<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>

from collections import defaultdict
class Solution:
    def findShortestSubArray(self, nums: List[int]) -> int:
        d = defaultdict(list)
        dCount = defaultdict(int)
        for i in range(len(nums)):
            if len(d[nums[i]]) < 2:
                d[nums[i]].append(i)
            else:
                d[nums[i]][1] = i
            dCount[nums[i]] += 1
        dCount = dict(sorted(dCount.items(), key = lambda x: x[1]*-1))
        maxCount = list(dCount.values())[0]
        numToCheck = [k for k,v in dCount.items() if v == maxCount]
        degree = float(+inf)
        for n in numToCheck:
            ls = d[n]
            diff = d[n][1]-d[n][0] if len(ls) == 2 else d[n][0]
            degree = min(degree, diff+1)
        return degree
18700