<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) {
        map<int, int> count;
        
        for(int num : nums){
            if(count.find(num) == count.end()){
                count[num] = 1;
            }else{
                count[num] += 1;
            }
        }
        
        int maxfreq, maxele;
        vector<int> candidates;
        
        map<int, int>::iterator it = max_element(count.begin(), count.end(), [] (const pair<int, int> & p1, const pair<int, int> & p2) {return p1.second < p2.second;});
        maxfreq = it->second;
        
        for(map<int, int>::iterator it = count.begin(); it!=count.end(); it++){
            if(it->second == maxfreq){
                candidates.push_back(it->first);
            }
        }
        
        int ans = INT_MAX;
        for(int candidate : candidates){
            vector<int> tofind = {candidate};
            int substart = find(nums.begin(), nums.end(), candidate) - nums.begin();
            int subend = find_end(nums.begin(), nums.end(), tofind.begin(), tofind.end()) - nums.begin();
            cout << maxele << " " << substart << " " << subend << endl;
            ans = min(ans, subend - substart + 1);
        }
        return ans;
    }
};
31162