<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:
typedef long long ll; 
typedef long double ld;
#define pb push_back
#define be(a) a.begin(), a.end()
#define vi vector<int> 
#define vl vector<ll>
#define vs vector<string>
#define vc vector<char>
#define yes cout<<"YES"<<endl;
#define no cout<<"NO"<<endl;
#define loop(i,n) for(int i=0;i<n;i++)
    int findShortestSubArray(vector<int>& nums) 
    {
        map<ll,vl> num;
        set<ll> cute;
        map<ll,ll> ct;
        ll maxi=-1;
        loop(i,nums.size())
        {
            cute.insert(nums[i]);
            ct[nums[i]]++;
            maxi=max(maxi,ct[nums[i]]);
            num[nums[i]].pb(i);
        }

        ll mini=1000000;
        for(auto& val:cute)
        {
            if(ct[val]==maxi)
            {
                mini=min(mini,abs(num[val][0]-num[val][num[val].size()-1])+1);
            }
        }
        return mini;
    }
};
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