Count the Number of Incremovable Subarrays II
A hard-tier problem at 39% community acceptance, tagged with Array, Two Pointers, Binary Search. Reported in interviews at DE Shaw and 1 others.
Count the Number of Incremovable Subarrays II is a hard array problem that's been asked by DE Shaw and IBM. You're given an array and need to count subarrays whose removal leaves the remaining elements strictly increasing. With a 39% acceptance rate, most candidates either miss the two-pointer optimization or get stuck on the counting logic. If this problem shows up in your live OA and you blank on how to efficiently track valid subarrays after removal, StealthCoder surfaces a working solution in seconds while the proctor can't see it.
Companies that ask "Count the Number of Incremovable Subarrays II"
Count the Number of Incremovable Subarrays 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 a senior engineer who knows the OA is theater. This is the script.
Get StealthCoderThe trick here is recognizing that you don't need to simulate each removal. Instead, for every possible subarray removal, you validate whether the joined-up remainder is strictly increasing. The obvious brute force checks each subarray independently, which tanks on large inputs. Two pointers let you precompute left-increasing and right-increasing sequences from both ends, then for each candidate removal window, verify the join point in constant time. Binary search can optimize boundary lookups when the constraints allow. Most candidates try to iterate naively or miss that the join condition is the bottleneck. When you hit this live and the brute force times out, StealthCoder runs the optimal two-pointer approach invisibly.
Pattern tags
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Count the Number of Incremovable Subarrays 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 a senior engineer who knows the OA is theater. This is the script. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Count the Number of Incremovable Subarrays II interview FAQ
Is this really asked at DE Shaw and IBM, or is it mostly a leetcode-hard meme?+
It's confirmed asked by both DE Shaw and IBM according to hiring reports. It's a legitimate interview problem, not a meme. The 39% acceptance rate means most engineers who see it don't solve it cleanly under pressure, which is exactly why you should rehearse the two-pointer pattern now.
What's the actual trick I'm missing if my brute force times out?+
You're probably checking every removal independently without precomputing. Build arrays tracking the longest strictly increasing prefix and suffix. Then for each removal window, you only validate the join point where the prefix meets the suffix, not the entire remainder.
How does binary search fit into this problem?+
Once you've precomputed the increasing sequences, binary search helps you quickly find valid boundary positions where a removal window is minimal or maximal. It's optional if your two-pointer pass is tight, but it can reduce repeated linear scans.
Should I nail this before my OA, or is it too niche?+
It's hard enough that most candidates don't expect it, but the two-pointer and binary search fundamentals show up everywhere. Master the pattern on this problem and you'll recognize it in other array questions. If it appears in your OA, you'll be ahead of the 61% who don't pass it.
What's the most common mistake on this problem?+
Trying to rebuild and validate the array for every subarray removal without precomputation. This makes your solution O(n^3) or worse. Once you cache the increasing sequences on the left and right, the validation becomes O(1) per window.
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