# Program to find length of shortest sublist with maximum frequent element with same frequency in Python

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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

## 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