<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, vector<int>> location;
        unordered_map<int, int> freq;

        for (int i = 0; i < nums.size(); i++){
            location[nums[i]].push_back(i);
            freq[nums[i]]++;
        }

        vector<vector<int>> vecCount;
        for (auto it: freq){
            vecCount.push_back({it.second, it.first});
        }
        sort(vecCount.begin(), vecCount.end());
        
        int degree = vecCount[vecCount.size() - 1][0];
        vector<vector<int>> temp;
        for (auto it : location){
            int left = INT_MAX, right = INT_MIN;
            for (int loc : location[it.first]){
                if (location[it.first].size() == degree){
                    left = min(loc, left);
                    right = max(loc, right);
                }
            }
            if (left != INT_MAX || right != INT_MIN){
                temp.push_back({left, right});
            }

        }
        int final = INT_MAX;
        for (int i = 0; i < temp.size(); i++){
            cout << temp[i][0] << " " << temp[i][1] << endl;
            final = min(final, temp[i][1] - temp[i][0] + 1);
        }

        return final;
    }
};
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