<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) {
        int maxEle = *max_element(nums.begin(), nums.end());
        vector<tuple<int, int, int>> freqfirstlast(maxEle + 1, {-1, -1, -1});
        int maxFreq = 1;
        for(int i=0; i<nums.size(); i++){
          if(get<0>(freqfirstlast[nums[i]]) == -1){
            freqfirstlast[nums[i]] = make_tuple(1, i, i);
          }else{
            int freq = ++get<0>(freqfirstlast[nums[i]]);
            get<2>(freqfirstlast[nums[i]]) = i;
            maxFreq = max(maxFreq, freq);
          }
        }
        int minLen = INT_MAX, first =0, last = 0;
        for(int i=0; i<maxEle + 1; i++){
          if(get<0>(freqfirstlast[i]) == maxFreq){
            if(get<2>(freqfirstlast[i]) - get<1>(freqfirstlast[i]) + 1 < minLen){
              first = get<1>(freqfirstlast[i]);
              last = get<2>(freqfirstlast[i]);
              minLen = last - first + 1;
            }
          }
        }
        cout<<first<<" "<<last<<endl;
        return last - first + 1;
    }
};
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