# Degree of an Array

## Description

Given a non-empty array of non-negative integers `nums`, the **degree** of this array is defined as the maximum frequency of any one of its elements.

Your task is to find the smallest possible length of a (contiguous) subarray of `nums`, that has the same degree as `nums`.

**Example 1:**

```
Input: [1, 2, 2, 3, 1]
Output: 2
Explanation: 
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.
```

**Example 2:**

```
Input: [1,2,2,3,1,4,2]
Output: 6
```

**Note:**

* `nums.length` will be between 1 and 50,000.
* `nums[i]` will be an integer between 0 and 49,999.

## Solutions

```cpp
class Solution {
public:
    int findShortestSubArray(vector<int>& nums) {
        unordered_map<int, int> freqs;
        unordered_map<int, pair<int, int>> ranges;
        for(int i = 0; i < nums.size(); ++i){
            if(++freqs[nums[i]] == 1){
                ranges[nums[i]].first = i;
            }
            ranges[nums[i]].second = i;
        }
        int max_freq = 0;
        for(auto it = freqs.begin(); it != freqs.end(); ++it){
            max_freq = max(max_freq, it->second);
        }
        int min_len = nums.size();
        for(auto it = freqs.begin(); it != freqs.end(); ++it){
            if(it->second == max_freq){
                auto p = ranges[it->first];
                min_len = min(min_len, p.second - p.first + 1);
            }
        }
        return min_len;
    }
};
```

```cpp
class Solution {
public:
    int findShortestSubArray(vector<int>& nums) {
        unordered_map<int, vector<int>> hash;
        for(int i = 0; i < nums.size(); ++i){
            hash[nums[i]].push_back(i);
        }
        int max_freq = 0;
        int min_len = nums.size();
        for(auto it = hash.begin(); it != hash.end(); ++it){
            if(it->second.size() == max_freq){
                min_len = min(min_len, it->second.back() - it->second.front() + 1);
            }else if(it->second.size() > max_freq){
                max_freq = it->second.size();
                min_len = it->second.back() - it->second.front() + 1;
            }
        }
        return min_len;
    }
};
```
