<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) {
        
        long long answer = 0;
        
        queue<int> numbers;
        int N = nums.size();
        int maxElement = -1;
        
        for(auto num: nums){
            maxElement = max(maxElement, num);
        }
        
        int maxCount = 0;
        
        for(int i=0; i<N; i++){
            
            numbers.push(nums[i]);
            
            if(nums[i] == maxElement) maxCount += 1;
            
            long long fwdSubarrays = (long long)(N-i);
            
            if(maxCount == k){
            
                while(maxCount == k){
                    
                    answer += fwdSubarrays;
                    
                    if(numbers.front() == maxElement){
                        maxCount -= 1;
                    }
                    
                    numbers.pop();
                }
            }
        }
        
        return answer;
        
    }
};
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