<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 {
    /*
    * TWO POINTER APPROACH
    * Ref: https://leetcode.com/problems/maximum-score-of-a-good-subarray/solutions/1108333/java-c-python-two-pointers/
    */
public:
    Solution() {
        ios::sync_with_stdio(false);
        cin.tie(nullptr);
        cout.tie(nullptr);
    }
    int maximumScore(vector<int>& nums, int k) {
        int n=nums.size();
        int i=k, j=k, res=nums[k], mini=nums[k];

        while(i-1>=0 || j+1<n) {
            int left = (i-1>=0) ? nums[i-1] : INT_MIN;
            int right = (j+1<n) ? nums[j+1] : INT_MIN;

            if(left >= right) {
                mini = min(mini, left);
                i--;
                res = max(res, mini*(j-i+1));
            } else {
                mini = min(mini, right);
                j++;
                res = max(res, mini*(j-i+1));
            }
        }
        return res;
    }
};
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