<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, vector<int>> m;
        int mx = INT_MIN;

        int i=0;
        for(auto& n: nums)
        {
            if(m[n].size() == 0)
            {
                vector<int> v{1, i, i};
                m[n] = v;
            }
            else
            {
                m[n][0] += 1;
                m[n][2] = i;
            }
            mx = max(mx, m[n][0]);
            i++;
        }

        int md = INT_MAX;
        for(auto& c: m)
        {
            cout << c.first<< " " << c.second[0] << " " << c.second[1] << " " << c.second[2] << endl;
            if(c.second[0] == mx) md = min(md, c.second[2]-c.second[1]+1);
        }


        return md;
    }
};
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