<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 TwoPointer1(self, MinArr, Right, k):
        SubLeft = k + 2
        Result = 0
        for i in range(k + 1, -1, -1):
            for j in range(SubLeft, Right):
                if MinArr[i] > MinArr[j]:
                    SubLeft = j
                    Result = max(Result, MinArr[i] * (j - i))
                    break
        return Result
    
    def TwoPointer2(self, MinArr, Right, k):
        SubLeft = k
        Result = 0
        for i in range(k + 1, Right):
            for j in range(SubLeft, -1, -1):
                if MinArr[i] > MinArr[j]:
                    SubLeft = j
                    Result = max(Result, MinArr[i] * (i - j))
                    break
        return Result


    def maximumScore(self, nums: list[int], k: int) -> int:
        Lenght = len(nums)
        Minimum = [10**5] * (Lenght + 2)
        Minimum[k + 1] = nums[k]
        Minimum[Lenght + 1] = Minimum[0] = -10**5

        for i in range(k - 1, -1, -1):
            Minimum[i + 1] = min(Minimum[i + 2], nums[i])
        for i in range(k + 1, Lenght):
            Minimum[i + 1] = min(Minimum[i], nums[i])

        return max(self.TwoPointer1(Minimum, Lenght + 2, k), 
                   self.TwoPointer2(Minimum, Lenght + 2, k))
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