Python Slicing | Reverse an array in groups of given size

Given an array, reverse every sub-array formed by consecutive k elements.

Examples:

Input: 
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9]
k = 3
Output:  
[3, 2, 1, 6, 5, 4, 9, 8, 7]

Input: 
arr = [1, 2, 3, 4, 5, 6, 7, 8]
k = 5
Output:  
[5, 4, 3, 2, 1, 8, 7, 6]

Input: 
arr = [1, 2, 3, 4, 5, 6]
k = 1
Output:  
[1, 2, 3, 4, 5, 6]

Input: 
arr = [1, 2, 3, 4, 5, 6, 7, 8]
k = 10
Output:  
[8, 7, 6, 5, 4, 3, 2, 1]

We have existing solution for this problem please refer Reverse an array in groups of given size link. We can solve this problem quickly in Python using list slicing and reversed() function.Below example will give you better understanding of approach.

Example:
Example

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# function to Reverse an array in groups of given size
  
def reverseGroup(input,k):
  
    # set starting index at 0
    start = 0
  
    # run a while loop len(input)/k times
    # because there will be len(input)/k number 
    # of groups of size k 
    result = []
    while (start<len(input)):
  
           # if length of group is less than k
           # that means we are left with only last 
           # group reverse remaining elements 
           if len(input[start:])<k:
                result = result + list(reversed(input[start:]))
                break
  
           # select current group of size of k
           # reverse it and concatenate 
           result = result + list(reversed(input[start:start + k]))
           start = start + k
    print(result)
  
# Driver program
if __name__ == "__main__":
    input = [1, 2, 3, 4, 5, 6, 7, 8]
    k = 5
    reverseGroup(input,k)

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Output:


[5, 4, 3, 2, 1, 8, 7, 6]





Using Direct Function











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# function to Reverse an array in groups of given size 


   


def reverseGroup(a, k): 


  


   # take an empty list 


   res = [] 


  


   # iterate over the list with increment of  


   # k times in each iteration 


   for i in range(0, len(a), k): 


       


       # reverse the list in each iteration over  


       # span of k elements using extend 


       res.extend((a[i:i + k])[::-1]) 


   print(res) 


   


# Driver program 


if __name__ == "__main__": 


    input = [1, 2, 3, 4, 5, 6, 7, 8] 


    k = 5


    reverseGroup(input, k) 



















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Output:


[5, 4, 3, 2, 1, 8, 7, 6]








    

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