<p>Given a non-empty array of non-negative integers <code>nums</code>, the <b>degree</b> of this array is defined as the maximum frequency of any one of its elements.</p>

<p>Your task is to find the smallest possible length of a (contiguous) subarray of <code>nums</code>, that has the same degree as <code>nums</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [1,2,2,3,1]
<strong>Output:</strong> 2
<strong>Explanation:</strong> 
The input array has a degree of 2 because both elements 1 and 2 appear twice.
Of the subarrays that have the same degree:
[1, 2, 2, 3, 1], [1, 2, 2, 3], [2, 2, 3, 1], [1, 2, 2], [2, 2, 3], [2, 2]
The shortest length is 2. So return 2.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,2,3,1,4,2]
<strong>Output:</strong> 6
<strong>Explanation:</strong> 
The degree is 3 because the element 2 is repeated 3 times.
So [2,2,3,1,4,2] is the shortest subarray, therefore returning 6.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>nums.length</code> will be between 1 and 50,000.</li>
	<li><code>nums[i]</code> will be an integer between 0 and 49,999.</li>
</ul>

class Solution {
public:
    int findShortestSubArray(vector<int>& nums) {
        array<int, 50000> frequencies {};
        array<int, 50000> starts {};
        array<int, 50000> ends {};
        starts.fill(-1);
        ends.fill(-1);

        vector<int> max_freqs {};
        int max_freq = 0;

        for (size_t i = 0; i < nums.size(); ++i) {
            int num = nums[i];
            int &frequence = frequencies[num];
            ++frequence;
            if (frequence == max_freq) {
                max_freqs.push_back(num);
            }
            else if (frequence > max_freq) {
                max_freqs.clear();
                max_freqs.push_back(num);
                max_freq = frequence;
            }

            if (starts[num] == -1) {
                starts[num] = i;
                ends[num] = i;
            }
            else {
                ends[num] = i;
            }
        }
        
        int min_len = numeric_limits<int>::max();

        for (int num : max_freqs) {
            int len = ends[num] - starts[num] + 1;
            min_len = min(len, min_len);
        }        

        return min_len;
    }
};
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