<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>& v) {
        int n=v.size();
        int k=0;
        map<int,int> m;
        for(int i=0;i<n;i++) m[v[i]]++;
        int num;
        for(auto it:m){
            if(k<it.second){
                k=it.second;
                num=it.first;
            }
        }
        int mx=0;
        for(int i=0;i<n;i++) mx=max(mx,v[i]);
        vector<int> leftPos(mx+1,-1);
        vector<int> rightPos(mx+1,-1);
        for(int i=0;i<n;i++){
            if(leftPos[v[i]]==-1) leftPos[v[i]]=i;
            rightPos[v[i]]=i;
        }
        int minLength=INT_MAX;
        for(int i=0;i<n;i++){
            if(m[v[i]]==k){
                ///cout<<v[i]<<endl;
              //  cout<<leftPos[i]<<" "<<rightPos[i]<<endl;
                minLength=min(minLength,rightPos[v[i]]-leftPos[v[i]]+1);
            }
        }
        return minLength;
    }

};
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