<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) {
        unordered_map<int, tuple<int, int, int>> occurrences;
        for (int i = 0; i < nums.size(); ++i) {
            auto res = occurrences.emplace(nums[i], make_tuple(1, i, i));
            if (!res.second) {
                get<0>(res.first->second)++;
                get<2>(res.first->second) = i;
            }
        }
        auto minLenIt = max_element(occurrences.begin(), occurrences.end(), [](const auto& el1, const auto& el2) {
            if (get<0>(el1.second) < get<0>(el2.second)) return true;
            if (get<0>(el1.second) > get<0>(el2.second)) return false;

            return get<2>(el1.second) - get<1>(el1.second) > get<2>(el2.second) - get<1>(el2.second);
        });
        // cout << minLenIt->first << ' ' << get<0>(minLenIt->second) << ' ' << get<1>(minLenIt->second) << ' ' << get<2>(minLenIt->second) << endl;
        return get<2>(minLenIt->second) - get<1>(minLenIt->second) + 1;
    }
};
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