Python | Append Odd element twice

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Given a list of numbers, the task is to create a new list from the initial list with the condition to append every odd element twice. Below are some ways to achieve the above task.

Method #1: Using list comprehension

Python3

# Python code to create a new list from initial list
# with condition to append every odd element twice.
 
# List initialization
Input = [1, 2, 3, 8, 9, 11]
 
# Using list comprehension 
Output = [elem for x in Input for elem in (x, )*(x % 2 + 1)]
 
# printing 
print("Initial list is:'", Input)
print("New list is:", Output)

Output

Initial list is:' [1, 2, 3, 8, 9, 11]
New list is: [1, 1, 2, 3, 3, 8, 9, 9, 11, 11]

Time Complexity: O(n), Here n is the size of the list.
Auxiliary Space: O(n)

Method #2: Using itertools

Python3

# Python code to create a new list from initial list
# with condition to append every odd element twice.
 
# Importing
from itertools import chain
 
# List initialization
Input = [1, 2, 3, 8, 9, 11]
 
# Using list comprehension and chain
Output = list(chain.from_iterable([i] 
            if i % 2 == 0 else [i]*2 for i in Input))
 
# printing 
print("Initial list is:'", Input)
print("New list is:", Output)

Output

Initial list is:' [1, 2, 3, 8, 9, 11]
New list is: [1, 1, 2, 3, 3, 8, 9, 9, 11, 11]

Time Complexity: O(n), Here n is the size of the list.
Auxiliary Space: O(n)

Method #3: Using Numpy array

Python3

# Python code to create a new list from initial list
# with condition to append every odd element twice.
 
# Importing
import numpy as np
 
# List initialization
Input = [1, 2, 3, 8, 9, 11]
Output = []
 
# Using Numpy repeat
for x in Input:
    (Output.extend(np.repeat(x, 2, axis = 0))
      if x % 2 == 1 else Output.append(x))
 
# printing 
print("Initial list is:'", Input)
print("New list is:", Output)

Output:

Initial list is: [1, 2, 3, 8, 9, 11]
New list is: [1, 1, 2, 3, 3, 8, 9, 9, 11, 11]

Time Complexity: O(n), Here n is the size of the list.
Auxiliary Space: O(n)

Method #4 : Using extend() method

Python3

# Python code to create a new list from initial list
# with condition to append every odd element twice.
 
# List initialization
Input = [1, 2, 3, 8, 9, 11]
 
 
Output = []
for i in Input:
    if(i%2!=0):
        Output.extend([i]*2)
    else:
        Output.append(i)
 
# printing
print("Initial list is:", Input)
print("New list is:", Output)

Output

Initial list is:' [1, 2, 3, 8, 9, 11]
New list is: [1, 1, 2, 3, 3, 8, 9, 9, 11, 11]

Time Complexity : O(N)
Auxiliary Space : O(N)

Method#5: Using Recursive method.

The algorithm creates a new list from the input list with the condition to append every odd element twice, using a recursive function.

Define a function append_odd_twice that takes an input list as its argument.
If the input list is empty, return an empty list (base case).
Otherwise, if the first element of the input list is odd, append it twice to the output list, otherwise append it once.
Recursively call the function with the rest of the input list and append the result to the output list.
Return the output list.

Python3

def append_odd_twice(input_list):
    # Base case: if the input list is empty, return an empty list
    if not input_list:
        return []
    else:
        # Recursive case: if the first element of the input list is odd,
        # append it twice to the output list, otherwise append it once.
        if input_list[0] % 2 == 1:
            return [input_list[0], input_list[0]] + append_odd_twice(input_list[1:])
        else:
            return [input_list[0]] + append_odd_twice(input_list[1:])
# List initialization
input_list = [1, 2, 3, 8, 9, 11]
 
# Call the recursive function to create the new list
output_list = append_odd_twice(input_list)
 
# Print the results
print("Initial list is:", input_list)
print("New list is:", output_list)

Output

Initial list is: [1, 2, 3, 8, 9, 11]
New list is: [1, 1, 2, 3, 3, 8, 9, 9, 11, 11]

The time complexity of the algorithm is O(n), where n is the length of the input list. The recursive function performs a constant amount of work for each element in the list.

The space complexity of the algorithm is also O(n), since the recursive function creates a new list for each recursive call.