<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:
        # the smallest number in the range times the size of the range
        # such that k is in the range 
        n = len(nums)
        def solve(nums, k):
            # we want to know the min value for each left/right value at that index
            # then bsearch the farthest we go such that the right value is the 
            # min value, the farther we go the higher our value
            left = [0] * k
            running_min = inf
            for i in range(k-1,-1,-1):
                running_min = min(running_min, nums[i])
                left[i] = running_min
            right =[]
            running_min = inf
            for i in range(k, n):
                running_min = min(running_min, nums[i])
                right.append(running_min)
            ret = 0
            for j in range(len(right)):
                min_value = right[j]
                i = bisect_left(left, min_value)
                size = (k + j) - i + 1
                ret = max(ret, size*min_value)
            return ret
        
        return max(solve(nums, k), solve(nums[::-1], n-k-1))
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