Count Increasing Quadruplets
A hard-tier problem at 34% community acceptance, tagged with Array, Dynamic Programming, Binary Indexed Tree. Reported in interviews at SAP and 0 others.
Count Increasing Quadruplets is a hard problem that's landed at SAP. You're counting ordered quadruples across an array where each element is strictly larger than the previous one. The acceptance rate is 33%, which means most people either miss the pattern or time out. The obvious nested-loop solution gets you fired. You need to recognize that this is a 4D counting problem that collapses into something tractable with dynamic programming or a data structure like a binary indexed tree. If you hit this during a live OA and freeze on the structure, StealthCoder surfaces a working solution in seconds, invisible to the proctor.
Companies that ask "Count Increasing Quadruplets"
Count Increasing Quadruplets is the kind of problem that decides whether you pass. StealthCoder reads the problem on screen and surfaces a working solution in under 2 seconds. Invisible to screen share. The proctor sees nothing. Built by an engineer who got tired of watching his cohort grind for six months and still get filtered at the OA stage.
Get StealthCoderThe trap is thinking you can brute force all quadruplets. You can't. The trick is building up counts incrementally: for each position, track how many valid triplets end before it, then multiply by the number of valid pairs before those, and so on. You're essentially doing DP where state i tracks increasing subsequences of length k ending at position i. A binary indexed tree or segment tree accelerates the range queries needed to count valid predecessors without recalculating. The other pitfall is off-by-one errors in the enumeration. When you sit down live and start coding the nested approach, you'll realize the quadratic or worse complexity won't pass. StealthCoder is your hedge for that moment: it delivers the DP or tree-based solution so you unblock and submit.
Pattern tags
You know the problem.
Make sure you actually pass it.
Count Increasing Quadruplets recycles across companies for a reason. It's hard-tier, and most candidates blank under the timer. StealthCoder is the hedge: an AI overlay invisible during screen share. It reads the problem and surfaces a working solution in under 2 seconds. Built by an engineer who got tired of watching his cohort grind for six months and still get filtered at the OA stage. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Count Increasing Quadruplets interview FAQ
Is this really just O(n^4) enumeration with pruning?+
No. You'd time out immediately. The problem requires either DP with state tracking for subsequence length, or a data structure like a binary indexed tree to count valid predecessors in O(log n) per element. Enumeration is the failure case most candidates hit.
Why is the acceptance rate so low at 33%?+
Most candidates don't recognize the subsequence-counting pattern or don't know how to apply a tree structure to speed up range queries. The problem feels array-heavy but demands DP or advanced indexing to pass time limits.
Do I need a binary indexed tree, or can pure DP work?+
Pure DP with O(n^2) or O(n^3) space and time can work if implemented tightly, depending on constraints. A binary indexed tree keeps you at O(n log n) and is safer. Both approaches require tracking partial counts by value range.
How does this relate to the other topics listed?+
Array is your input format. Dynamic Programming tracks how many k-length increasing subsequences exist. Binary Indexed Tree accelerates the range count queries. Enumeration means the final loop over candidate positions. Prefix Sum is often a simpler relative of the tree approach.
If SAP asked this, should I expect it at other companies?+
Not necessarily. It's a niche hard problem. One company report doesn't mean it's on the FAANG circuit. Prepare for it if you're targeting SAP specifically, or if you're drilling hard array-DP hybrids.
Want the actual problem statement? View "Count Increasing Quadruplets" on LeetCode →