<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, int> times;
        vector<int> candidate, length;

        int degree = 0, number = 0;

        for (int i : nums) {

            auto search = times.find(i);

            if (search == times.end())
                times.insert({i, 1});
            else
                times[i]++;
        }

        for (auto pair : times)
            if (pair.second >= degree)
                degree = pair.second;

        if(degree == 1)
            return 1;


        for (auto pair : times)
            if (pair.second == degree)
                candidate.push_back(pair.first);

        for (int j = 0; j < candidate.size(); j++) {

            int count = 1, firstPos = 0, lastPos = 0;

            number = candidate[j];

            for (int i = 0; i < nums.size(); i++) {

                if (nums[i] == number && count == 1) {
                    firstPos = i;
                    count++;
                }

                else if (nums[i] == number && count == degree) {
                    lastPos = i;
                    break;
                }

                else if (nums[i] == number)
                    count++;
            }
            length.push_back(lastPos - firstPos + 1);
        }

        int min = INT_MAX;

        for (int i : length)
            if (i < min)
                min = i;

        return min;
    }
};
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