Adjacent Increasing Subarrays Detection I

Adjacent Increasing Subarrays Detection I

What is this problem about?

The Adjacent Increasing Subarrays Detection I interview question asks you to analyze an array of integers and determine if there exist two "adjacent" subarrays, both of length $k$, that are strictly increasing. Adjacent means the second subarray starts immediately after the first one ends. Specifically, if the first subarray covers indices $[i, i + k - 1]$, the second must cover $[i + k, i + 2k - 1]$.

Why is this asked in interviews?

Companies like Microsoft and Amazon use this Adjacent Increasing Subarrays Detection I coding problem to test a candidate's ability to handle array indexing and window-based logic. It’s a foundational problem that evaluates how carefully you manage boundary conditions and whether you can solve a problem in a single linear pass rather than using nested loops.

Algorithmic pattern used

This problem primarily uses the Sliding Window or Linear Scan interview pattern. You need to keep track of the lengths of increasing sequences as you traverse the array. The core logic involves identifying the "breaks" in increasing order and checking if the current and previous increasing stretches are long enough to satisfy the $k$ requirement.

Example explanation

Imagine an array nums = [2, 5, 8, 3, 7, 10] and $k = 3$.

1. Check the first subarray of length 3: [2, 5, 8]. It is strictly increasing.
2. Check the immediately following subarray of length 3: [3, 7, 10]. It is also strictly increasing.
3. Since both are length 3 and are strictly increasing, the function returns true. If $k$ was 4, this would return false because we don't have enough elements to form two adjacent subarrays of that size.

Common mistakes candidates make

• Off-by-one errors: Miscalculating the starting index of the second subarray or the ending boundary of the array.
• Non-Strict Increases: Failing to check for "strictly" increasing conditions (e.g., treating [2, 2, 3] as increasing when it is actually non-decreasing).
• Inefficiency: Re-checking the "increasing" property for every possible starting index, which leads to $O(N \times k)$ time instead of the optimal $O(N)$.

Interview preparation tip

When dealing with "adjacent" structures, try to pre-calculate the length of the increasing sequence ending at or starting from each index. This often simplifies the logic from complex nested loops to a single comparison.