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.

这个题目思路就是用Counter去计算数量, 然后用d1, d2 分别存每个element存在的最左端和最右端, 然后min(ans, d2[k] - d1[k] + 1), 这里ans初始化为len(ans).

Code

class Solution:

    def degreeArray(self, nums):

        d = collections.Counter(nums)

        fre = max(d.values())

        d1, d2, ans = {}, {}, len(nums)

        for index, num in enumerate(nums):

            if num not in d1:

                d1[num] = index

            d2[num] = index

        for k in d.keys():

            if d[k] == fre:

                ans = min(ans, d2[k] - d1[k] + 1)

        return ans