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
# 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
# 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
# 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
# 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.
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.