<p>You are given an integer array <code>nums</code> and a <strong>positive</strong> integer <code>k</code>.</p>

<p>Return <em>the number of subarrays where the <strong>maximum</strong> element of </em><code>nums</code><em> appears <strong>at least</strong> </em><code>k</code><em> times in that subarray.</em></p>

<p>A <strong>subarray</strong> is a contiguous sequence of elements within an array.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,3,2,3,3], k = 2
<strong>Output:</strong> 6
<strong>Explanation:</strong> The subarrays that contain the element 3 at least 2 times are: [1,3,2,3], [1,3,2,3,3], [3,2,3], [3,2,3,3], [2,3,3] and [3,3].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,4,2,1], k = 3
<strong>Output:</strong> 0
<strong>Explanation:</strong> No subarray contains the element 4 at least 3 times.
</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;= 10<sup>6</sup></code></li>
	<li><code>1 &lt;= k &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    long long countSubarrays(vector<int>& nums, int k) {
        int max_num = 0;
        for(int i = 0; i < nums.size(); ++i){
            if(max_num < nums[i]){
                max_num = nums[i];
            }
        }
        vector<int> cnts(k + 1);
        long long tot_cnts = 0;
        for(int i = nums.size() - 1; i >= 0; --i){
            if(nums[i] == max_num){
                int tmp = cnts[k];
                for(int j = k; j > 0; --j){
                    cnts[j] = cnts[j-1];
                }
                cnts[k] += tmp;
                cnts[1] += 1;
                cnts[0] = 0;
            }else{
                cnts[0] +=1;
            }
            // std::cout << cnts[k] << std::endl;
            tot_cnts += cnts[k];
        }
        return tot_cnts;
    }
};
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