<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 helper(self, nums, target):
        left = right = None
        for idx, num in enumerate(nums):
            if num != target: continue
            if left == None: left = idx
            right = idx
        
        return right - left + 1
        
    def findShortestSubArray(self, nums: List[int]) -> int:
        # find the degree, and corresponding number
        # find the leftmost instance of said number
        # find the rightmost instance of said number
        # return rightmost - leftmost
        numCt = defaultdict(int)
        
        for num in nums:
            numCt[num]+=1
            
        choices = list(reversed(sorted(numCt.items(), key=lambda x: x[1])))
        threshold = choices[0][1]
        choices = [x for x in choices if x[1]==threshold]
        
        return min([self.helper(nums, x[0]) for x in choices])
        

17800