Special Behavior Patterns (Sliding Window) — Reconstructed Version

Note: The exact wording of Amazon’s question titled “Special Behavior Patterns (Sliding Window)” does not appear to be publicly available.
This is a reconstructed version, based on partial public references to an Amazon OA/phone-screen problem involving “special behavior patterns” and a sliding-window/monotonic-stack style solution. The original text, examples, and constraints may differ.[1][2][3]

Problem Description

You are given an integer array behavior of length n and an integer k representing a window size.

For any contiguous subarray (window) of length k, say behavior[i..i + k - 1], define a position j (where i ≤ j ≤ i + k - 1 ) to be special within this window if:

• behavior[j] is strictly greater than every element to its right inside the same window, i.e.
  for all t such that j < t ≤ i + k - 1, behavior[j] > behavior[t].

Intuitively, within each window, a “special” index is a point where the current value is a new high relative to everything that comes after it in that window. The collection of all special indices within a given window forms that window’s behavior pattern.

Your task is to:

• Consider every contiguous window of length k in the array.
• For each window, count how many indices in that window are special.
• Return the sum of these counts over all windows.

Formally, if S(i) denotes the number of special positions in window behavior[i..i + k - 1], you must compute:

$$ \text{answer} = \sum_{i=0}^{n-k} S(i) $$

Input / Output Format

You may assume the problem is exposed as a function in your language of choice.

Function signature (one possible form):

• Input

  • behavior: integer array (e.g., int[] behavior in Java, vector<int>& behavior in C++, List[int] behavior in Python)
  • k: integer window size

• Output

  • An integer (typically 64-bit, e.g. long long / long) representing the sum of counts of special positions over all length‑k windows.

Example signatures

• C++:
  long long countSpecialBehaviorPatterns(const vector<int>& behavior, int k);

• Java:
  long countSpecialBehaviorPatterns(int[] behavior, int k);

• Python:
  def count_special_behavior_patterns(behavior: list[int], k: int) -> int:

Examples

Example 1

Input

• behavior = [3, 1, 4, 2]
• k = 3

All windows of size 3:

1. Window i = 0: behavior[0..2] = [3, 1, 4]

   • Index 0 (value 3): right side in window is [1, 4].
     • 3 > 1 is true, but 3 > 4 is false → not special.
   • Index 1 (value 1): right side is [4]; 1 > 4 is false → not special.
   • Index 2 (value 4): right side is empty; vacuously greater than all → special.

   So S(0) = 1.

2. Window i = 1: behavior[1..3] = [1, 4, 2]

   • Index 1 (value 1): right side [4, 2]; 1 > 4 is false → not special.
   • Index 2 (value 4): right side [2]; 4 > 2 is true → special.
   • Index 3 (value 2): right side empty → special.

   So S(1) = 2.

Total answer:

$$ \text{answer} = S(0) + S(1) = 1 + 2 = 3 $$

Output

• 3

Example 2

Input

• behavior = [5, 4, 3, 2, 1]
• k = 2

All windows of size 2:

1. [5, 4]

   • 5 > 4 → special
   • 4 has empty right side → special
     S(0) = 2

2. [4, 3]

   • 4 > 3 → special
   • 3 has empty right side → special
     S(1) = 2

3. [3, 2]

   • 3 > 2 → special
   • 2 has empty right side → special
     S(2) = 2

4. [2, 1]

   • 2 > 1 → special
   • 1 has empty right side → special
     S(3) = 2

Total:

$$ \text{answer} = 2 + 2 + 2 + 2 = 8 $$

Output

• 8

Example 3

Input

• behavior = [2, 7, 7, 3, 5]
• k = 3

Windows:

1. i = 0: [2, 7, 7]

   • Index 0 (2): right [7, 7]; 2 > 7 is false → not special.
   • Index 1 (7): right [7]; 7 > 7 is false (strict inequality) → not special.
   • Index 2 (7): right empty → special.
     S(0) = 1

2. i = 1: [7, 7, 3]

   • Index 1 (7): right [7, 3]; 7 > 7 is false → not special.
   • Index 2 (7): right [3]; 7 > 3 is true → special.
   • Index 3 (3): right empty → special.
     S(1) = 2

3. i = 2: [7, 3, 5]

   • Index 2 (7): right [3, 5]; 7 > 3 and 7 > 5 → special.
   • Index 3 (3): right [5]; 3 > 5 is false → not special.
   • Index 4 (5): right empty → special.
     S(2) = 2

Total:

$$ \text{answer} = 1 + 2 + 2 = 5 $$

Output

• 5

Constraints

These constraints are chosen to reflect a typical Amazon medium‑difficulty OA / phone-screen problem requiring an $$O(n)$$ or $$O(n \log n)$$ solution.

• 1 ≤ n ≤ 2 * 10^5
• 1 ≤ k ≤ n
• -10^9 ≤ behavior[i] ≤ 10^9 (or a similar 32‑bit integer range)
• The sum may require 64‑bit storage, so use a 64‑bit integer type for the result.

Solution Hint (High-Level)

Maintain a sliding window over the array and, for each window, keep a monotonic decreasing deque (or stack-like deque) of indices whose values are strictly decreasing from front to back:

• As you slide the window to the right:
  • When adding a new element at position r, pop from the back of the deque while behavior[r] ≥ behavior[deque.back()], then push r.
    This removes any element that is no longer greater than all elements to its right within the window.
  • Before moving the left boundary l forward, if deque.front() == l, pop from the front, because that index leaves the window.
• At any step, the deque contains exactly those indices in the current window whose values are strictly greater than everything to their right within that window—i.e., exactly the special positions for that window.
• Therefore, the contribution of the current window is simply deque.size().
  Summing this over all windows gives the answer in overall O(n) time.