Count Minimum-Difference Pairs
Reported by candidates from Shield AI's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
The data structure this one hinges on is just a sorted array, and Shield AI's July 2026 OA report treats it like a trap for people who reach for a hash map or a nested loop. You get distinct integers, find the smallest gap between any two, then count how many unordered pairs hit that gap. With 100000 elements, brute force dies fast. If you've got an OA in the next couple of days, this is a sort-then-scan problem, and it's quicker than it looks. StealthCoder is there as a safety net if you blank mid-assessment.
The problem
Given an array values of distinct integers, find the minimum absolute difference between any two values and return the number of unordered pairs that have that difference. Function countMinimumDifferencePairs(values: int[]) → int Examples Example 1 values = [4,2,1,3] return = 3 The minimum difference is 1, achieved by (1,2), (2,3), and (3,4). Example 2 values = [1,5,9] return = 2 Constraints 2 <= values.length <= 100000. All values are distinct and fit in a signed 32-bit integer.
Reported by candidates. Source: FastPrep
Pattern and pitfall
Sort the array. The minimum absolute difference between any two values must occur between adjacent elements in sorted order, so you never compare non-neighbors. One pass computes the smallest adjacent gap. A second pass (or the same pass with a reset) counts how many adjacent pairs equal that gap. Because values are distinct, the gap is always at least 1, and each pair at the minimum gap is counted exactly once. Pitfalls: using int for the difference can overflow when values span the signed 32-bit range, so use a long. Also don't count pairs with a double loop, that's O(n^2) and times out at 100000. Total cost is O(n log n) for the sort and O(n) for the scan. If the logic slips under pressure, StealthCoder can surface the sort-and-scan solution live during the OA.
Memorize the pattern. If you can't, run StealthCoder. The proctor sees the IDE. They don't see what's behind it.
You can drill Count Minimum-Difference Pairs 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 by an engineer who treats the OA as theater. If yours is tonight, you don't have time to grind. You have time to hedge.
Get StealthCoderThis OA pattern shows up on LeetCode as minimum absolute difference. If you have time before the OA, drill that.
You've seen the question.
Make sure you actually pass Shield AI's OA.
Shield AI reuses patterns across OAs. Made by an engineer who treats the OA as theater. If yours is tonight, you don't have time to grind. You have time to hedge. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Count Minimum-Difference Pairs FAQ
How hard is the Shield AI Count Minimum-Difference Pairs question really?+
Easy to medium. The whole problem is one observation: the minimum gap lives between adjacent sorted elements. Once you see that, the code is about ten lines. The only real risk is overflow and an accidental O(n^2) approach.
What's the trick to solve it fast?+
Sort first. After sorting, only adjacent pairs can produce the minimum difference. Scan once to find the smallest gap, then count how many adjacent pairs match it. No hash map, no nested loop needed.
Do I need to worry about integer overflow?+
Yes. Values fit in signed 32-bit, so subtracting a large negative from a large positive can exceed the int range. Store differences in a 64-bit type like long in Java or long long in C++. Python handles it automatically.
What's the time complexity I should aim for?+
O(n log n) from sorting, plus O(n) for the scan. With n up to 100000 that's comfortably fast. Anything quadratic will likely time out, so skip any pair-by-pair comparison.
How do I prepare for this in 48 hours?+
Practice the sort-then-scan-adjacent pattern on a couple of minimum-difference problems. Write it once from memory, test on the two examples, then add an edge case with two elements and one with extreme 32-bit values.