Get Min Difference
Reported by candidates from IXL Learning's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
IXL Learning sent you a problem called Get Min Difference in March 2024, and you've got maybe 48 hours to lock down the pattern. This one trips candidates because it sounds simple but hinges on a sorting insight that separates fast solutions from slow ones. The trick is recognizing you're not comparing every pair. If you blank on approach during the live OA, StealthCoder will feed you the structure in real time so you don't tank the problem.
Pattern and pitfall
Get Min Difference is fundamentally a sorting problem disguised as a search problem. Once you sort the input array, the minimum difference between any two elements will always be between adjacent elements in the sorted array. No need to check all pairs. The algorithmic leap is that sorting reduces the problem from O(n²) brute force to O(n log n), and then a single pass through the sorted array gives you the answer in O(n). The common pitfall is overthinking it as a two-pointer or hash-table problem when the pattern is just sort-and-scan. If you freeze on the live OA, StealthCoder will show you the sorted array approach and you'll execute it cleanly.
If this hits your live OA and you blank, StealthCoder solves it in seconds, invisible to the proctor.
You can drill Get Min Difference 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. Built by an Amazon engineer who would have shipped this the night before his JPMorgan OA if he'd had it.
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IXL Learning reuses patterns across OAs. Built by an Amazon engineer who would have shipped this the night before his JPMorgan OA if he'd had it. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Get Min Difference FAQ
Why does sorting solve this?+
After sorting, the minimum difference must occur between adjacent elements. Any non-adjacent pair will have a larger difference. This insight collapses O(n²) brute force into O(n log n) + O(n) single pass.
What's the edge case everyone misses?+
Empty array or array with one element. Handle those before you enter the scan loop. Also watch for integer overflow if you're subtracting large values, though IXL likely keeps numbers reasonable.
Do I need two pointers here?+
No. After sorting, just iterate through and track the minimum difference as you go. Two pointers would be overengineering. Simple loop wins.
How do I prep for this in 48 hours?+
Run through the sorting insight on paper. Write out a small example, sort it, verify the answer is between adjacent elements. Code it once cleanly. That's it. Don't memorize.
Is this still asked after March 2024?+
Yes. Sorting-based min/max problems are evergreen at IXL and other ed-tech companies. The pattern recycles. Master it now and you'll recognize it again.