<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:
    def findShortestSubArray(self, nums: List[int]) -> int:
        count = Counter(nums)
        mx = max(count.values())
        maxs = set()
        for k,v in count.items():
            if v == mx:
                maxs.add(k)
        # print(maxs)
        j = len(nums)-1
        i = 0
        store = defaultdict(list)
        while i <= j:
            if nums[i] in maxs:
                store[nums[i]].append(i)
            if nums[j] in maxs:
                store[nums[j]].append(j)
            i +=1
            j -=1

        # print(store)
        ans = len((nums))
        for k,lis in store.items():
            diff = max(lis) - min(lis)
            ans = min(ans,diff)
        return ans+1
18400