<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>& a, int k) {
        int n = a.size();
        int next[n];
        int prev[n];
        stack<int> s;
        int l, idx, ans = -1 ;

        // ONE PASS SOLUTION 

        for(int i=0; i<n; i++) {
            while(!s.empty() && a[s.top()] > a[i]){
                idx = s.top();
                s.pop();
                l = s.empty() ? 0 : s.top();
                next[idx] = i;
                prev[idx] = l;
                if(a[prev[idx]] < a[idx]) prev[idx]++;
                if(a[next[idx]] < a[idx]) next[idx]--;

                if(next[idx]>=k && prev[idx]<=k) ans = max(ans, a[idx]*(next[idx]-prev[idx] + 1));
            }
            
            s.push(i);
        }

        while(!s.empty()){
            idx = s.top();
            s.pop();
            l = s.empty() ? 0 : s.top();
            next[idx] = n-1;
            prev[idx] = l;
            if(a[prev[idx]] < a[idx]) prev[idx]++;
            if(a[next[idx]] < a[idx])next[idx]--;
            if(next[idx]>=k && prev[idx]<=k)ans = max(ans, a[idx]*(next[idx]-prev[idx] + 1));
        }
        
        // for(int i=0; i<n; i++) cout << i << " : " << a[i] << "  " << next[i] << " " << prev[i] << endl;
        
        return ans;
    }
};

auto init = []() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  return '$';
}();
94000