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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

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))

[10, 20, 30, 40, 30, 10]

3

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