<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 {
public:
    int findShortestSubArray(vector<int>& nums) {
        unordered_map<int, int> freq, first, last;
        int degree = 0, minLength = nums.size();

        // Collect frequency, first occurrence, and last occurrence of each element
        for (int i = 0; i < nums.size(); i++) {
            int n = nums[i];
            freq[n]++;
            if (first.find(n) == first.end()) {
                first[n] = i;  // Mark the first occurrence
            }
            last[n] = i;  // Continuously update the last occurrence
            degree = max(degree, freq[n]);  // Update the degree
        }

        // Find the minimum length of subarrays with the same degree
        for (auto& [num, count] : freq) {
            if (count == degree) {
                minLength = min(minLength, last[num] - first[num] + 1);
            }
        }

        return minLength;
    }
};
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