Min Diff
Reported by candidates from Oracle's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
Oracle's "Min Diff" question landed on candidates in October 2024, and it's testing whether you can find the minimum difference between two elements in an array or similar structure. The trap is overthinking it. You don't need a fancy algorithm here, just the right data structure and a sorted pass. If you blank on the live OA, StealthCoder will feed you the pattern instantly so you don't tank the score.
Pattern and pitfall
Min Diff boils down to sorting and linear traversal. Once sorted, the minimum difference between any two elements always lives in adjacent pairs. Walk through once, track the smallest gap. The common miss: candidates try hash tables or nested loops when the answer is just O(n log n) sort plus O(n) comparison. If you're taking the OA live and the exact problem wording throws you, StealthCoder reads it off your screen and surfaces the solution without the proctor seeing a thing. That's your insurance policy.
Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.
You can drill Min Diff 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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Oracle reuses patterns across OAs. Made for the candidate who got the OA invite this morning and has 72 hours, not six months. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Min Diff FAQ
Is this about sorting or searching?+
It's sorting. Once you sort the array, you only need one pass to find the minimum difference between adjacent elements. That's the whole trick. No binary search needed.
What if there are negative numbers or duplicates?+
Doesn't matter. Sort still works. Duplicates give you a difference of zero, which is the minimum possible. Negatives don't change the algorithm at all.
How do I optimize if the array is already sorted?+
You skip the sort and go straight to O(n) linear scan. If the problem statement says it's sorted, use that. Most don't, so assume you sort.
Is there a trick to avoid sorting the entire array?+
Not really. O(n log n) is clean and fast enough for most constraints. If you're under brutal time limits, partial sorting exists but it's overkill for this problem.
What's the most common failure mode candidates hit?+
Overcomplicating it. They use extra space or nested loops when sorted adjacent pairs is all you need. Keep it simple, sort, scan, done.