<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) {
        // finding the largest rectangle of the histogram

        // to store index
        stack<int> stk;

        // only compute the are when current element is smaller than stk.top()
        // only take the area range which involves k 
        int nMax = 0;
        for( int i = 0; i <= nums.size(); i++ ) {
            int nNum = i < nums.size() ? nums[i] : 0;
            while( !stk.empty() && nNum < nums[stk.top()] ) {
                int ind = stk.top();
                int value = nums[ind];
                stk.pop();
                int leftInd = stk.empty() ? -1 : stk.top();

                //printf( "%d, %d, %d\n", i, nNum, leftInd );

                if( k >= leftInd + 1 && k <= i - 1 ) {
                    int len = i - leftInd - 1;      // (range) leftInd+1, i-1
                    int area = len * value;
                    //printf( "area: %d, %d, %d\n", len, area, value );
                    nMax = max( area, nMax );
                }
            }

            stk.push(i);
        }

        return nMax;
    }
};
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