<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();
        if(n==1)
            return 1;

        unordered_map<int, int> mp;
        int val = INT_MIN;

        for(auto it : nums)
        {
            mp[it]++;
            val = max(mp[it] , val);
        }

        vector<int> vec;

        for(auto it : mp)
        {
            if(it.second == val)
                vec.push_back(it.first);
        }

        int result = INT_MAX;

        for(auto it : vec)
        {
            int i=0 , j=n-1;
            while(nums[i]!=it)
                i++;

            while(nums[j]!=it)
                j--;
            
            result = min(result , j-i);
        }

        return ++result;
    }
};
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