Find the Count of Monotonic Pairs II
A hard-tier problem at 23% community acceptance, tagged with Array, Math, Dynamic Programming. Reported in interviews at BNY Mellon and 0 others.
Find the Count of Monotonic Pairs II is a hard combinatorics problem that shows up in BNY Mellon interviews, and it hits different from typical array work. You're not just iterating or sorting. The problem asks you to count valid pairs based on monotonic constraints, which means you need to understand how mathematical relationships between elements constrain your solution space. At 23% acceptance, most candidates either miss the combinatorial structure entirely or build a solution that times out on larger inputs. If this lands on your OA and you blank on the counting trick, StealthCoder surfaces a working approach in seconds while staying invisible to the proctor.
Companies that ask "Find the Count of Monotonic Pairs II"
Find the Count of Monotonic Pairs II 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 realized the OA tests how well you memorized 200 problems, not how well you code.
Get StealthCoderThe core insight is recognizing that monotonic pair constraints aren't solvable by brute force enumeration. You'll need to reformulate the problem using dynamic programming or combinatorial math, often with prefix sums to avoid recalculating overlapping ranges. The trap is trying to iterate through all pairs and check each one, which fails on scale. Most candidates either misunderstand how the monotonic property applies across both elements of the pair, or they implement a correct DP but forget to optimize the state transitions. The trick often involves recognizing that valid pair counts reduce to a math formula once you fix one variable, letting you use combinations or prefix calculations to jump to the answer. StealthCoder is your hedge if the combinatorial pattern doesn't click during the live assessment, giving you the code structure and optimizations without seconds wasted on trial and error.
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Find the Count of Monotonic Pairs II 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 Amazon engineer who realized the OA tests how well you memorized 200 problems, not how well you code. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Find the Count of Monotonic Pairs II interview FAQ
What's the actual trick to counting monotonic pairs?+
You can't brute-force check all pairs. Reformulate: fix one variable, derive the constraint on the other as a math inequality or range, then count valid values using combinatorics or prefix sums. This collapses what looks like O(n^2) enumeration into a faster traversal with mathematical lookups.
Is this still asked at finance companies like BNY Mellon?+
Yes. It's a BNY Mellon problem. Quant and finance firms value candidates who think in terms of mathematical constraints and efficient counting. This problem tests both. Expect it in quantitative or systematic trading roles.
Why does the greedy or obvious array approach fail here?+
Monotonic pair counting isn't about selecting or ordering elements. It's about counting how many (x, y) tuples satisfy both the monotonic property and value bounds. Greedy won't enumerate the solution space correctly. You need DP or combinatorics.
How do prefix sums help with this problem?+
Once you derive the range of valid values for one pair element given the other, prefix sums let you count how many valid values exist in O(1) instead of O(n). This is the optimization that keeps your solution under time limits.
Is combinatorics knowledge required to solve this?+
Not formally. Understanding that you're counting distinct outcomes under constraints is enough. You'll use combinations or factorial math to evaluate your DP results, but the main barrier is setting up the recurrence or formula correctly, not combinatorics theory.
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