# Python code to demonstrate
# method to remove i'th character
# Naive Method

# Initializing String
test_str = "CodeCatch"

# Printing original string
print ("The original string is : " + test_str)

# Removing char at pos 3
# using loop
new_str = ""

for i in range(len(test_str)):
	if i != 2:
		new_str = new_str + test_str[i]

# Printing string after removal
print ("The string after removal of i'th character : " + new_str)

from time import sleep

def delay(fn, ms, *args):
  sleep(ms / 1000)
  return fn(*args)

delay(lambda x: print(x), 1000, 'later') # prints 'later' after one second

my_list = ["blue", "red", "green"]

#1- Using sort or srted directly or with specifc keys
my_list.sort() #sorts alphabetically or in an ascending order for numeric data 
my_list = sorted(my_list, key=len) #sorts the list based on the length of the strings from shortest to longest. 
# You can use reverse=True to flip the order

#2- Using locale and functools 
import locale
from functools import cmp_to_key
my_list = sorted(my_list, key=cmp_to_key(locale.strcoll))

{"output":{"name":"format_final_json_response","arguments":{"output":{}}}}

my_list = [1, 2, 3, 4, 5]
removed_element = my_list.pop(2)  # Remove and return element at index 2
print(removed_element)  # 3
print(my_list)  # [1, 2, 4, 5]

last_element = my_list.pop()  # Remove and return the last element
print(last_element)  # 5
print(my_list)  # [1, 2, 4]

from collections import Counter

def find_parity_outliers(nums):
  return [
    x for x in nums
    if x % 2 != Counter([n % 2 for n in nums]).most_common()[0][0]
  ]

find_parity_outliers([1, 2, 3, 4, 6]) # [1, 3]