<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 {
    struct Entry {
        int degree;
        int start;
        int end;

        int length() {
            return this->end - this->start + 1;
        }
    };

public:
    int findShortestSubArray(vector<int>& nums) {
        map<int, Entry> entries;

        // Iterate over nums and update entries for each number to keep track of start, end, and degree
        for (int i = 0; i < size(nums); i++) {
            int n = nums[i];
            auto it = entries.find(n);
            if (it != entries.end()) {
                it->second.end = i;
                it->second.degree += 1;
            }
            else {
                entries[n] = Entry(1, i, i);
            }
        }

        Entry max(0, 0, 0);
        // Iterate over the values in the entries and find the one with the highest degree and the smallest range
        for (auto& [key, entry] : entries) {
            if (entry.degree > max.degree || (entry.degree == max.degree && entry.length() < max.length())) {
                max = entry;
            }
        }

        return max.length();
    }
};
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