<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, pair<int, pair<int, int>>> mp;
        int len = nums.size();
        int max_freq = 0;
        int globalMinimum = len;
        for(int i=0; i<len; i++)
        {
            if(mp[nums[i]].first == 0)
            {
                mp[nums[i]].first++;
                mp[nums[i]].second.first = i;
                // corner case - max frequency is 1
                mp[nums[i]].second.second = i;
            }
            else
            {
                mp[nums[i]].first++;
                mp[nums[i]].second.second = i;
            }

            max_freq = max(max_freq, mp[nums[i]].first);
        }

        for(int i=0; i<len; i++)
        {
            if(mp[nums[i]].first == max_freq)
            {
                globalMinimum = min(globalMinimum, (mp[nums[i]].second.second - mp[nums[i]].second.first) + 1);
            }
        }

        return globalMinimum;
    }
};

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