<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 number;
        unordered_map<int,pair<int,vector<int>>> hashmap;
        for(int i=0;i<nums.size();i++)
        {
            number  = nums[i];
            hashmap[number].first++;
            hashmap[number].second.push_back(i);
        }
        int degree = -1;
        for(auto it:hashmap)
        {
            if(it.second.first>degree){
                degree = it.second.first;
            }
        }
        
        int ans = nums.size();
            
        for(auto it:hashmap)
        {
            if(it.second.first==degree){
                int minimum = *min_element(it.second.second.begin(),it.second.second.end());
                int maximum = *max_element(it.second.second.begin(),it.second.second.end());
                
                if(maximum-minimum<ans){
                    ans = maximum - minimum;
                }
                
            }
        }
        
        return ans+1;
    }
};
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