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
"""
count={}
left={}
right={}
for i,x in enumerate(nums):
if x not in left: left[x]=i
right[x]=i
count[x]=count.get(x,0)+1 degree=max(count.values())
ans=len(nums) for x in count:
if count[x]==degree:
if ans>right[x]-left[x]+1:
ans=right[x]-left[x]+1
return ans

  

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