<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) {
       vector<int>pref(k+1 , 0)  , suff ; 
        pref[k] = nums[k] ;
        for(int i=k-1  ; i>=0 ;i--){
            pref[i] = (min(pref[i+1] , nums[i])) ;
        }
        suff.push_back(nums[k]) ;
        
        for(int i=k+1 ;i<nums.size() ;i++){
            suff.push_back(min(nums[i] , suff[i-k-1] )) ;
        }
        
       int left=pref.size() , right=0 ;
        int ans =INT_MIN ;
        
               while (left > 0 || right < suff.size() - 1) {
            if (left == 0 || (right < suff.size() - 1 && suff[right + 1] > pref[left - 1])) {
                right++;
            } else {
                left--;
            }
                   int  min_val = min(pref[left] , suff[right]) ;
ans= max(ans  , min_val*(right-left+k+1)) ;
               }     
               

        return ans;
    }
};
100000