Count Triples with Diff
Reported by candidates from DataBricks's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
DataBricks hit you with Count Triples with Diff in August. You've got an array and need to count triples where elements satisfy a difference condition. It's a counting problem disguised as a brute force check. The naive approach works if the array is small, but if they're testing your optimization instinct, you'll need to think about hash tables or sorting. StealthCoder reads the exact constraint and feeds you the pattern in real time during the OA.
Pattern and pitfall
The problem asks you to count triples (i, j, k) where some difference constraint holds. The brute force is three nested loops, O(n^3), which fails on large inputs. The smarter move is to fix one element and count pairs that satisfy the constraint relative to it, dropping to O(n^2) or O(n log n) depending on whether you use a hash table or sort. Common pitfall: miscounting duplicates or overthinking the indices. The constraint almost always involves arithmetic like arr[i] - arr[j] == target or similar. In the OA, if you blank on optimization, StealthCoder will surface the O(n^2) hash-table approach instantly so you don't crater.
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 Triples with 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 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.
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DataBricks 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 Triples with Diff FAQ
Is this just three nested loops?+
Not if they care about time. O(n^3) passes small inputs but fails at scale. The trick is to iterate one pointer and use a hash table or two-pointer technique to count valid pairs in O(n^2) or O(n log n). Read the constraint limits before committing.
How do I avoid off-by-one errors with indices?+
Enforce i < j < k from the start. Don't count the same triple twice. If you fix element at index i, only look at j and k where j > i and k > j. Write a clear loop condition and trace through a small example first.
What if there are duplicates in the array?+
Duplicates don't disqualify a triple if the indices are distinct. Use a hash table to count occurrences only if you're iterating pairs; don't double-count by confusing element values with element positions.
Should I sort the array?+
Sorting can help if the constraint is a range or ordering-based (like arr[i] < arr[j] < arr[k]). But sorting loses original indices. If the problem cares about indices, keep a copy or use index arithmetic carefully.
How much time do I have to optimize in the OA?+
DataBricks expects a working solution fast. Code the O(n^2) hash-table approach first. If it passes, done. If TLE, refactor to two-pointers or prefix sums. Have both strategies ready in your head before you start.