Python – Count Maximum consecution of K in N consecutive batches

Given a list, the task is to write a Python Program to count a maximum number of times K occurs consecutively in each batch of size N.

Example:

Input : test_list = [4, 4, 5, 4, 4, 4, 1, 2, 3, 4, 4, 4, 4, 4, 4, 4, 5, 6, 7, 4, 4, 5, 3, 4, 4, 4, 7, 4, 4, 4,  5, 6, 3, 5, 4, 4, 4, 6, 4, 4, 1, 4, 2, 4, 4], N = 15, K = 4

Output : [6, 3, 3]

Explanation : For first batch of 15 elements, 4 occurs in consecution, 2, 3 and 6. times. Since 6 is maximum hence one of output.

Input : test_list = [4, 4, 5, 4, 4, 4, 1, 2, 3, 4, 4, 4, 4, 4, 4, 4, 5, 6, 7, 4, 4, 5, 3, 4, 4, 4, 7, 4, 4, 4], N = 15, K = 4

Output : [6, 3]

Explanation : For first batch of 15 elements, 4 occurs in consecution, 2, 3 and 6. times. Since 6 is maximum hence one of output.

Method : Using groupby() + max() + list comprehension + slicing 

In this, we get each consecution using slicing and using range to jump K elements to start grouping for each batch. The max() gets the maximum length of K consecutive group in each batch.

Python3

# Python3 code to demonstrate working of
# Maximum consecution of K in N consecutive batches
# Using groupby() + max() + list comprehension + slicing
from itertools import groupby

# initializing list
test_list = [4, 4, 5, 4, 4, 4, 1, 2, 3, 4, 4, 4, 4, 4, 4,
             4, 5, 6, 7, 4, 4, 5, 3, 4, 4, 4, 7, 4, 4, 4,
             5, 6, 3, 5, 4, 4, 4, 6, 4, 4, 1, 4, 2, 4, 4]

# printing original list
print("The original list is : " + str(test_list))

# initializing N
N = 15

# initializing K
K = 4

# max() getting max. consecution of K, ranges being evaluated using slices
# and skips using range()
res = [max(len(list(group)) for ele, group in groupby(sub) if ele == K)
          for sub in (test_list[idx : idx + N]
          for idx in range(0, len(test_list), N))]

# printing result
print("Maximum consecution of K for each batch : " + str(res))

Output:

The original list is : [4, 4, 5, 4, 4, 4, 1, 2, 3, 4, 4, 4, 4, 4, 4, 4, 5, 6, 7, 4, 4, 5, 3, 4, 4, 4, 7, 4, 4, 4, 5, 6, 3, 5, 4, 4, 4, 6, 4, 4, 1, 4, 2, 4, 4]

Maximum consecution of K for each batch : [6, 3, 3]

Time Complexity: O(n*n)
Auxiliary Space: O(n)