<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> mpp;   
    unordered_map<int, int> first; 
    int n = nums.size();
    
    
    for (int i = 0; i < n; i++) {
        if (mpp.find(nums[i]) == mpp.end()) {
            first[nums[i]] = i;
        }
        mpp[nums[i]]++;
    }
    
    int degree = 0;
    for (auto it : mpp) {
        degree = max(degree, it.second); 
    }
    
    int ans = n; 
    
    unordered_map<int, int> count; 
    
    for (int i = 0, j = 0; j < n; j++) {
        count[nums[j]]++; 
        
        while (count[nums[j]] == degree) {
            ans = min(ans, j - i + 1);
            count[nums[i]]--; 
            if (count[nums[i]] == 0) {
                count.erase(nums[i]); 
            }
            i++; 
        }
    }
    
    return ans; 
    }
};
30397