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# Python | Print list elements in circular range

Given a list of elements, the task is to print element in group of k till n-iteration in circular range. Examples:

Input: list = [1, 2, 3, 4, 5, 6, 7], k = 3, n =10
Output:
[1, 2, 3]
[4, 5, 6]
[7, 1, 2]
[3, 4, 5]
[6, 7, 1]
[2, 3, 4]
[5, 6, 7]
[1, 2, 3]
[4, 5, 6]
[7, 1, 2]

Input: list = [10, 20, 30, 40, 50, 60, 70], k = 4, n = 5
Output:
[10, 20, 30, 40]
[50, 60, 70, 10]
[20, 30, 40, 50]
[60, 70, 10, 20]
[30, 40, 50, 60]

We can use itertools with zip to do this task. Example #1:

## Python3

 # Python code to print element in group# of 5 till 9 iteration in circular range. # Importingfrom itertools import cycle # list initializationList = [90, 99, 192, 0, 43, 55] # Defining no of iterationsn = 9 # Defining no of groupingk = 5 for index, *ans in zip(range(n), *[cycle(List)] * k):    # printing ans    print(ans)

Output:

[90, 99, 192, 0, 43]
[55, 90, 99, 192, 0]
[43, 55, 90, 99, 192]
[0, 43, 55, 90, 99]
[192, 0, 43, 55, 90]
[99, 192, 0, 43, 55]
[90, 99, 192, 0, 43]
[55, 90, 99, 192, 0]
[43, 55, 90, 99, 192]

Time complexity: O(n), where n is the length of the input list.
Auxiliary space: O(n), where n is the length of the output list.

Example #2:

## Python3

 # Python code to print element in group# of 6 till 4 iteration in circular range. # Importingfrom itertools import cycle # list initializationList = ['Geeks', 'for', 'geeks', 'is', 'portal'] # Defining no of iterationsn = 4 # Defining no of groupingk = 6 for index, *ans in zip(range(n), *[cycle(List)] * k):    # printing ans    print(ans)

Output:

['Geeks', 'for', 'geeks', 'is', 'portal', 'Geeks']
['for', 'geeks', 'is', 'portal', 'Geeks', 'for']
['geeks', 'is', 'portal', 'Geeks', 'for', 'geeks']
['is', 'portal', 'Geeks', 'for', 'geeks', 'is']

Example #3:  Using slicing

This approach works by calculating the start and end indices for each iteration using the modulus operator to ensure that the indices wrap around to the beginning of the list when they exceed the length of the list.

## Python3

 def print_circular_range(lst, k, n):    for i in range(n):        start = i * k % len(lst)        end = (start + k) % len(lst)        if end > start:            print(lst[start:end])        else:            print(lst[start:] + lst[:end]) # Example usagelst = [1, 2, 3, 4, 5, 6, 7]print_circular_range(lst, 3, 10)#This code is contributed by Edula Vinay Kumar Reddy

Output

[1, 2, 3]
[4, 5, 6]
[7, 1, 2]
[3, 4, 5]
[6, 7, 1]
[2, 3, 4]
[5, 6, 7]
[1, 2, 3]
[4, 5, 6]
[7, 1, 2]

Time complexity: O(n)

Auxiliary Space: O(1)

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