<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>

#define FOR(i, n) for(int i=0; i<n; i++)
#define IFOR(i, a, b) for(int i=a; i<b; i++)

class Solution {
public:
    int maximumScore(vector<int>& nums, int k) {
        int n = nums.size();
        int ans = nums[k];
        // Itertaion 1
        stack<int> st1;
        FOR(i, k+1) st1.push(i);
        int mn1 = nums[k];
        IFOR(i, k, n) {
            mn1 = min(mn1, nums[i]);
            while(!st1.empty()) {
                int tp = st1.top();
                if(nums[tp] >= mn1) st1.pop();
                else break;
            }
            if(st1.size() == 0) {
                ans = max(ans, mn1*(i+1));
            } else {
                ans = max(ans, mn1*(i-st1.top()));
            }
        }

        // Itertaion 2
        stack<int> st2;
        for(int i=n-1; i>=k; i--) st2.push(i);
        int mn2 = nums[k];
        for(int i=k; i>=0; i--) {
            mn2 = min(mn2, nums[i]);
            while(!st2.empty()) {
                int tp = st2.top();
                if(nums[tp] >= mn2) st2.pop();
                else break;
            }
            if(st2.size() == 0) {
                ans = max(ans, mn2*(n-i));
            } else {
                ans = max(ans, mn2*(st2.top()-i));
            }
        }

        return ans;
    }
};
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