<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 Counter

# track the first location and the counter when go through the list
# a good example to use setdefault() and get() in dictionary
# there are two factors to query the degree of an array. 
class Solution:
    def findShortestSubArray(self, A):
        first, count, res, degree = {}, {}, 0, 0
        
        for i, a in enumerate(A):
            first.setdefault(a, i)  # setdefault modifies the dictionary, while get() does not modify the dictionary
            count[a] = count.get(a, 0) + 1 # counter
            
            if count[a] > degree:
                # get a new one, with larger degree.
                degree = count[a]
                res = i - first[a] + 1
            elif count[a] == degree:
                # the one candiate to consider the shortest one
                res = min(res, i - first[a] + 1)
                
        return res
17500