697 Degree of an Array

697. Degree of an Array

1. Question

Given a non-empty array of non-negative integersnums, 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 ofnums, that has the same degree asnums.

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.lengthwill be between 1 and 50,000.

nums[i]will be an integer between 0 and 49,999.

2. Implementation

(1) Hash Table

class Solution {
    public int findShortestSubArray(int[] nums) {
        Map<Integer, int[]> range = new HashMap<>();
        Map<Integer, Integer> count = new HashMap<>();

        int maxFreq = 0;

        for (int i = 0; i < nums.length; i++) {
            count.put(nums[i], count.getOrDefault(nums[i], 0) + 1);
            maxFreq = Math.max(maxFreq, count.get(nums[i]));

            range.putIfAbsent(nums[i], new int[] {-1, -1});
            int[] temp = range.get(nums[i]);
            if (temp[0] == -1) {
                temp[0] = i;
            }
            temp[1] = i;
        }

        int minLen = nums.length;

        for (int key : count.keySet()) {
            if (count.get(key) == maxFreq) {
                int[] temp = range.get(key);
                minLen = Math.min(minLen, temp[1] - temp[0] + 1);
            }
        }
        return minLen;
    }
}

3. Time & Space Complexity

Hash Table: 时间复杂度O(n), 空间复杂度O(n)