Is Adjacent Increasing Subarrays Detection II easy or hard?

Adjacent Increasing Subarrays Detection II is generally considered a Medium difficulty problem. The challenge is recognizing that strictly increasing segments form runs and that the answer can be derived from run lengths rather than checking every possible subarray.

Adjacent Increasing Subarrays Detection II Python/Java solution

Python and Java implementations usually compute increasing run lengths while iterating through the array. The optimized solution maintains current run length and previous run length, updating the maximum possible k during the scan. Both implementations run in O(n) time.

How to solve Adjacent Increasing Subarrays Detection II in O(n)?

Traverse the array and measure lengths of strictly increasing segments. For each segment of length L, you can place two adjacent increasing subarrays inside it with k = L / 2. For two consecutive segments with lengths L and R, you can place them across the boundary with k = min(L, R). Track the maximum k while scanning.

What is the best approach for Adjacent Increasing Subarrays Detection II?

The best approach is a single-pass run-length method. Treat the array as a series of strictly increasing runs and compute the maximum k using two cases: placing both subarrays inside one run (k = runLength / 2) or across two consecutive runs (k = min(leftRun, rightRun)). This runs in O(n) time and O(1) space.

Is Adjacent Increasing Subarrays Detection II asked at Google/Amazon/Meta?

Problems involving increasing runs and subarray boundaries frequently appear in interviews at companies like Amazon, Google, and Meta. While the exact problem ID may vary, recognizing increasing segments and optimizing to O(n) is a common interview pattern.

What data structure is used in Adjacent Increasing Subarrays Detection II?

The problem mainly uses array traversal and run-length tracking. A dynamic programming variant uses two arrays to store increasing lengths from the left and right, while the optimized solution only keeps integer counters.

What is the time complexity of Adjacent Increasing Subarrays Detection II?

The optimal solution runs in O(n) time because the array is scanned once while tracking lengths of increasing runs. A DP version using prefix and suffix arrays also runs in O(n) time but uses O(n) additional space.

Description Hints 2 Solution Video 15:51

3350. Adjacent Increasing Subarrays Detection II

Medium58.9% acceptance1 company

Given an array nums of n integers, your task is to find the maximum value of k for which there exist two adjacent subarrays of length k each, such that both subarrays are strictly increasing. Specifically, check if there are two subarrays of length k starting at indices a and b (a < b), where:

Both subarrays nums[a..a + k - 1] and nums[b..b + k - 1] are strictly increasing.
The subarrays must be adjacent, meaning b = a + k.

Return the maximum possible value of k.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [2,5,7,8,9,2,3,4,3,1]

Output: 3

Explanation:

The subarray starting at index 2 is [7, 8, 9], which is strictly increasing.
The subarray starting at index 5 is [2, 3, 4], which is also strictly increasing.
These two subarrays are adjacent, and 3 is the maximum possible value of k for which two such adjacent strictly increasing subarrays exist.

Example 2:

Input: nums = [1,2,3,4,4,4,4,5,6,7]

Output: 2

Explanation:

The subarray starting at index 0 is [1, 2], which is strictly increasing.
The subarray starting at index 2 is [3, 4], which is also strictly increasing.
These two subarrays are adjacent, and 2 is the maximum possible value of k for which two such adjacent strictly increasing subarrays exist.

Constraints:

2 <= nums.length <= 2 * 10⁵
-10⁹ <= nums[i] <= 10⁹

Asked at 1 company

Google

Topics

Array Binary Search

Hints open one at a time

Hint 1

Find the boundaries between strictly increasing subarrays.

Hint 2

Can we use binary search?

15:51

Still stuck?Read 3 approaches, or watch a walkthrough.

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1class Solution:
2    def maxIncreasingSubarrays(self, nums: List[int]) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

[2,5,7,8,9,2,3,4,3,1]

Input

[1,2,3,4,4,4,4,5,6,7]

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