<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){
       
       int n = nums.size();
       map<int, int>freq;         //[number , freq]      
       int maxFreq = INT_MIN;

       for(int i=0 ; i<n ; i++)
       {
           freq[nums[i]]++;
           maxFreq = max(maxFreq, freq[nums[i]]);
       }

       set<int>st;   //contains elements with maximum freq

       for(auto it : freq)
       {
           if(it.second == maxFreq)
           {
               st.insert(it.first);
           }
       }

       map<int, int>startInd, endInd;    //[element , starting index]

       for(int i=0 ; i<n ; i++)
       {
           startInd[nums[i]] = -1;
           endInd[nums[i]] = -1;

       }

       for(int i=0 ; i<n; i++)
       {
           if(startInd[nums[i]] == -1)
           {
               //it's the first time the element has occured
               startInd[nums[i]] = i;
           }
       }

       for(int i=n-1 ; i>=0 ; i--)
       {
           if(endInd[nums[i]] == -1)
           {
               //last index
               endInd[nums[i]] = i;
           }
       }

       int len = INT_MAX;

       for(int i=0 ; i<n ; i++)
       {
           if(st.find(nums[i]) != st.end())     //only maxFreq element 
           {

              len = min(len , endInd[nums[i]] - startInd[nums[i]] + 1);
           }
       }

       return len;
        
    }
};
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