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. # Importing from itertools import cycle # list initialization List = [ 90 , 99 , 192 , 0 , 43 , 55 ] # Defining no of iterations n = 9 # Defining no of grouping k = 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. # Importing from itertools import cycle # list initialization List = [ 'Geeks' , 'for' , 'geeks' , 'is' , 'portal' ] # Defining no of iterations n = 4 # Defining no of grouping k = 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 usage lst = [ 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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