Degree of an Array.java

/*
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.

time = O(n) linear scan
space = O(n) for the map
*/

class Solution {
    public int findShortestSubArray(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        Map<Integer, Integer> left = new HashMap<>(), right = new HashMap<>(), count = new HashMap<>();
        int maxCount = 0;
        for (int i = 0; i < nums.length; i++) {
            if (!left.containsKey(nums[i])) {
                left.put(nums[i], i);
            }
            right.put(nums[i], i);
            count.put(nums[i], count.getOrDefault(nums[i], 0) + 1);
            maxCount = Math.max(maxCount, count.get(nums[i]));
        }
        
        int minLength = Integer.MAX_VALUE;
        for (Integer key : count.keySet()) {
            if (count.get(key) == maxCount) {
                minLength = Math.min(minLength, right.get(key) - left.get(key) + 1);
            }
        }
        return minLength;
    }
}