<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>

int f[50000];
class Solution {
public:
    int findShortestSubArray(vector<int>& v) {
        int n=v.size(),d=1,m=1;
        //map<int,int> f;
        vector<int> a;
        memset(f,0,sizeof(int)*50000);
        for(int i=0;i<n;++i){
            //if(v[i-1]==v[i]) d++;
            //else d=1;
            //m=max(d,m);
            f[v[i]]++;
            m=max(m,f[v[i]]);
            //if(m==f[v[i]]){
                //a.push_back(v[i]);
           // }
        }
        for(int i=0;i<n;i++)
            if(m==f[v[i]]) 
                a.push_back(v[i]);
        //cout<<d;
        int j=0,k=n-1;
        int ans=1e9;
        //cout<<'a'<<endl;
        for(int i=0;i<a.size();++i){
            j=0;
            k=n-1;
            d=a[i];
            while(v[j]!=d) {
                j++;
            }
            while(v[k]!=d) k--;
            //cout<<j<<' '<<k<<endl;
            ans=min(ans,k-j+1);
        }
        
        
        return ans;
    }
};
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