text: # Program to find length of shortest sublist with maximum frequent element with same frequency in Python
  
  Suppose we have a list of numbers called nums. If the frequency of a most frequent number in nums is k. We have to find the length of a shortest sublist such that the frequency of its most frequent item is also k.
  
  So, if the input is like nums = [10, 20, 30, 40, 30, 10], then the output will be 3, because here the most frequent numbers are 10 and 30 , here k = 2. If we select the sublist [30, 40, 30] this is the shortest sublist where 30 is present and its frequency is also 2.
  
  To solve this, we will follow these steps −
  
  • L := size of nums
  • rnums := reverse of nums
  • d := a map containing frequencies of each elements present in nums
  • mx := maximum of list of all values of d
  • vs := a list of k for each k in d if d[k] is same as mx
  • mn := L
  • for each v in vs, do
  • mn := minimum of mn and ((L - (index of v in rnums) - (index of v in nums))
  • return mn
  
  ## Example
  
  Let us see the following implementation to get better understanding −
  
  from collections import Counter
  def solve(nums):
  L = len(nums)
  rnums = nums[::-1]
  
  d = Counter(nums)
  mx = max(d.values())
  vs = [k for k in d if d[k] == mx]
  
  mn = L
  for v in vs:
  mn = min(mn, (L - rnums.index(v)) - nums.index(v))
  return mn
  
  nums = [10, 20, 30, 40, 30, 10]
  print(solve(nums))
  
  ## Input
  
  [10, 20, 30, 40, 30, 10]
  
  
  ## Output
  
  3
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