Degree of an array

Time: O(N); Space: O(N); easy

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: nums = [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: nums = [1,2,2,3,1,4,2]

Output: 6

Notes:

• nums.length will be between 1 and 50,000.

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

import collections

class Solution1(object):
    def findShortestSubArray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        counts = collections.Counter(nums)
        left, right = {}, {}
        for i, num in enumerate(nums):
            left.setdefault(num, i)
            right[num] = i
        degree = max(counts.values())
        return min(right[num] - left[num] + 1 \
                   for num in counts.keys() \
                   if counts[num] == degree)

s = Solution1()
nums = [1, 2, 2, 3, 1]
assert s.findShortestSubArray(nums) == 2
nums = [1,2,2,3,1,4,2]
assert s.findShortestSubArray(nums) == 6

See also:

https://leetcode.com/problems/degree-of-an-array

https://www.lintcode.com/problem/degree-of-an-array