Count Strictly Increasing Subarrays
A medium-tier problem at 71% community acceptance, tagged with Array, Math, Dynamic Programming. Reported in interviews at Bridgewater Associates and 1 others.
Count Strictly Increasing Subarrays hits you when you expect a standard DP problem but the trick isn't about memoization. You're asked to count all contiguous subarrays where every element is strictly greater than the one before it. Bridgewater and J.P. Morgan both ask this, and the acceptance rate sits at 71%, which means a solid chunk of candidates miss the optimization that separates brute force from elegant. The problem looks simple until you realize that naive counting of every subarray explodes your runtime.
Companies that ask "Count Strictly Increasing Subarrays"
Count Strictly Increasing Subarrays 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 Amazon engineer who used it to pass JPMorgan's OA and system design loop.
Get StealthCoderThe pattern here is realizing you don't need to iterate through every possible subarray. Instead, track the current streak of strictly increasing elements. Every time the streak ends, you calculate how many subarrays that streak contains using a math trick: a streak of length n gives you n*(n+1)/2 total subarrays. Most candidates fall into the trap of nested loops, checking each subarray individually. The real insight is that you can solve this in a single pass. You track the length of the current increasing run, reset it when you hit a non-increasing pair, and accumulate the count formula each time. If this pattern doesn't click during the live assessment and you hit a wall, StealthCoder surfaces the optimized one-pass solution in seconds, invisible to the proctor.
Pattern tags
You know the problem.
Make sure you actually pass it.
Count Strictly Increasing Subarrays recycles across companies for a reason. It's medium-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 Amazon engineer who used it to pass JPMorgan's OA and system design loop. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Count Strictly Increasing Subarrays interview FAQ
Is this really just about the math formula for counting subarrays?+
Yes. Once you see that a strictly increasing streak of length n contains exactly n*(n+1)/2 subarrays, the problem becomes trivial. The hard part is recognizing that pattern applies here and abandoning the nested-loop instinct. Most wrong submissions come from candidates who code the brute force correctly but run out of time.
Why do Bridgewater and J.P. Morgan ask this?+
It tests whether you can move past the obvious solution to spot a mathematical shortcut. Both firms value pattern recognition and efficiency, not just correct code. A candidate who jumps to nested loops without thinking shows weaker algorithmic maturity than one who pauses and spots the streak approach.
Does Array, Math, and Dynamic Programming mean I need DP here?+
No. The Dynamic Programming tag is misleading for this problem. You don't need state tables or memoization. The math component is the real hint: the formula for counting subarrays from a streak length. Array is just the input type. Don't overthink the tags.
What's the common mistake that tanks submissions?+
Writing O(n^2) or O(n^3) code that checks every subarray explicitly. With large inputs, this times out. The correct approach is O(n) single pass. Candidates who code carefully but don't optimize for the counting trick pass smaller test cases but fail hidden ones.
How do I practice this if I can't find it on problem banks?+
Understand the core insight: how to count subarrays from a contiguous segment without enumerating them. Then apply it to this specific problem. If you blank during the real OA, StealthCoder handles it instantly. That's the safety net.
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