<p>You are given an array of integers <code>nums</code> <strong>(0-indexed)</strong> and an integer <code>k</code>.</p>

<p>The <strong>score</strong> of a subarray <code>(i, j)</code> is defined as <code>min(nums[i], nums[i+1], ..., nums[j]) * (j - i + 1)</code>. A <strong>good</strong> subarray is a subarray where <code>i &lt;= k &lt;= j</code>.</p>

<p>Return <em>the maximum possible <strong>score</strong> of a <strong>good</strong> subarray.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,4,3,7,4,5], k = 3
<strong>Output:</strong> 15
<strong>Explanation:</strong> The optimal subarray is (1, 5) with a score of min(4,3,7,4,5) * (5-1+1) = 3 * 5 = 15. 
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [5,5,4,5,4,1,1,1], k = 0
<strong>Output:</strong> 20
<strong>Explanation:</strong> The optimal subarray is (0, 4) with a score of min(5,5,4,5,4) * (4-0+1) = 4 * 5 = 20.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= k &lt; nums.length</code></li>
</ul>

class Solution:
    def maximumScore(self, nums: List[int], k: int) -> int:
        n = len(nums)

        leftBoundary = [-1] * n  # Store the next lower element from the left.
        rightBoundary = [n] * n  # Store the next lower element from the right.

        leftStack = []  # Monotonic Stack to help calculate left array.
        rightStack = []  # Monotonic Stack to help calculate right array.

        # Calculate next lower element from the left.
        for i in range(n - 1, -1, -1):
            while leftStack and nums[leftStack[-1]] > nums[i]:
                # If the current element is smaller than elements in the stack,
                # set the left boundary for those elements to the current element's index.
                leftBoundary[leftStack[-1]] = i
                leftStack.pop()
            leftStack.append(i)

        # Calculate next lower element from the right.
        for i in range(n):
            while rightStack and nums[rightStack[-1]] > nums[i]:
                # If the current element is smaller than elements in the stack,
                # set the right boundary for those elements to the current element's index.
                rightBoundary[rightStack[-1]] = i
                rightStack.pop()
            rightStack.append(i)

        maxScore = 0  # Initialize the maximum score to 0.

        # Calculate the maximum score for good subarrays.
        for i in range(n):
            if leftBoundary[i] < k and rightBoundary[i] > k:
                # Calculate the score for the subarray that contains element 'k'.
                subarrayScore = nums[i] * (rightBoundary[i] - leftBoundary[i] - 1)
                maxScore = max(maxScore, subarrayScore)  # Update the maximum score.

        return maxScore  # Return the maximum score for a good subarray.
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