflow. Values reach 1e9 in magnitude, so one difference can hit 2e9 and the sum can reach 4e12. Use a 64-bit long for the difference and the accumulator, never int. Another miss is rotating the array physically each time, which wastes work and invites off-by-one bugs. Use index math instead. A faster approach exists in theory, but it's unnecessary here. If the modulo indexing or the overflow slips your mind mid-assessment, StealthCoder is the hedge that catches it live.
Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.
You can drill Minimum Difference Across Cyclic Rotations cold, or you can hedge it. StealthCoder runs invisibly during screen share and surfaces a working solution in under 2 seconds. The proctor sees the IDE. They don't see what's behind it. Made for the candidate who got the OA invite this morning and has 72 hours, not six months.
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Minimum Difference Across Cyclic Rotations FAQ
How hard is the ZipRecruiter minimum rotation difference problem really?+
Easy to medium. The brute force is the intended solution given n is at most 2000. The difficulty is noticing that, then handling 64-bit overflow and the modulo indexing correctly. If you overthink it, you'll waste time on an optimization nobody asked for.
What's the trick to solving it?+
Skip the physical rotation. For each shift from 0 to n-1, compute the sum of abs(first[(i + shift) % n] - second[i]) across all i. Track the minimum. That's O(n^2), about 4 million steps at the max size, which is fine.
Do I need a hash map or heap here?+
No. The problem is about cyclic alignment, not lookups or ordering. Two arrays and index arithmetic are enough. Reaching for extra structures adds bugs without improving the complexity you actually need.
What edge cases break most solutions?+
Integer overflow is the big one. Each difference can reach 2e9 and the total can reach 4e12, so use long everywhere. Also check length 1, where the only rotation is the identity, and negative values, where abs must wrap the full difference.
How do I prepare for this in 48 hours?+
Write the nested loop version once from memory, using modulo indexing and a long accumulator. Test it on both examples, where expected outputs are 0 and 5. Then practice a couple of other cyclic-shift problems so the modulo trick feels automatic.