Find the Number of Subarrays Where Boundary Elements Are Maximum

Find the Number of Subarrays Where Boundary Elements Are Maximum

What is this problem about?

In the Find the Number of Subarrays Where Boundary Elements Are Maximum coding problem, you are given an array of integers. You need to count the number of subarrays where both the first and the last elements are equal to the maximum element within that specific subarray. For example, in [3, 1, 3], the subarray is valid because 3 is the max and both boundaries are 3. In [3, 4, 3], the subarray is invalid because 4 is the max.

Why is this asked in interviews?

LinkedIn and Amazon ask this "Hard" problem to test your understanding of Monotonic Stack interview patterns. It evaluations whether you can maintain a running state of "active" ranges while ensuring that no larger element "breaks" those ranges. It’s a test of efficient stack-based searching and counting.

Algorithmic pattern used

This problem is solved using a Monotonic Stack and Frequency Counting.

1. Iterate through the array. Maintain a stack of [value, count] pairs where the values are in strictly decreasing order.
2. For each element $X$:
   • While the stack is not empty and the top value is smaller than $X$:
     • Pop the stack. These smaller values can no longer be part of a valid subarray boundary because $X$ will now be the maximum in any subarray spanning them.
   • If the stack's top value equals $X$:
     • The number of valid subarrays ending here is top.count + 1.
     • Increment the count of the top element.
   • If the stack is empty or the top value is larger than $X$:
     • Push [X, 1] onto the stack. One valid subarray exists (just the element $X$ itself).
3. The total count is the sum of all "contribution" values found during the process.

Example explanation

nums = [1, 3, 2, 3, 3]

1. 1: Push [1, 1]. Total = 1.
2. 3: 3 > 1. Pop [1, 1]. Push [3, 1]. Total = $1+1 = 2$.
3. 2: 2 < 3. Push [2, 1]. Total = $2+1 = 3$.
4. 3: 3 > 2. Pop [2, 1]. Stack top is 3. top.count = 1. New count = 2. Total = $3+2 = 5$.
5. 3: Stack top is 3. top.count = 2. New count = 3. Total = $5+3 = 8$. Result: 8.

Common mistakes candidates make

• Brute Force: Checking every subarray ($O(N^2)$), which is too slow.
• Missing intermediate elements: Not realizing that any element in between the boundaries must be $\le$ the boundaries.
• Stack order: Using a non-monotonic stack, which makes it hard to know when a boundary is "invalidated" by a larger number.

Interview preparation tip

When a problem involves "Maximum/Minimum in a subarray," the Monotonic Stack is almost always the key. It allows you to process elements in a way that respects their "area of dominance" in linear time.