Degree of an Array
<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>
Array; Hash Table
class Solution {
    public int findShortestSubArray(int[] nums) {
        Map<Integer, Integer> cnt = new HashMap<>();
        Map<Integer, Integer> left = new HashMap<>();
        Map<Integer, Integer> right = new HashMap<>();
        int degree = 0;
        for (int i = 0; i < nums.length; ++i) {
            int v = nums[i];
            cnt.put(v, cnt.getOrDefault(v, 0) + 1);
            degree = Math.max(degree, cnt.get(v));
            if (!left.containsKey(v)) {
                left.put(v, i);
            }
            right.put(v, i);
        }
        int ans = 1000000;
        for (int v : nums) {
            if (cnt.get(v) == degree) {
                int t = right.get(v) - left.get(v) + 1;
                if (ans > t) {
                    ans = t;
                }
            }
        }
        return ans;
    }
}