Find Num of Interesting Pairs
Reported by candidates from DE Shaw's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
You're seeing a pair-counting problem from DE Shaw in May 2024, and the title alone tells you they want you to optimize past brute force. Interesting pairs usually means some property holds between two elements, and the naive approach will time out. You'll need to recognize the underlying structure, whether that's a math trick, a hash-based lookup, or a sorting advantage. This is exactly the kind of problem where a blank moment during the OA can cost you, so knowing the pattern beforehand is your safety net.
Pattern and pitfall
Without the full problem text, the pattern isn't explicit, but pair-counting problems at DE Shaw typically lean on hash-tables or sorting. The word 'interesting' is the signal: it means a specific relationship between elements i and j that you have to check. The trap is iterating all pairs and checking each one, which is O(n^2) and won't pass. The real solution usually involves either hashing to avoid redundant checks, sorting to skip impossible pairs, or a mathematical property that reduces the search space. StealthCoder can read the constraint and spot the pattern in seconds during your OA, so you won't be guessing between three different approaches under pressure.
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Find Num of Interesting Pairs FAQ
What makes a pair 'interesting' in this context?+
The problem text would define it explicitly, but it's typically a condition on two indices i and j, like their values satisfy an equation, their difference is within bounds, or a formula evaluates to true. The 'interesting' framing is a hint that naive checking will be slow. You need to invert the problem.
Will hash-table or sorting work for this?+
Both can work depending on the constraint. If the condition is checkable via a lookup (e.g., 'does value Y exist'), hash-table is fast. If the condition depends on order or ranges, sorting plus two-pointers or binary search is cleaner. Read the constraint carefully and pick the one that avoids O(n^2).
How do I avoid double-counting pairs?+
Only iterate i from 0 to n-1 and j from i+1 to n-1, or use a two-pointer approach after sorting. Don't loop both indices freely or you'll count (i,j) and (j,i) separately. The problem likely specifies ordered or unordered pairs, so check that first.
What's the common pitfall here?+
Writing O(n^2) code and assuming it'll pass because the title sounds simple. DE Shaw expects optimized solutions. If your first instinct is nested loops, stop and think about whether you can precompute, hash, or sort to skip redundant work.
Can I solve this in 48 hours of prep?+
If you already know hash-tables and sorting, yes. The key is recognizing the pattern when you see the full problem text in the OA. Spend your time understanding what 'interesting' means, then apply the right tool. That's it.