<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>

# My idea: DP[i] = "Using current element as the min item, the score I can get"
# i.e., find where the boundary of my "min" can extend, left and right 

from collections import deque

class Solution:
    def maximumScore(self, nums: List[int], k: int) -> int:
        n = len(nums)
        leftStack = [(float("-inf"), -1)]
        leftSmaller = [-1 for _ in range(n)]

        for i in range(n):
            while leftStack[-1][0] >= nums[i]:
                leftStack.pop()
            leftSmaller[i] = leftStack[-1][1] # index of left larger
            leftStack.append((nums[i], i))
        # print(leftSmaller)

        rightStack = [(float("-inf"), n)]
        rightSmaller = [n for _ in range(n)]
        
        for i in range(n-1, -1, -1):
            while rightStack[-1][0] >= nums[i]:
                rightStack.pop()
            rightSmaller[i] = rightStack[-1][1]
            rightStack.append((nums[i], i))
        # print(rightSmaller)

        maxScore = float("-inf")
        for i in range(n):
            currLen = rightSmaller[i] - leftSmaller[i] - 1
            currScore = currLen * nums[i]

            # And don't forget the constraint that must have k inside the range!
            leftBound = leftSmaller[i] + 1
            rightBound = rightSmaller[i] - 1
            if k >= leftBound and k <= rightBound:
                maxScore = max(maxScore, currScore)

        return maxScore


        
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