# Time:  O(n)
# Space: O(n)

# 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.

class Solution(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)