<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 {
public:
    int maximumScore(vector<int>& nums, int k) {
        int n = nums.size();
        
        // Arrays to store the left and right boundaries for each element
        vector<int> left(n), right(n);
        
        // Initialize left boundaries
        for (int i = 0; i < n; i++) {
            left[i] = i;
            while (left[i] > 0 && nums[left[i] - 1] >= nums[i]) {
                left[i] = left[left[i] - 1];  // Move leftwards while nums are greater than or equal
            }
        }
        
        // Initialize right boundaries
        for (int i = n - 1; i >= 0; i--) {
            right[i] = i;
            while (right[i] < n - 1 && nums[right[i] + 1] >= nums[i]) {
                right[i] = right[right[i] + 1];  // Move rightwards while nums are greater than or equal
            }
        }
        
        // Calculate the maximum score for subarrays that include index k
        int maxScore = 0;
        for (int i = 0; i < n; i++) {
            if (left[i] <= k && right[i] >= k) {  // Check if the subarray includes k
                int score = nums[i] * (right[i] - left[i] + 1);  // Calculate score for subarray
                maxScore = max(maxScore, score);
            }
        }
        
        return maxScore;
    }
};

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