<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, int> m;
        map<int, vector<string>> n;
        map<int, int> n2;
        int maxi = 0;
        for (int num : nums)
        {
            m[num]++;
            maxi = max(maxi, m[num]);
        }

        for (int i = 0; i < nums.size(); i++)
        {
            if (m[nums[i]] == maxi)
            {

                n[nums[i]].push_back(to_string(i));
            }
        }

         for(auto it:n){
            cout<<it.first<<": "<<it.second[0]<<endl;
        }

        int mini = INT_MAX;

        
        for (auto it : n)
        {
            // n2[it.first] = (it.second.back() - '0') - (it.second[0].front() - '0') + 1;
            // mini = min(mini, n2[it.first]);
            int sum = 0;
            string st1 = "";
            string st2 = "";
            for(auto in : it.second.back()){
                st1+= in;
            }

             for(auto in : it.second.front()){
                st2+= in;
            }
            sum = stoi(st1) - stoi(st2) + 1;
            mini = min(mini,sum);
        }

        return mini;
    }
};
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