<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) {
        // map<int, array<size_t, 3>> mp;
        unordered_map<int, array<size_t, 3>> mp;
        size_t maxx = 0;
        for(size_t i = 0; i < nums.size(); i++) {
            auto [it, ok] = mp.try_emplace(nums[i], array<size_t, 3>{1, i, i});
            auto& arr = it->second;
            if(!ok) {
                arr[0]++;
                arr[2] = i;
            }
            maxx = max(maxx, arr[0]);
        }

        size_t ans = SIZE_MAX;
        for(auto [num, arr] : mp) {
            if(arr[0] == maxx) {
                ans = min(ans, arr[2] - arr[1] + 1);
            }
        }
        return ans;
    }
};
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