<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:
        
        n = len(nums)

        num_frequency = {}

        for i in range(n):
            if nums[i] not in num_frequency:
                num_frequency[nums[i]] = {"start":i, "end":i, "freq":1}
            else:
                num_frequency[nums[i]]["end"] = i
                num_frequency[nums[i]]["freq"] += 1

        smallest_subarray_len = float("inf")
        max_frequency = float("-inf")

        for num in num_frequency:
            
            freq = num_frequency[num]["freq"]
            sub_array_len = num_frequency[num]["end"] - num_frequency[num]["start"] + 1

            if freq > max_frequency:
                max_frequency = freq
                smallest_subarray_len = sub_array_len
            elif freq == max_frequency and sub_array_len < smallest_subarray_len:
                max_frequency = freq
                smallest_subarray_len = sub_array_len

        return smallest_subarray_len
18600