# 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)

import itertools
import string
import time

def guess_password(real):
    chars = string.ascii_lowercase + string.ascii_uppercase + string.digits + string.punctuation
    attempts = 0
    for password_length in range(1, 9):
        for guess in itertools.product(chars, repeat=password_length):
            startTime = time.time()
            attempts += 1
            guess = ''.join(guess)
            if guess == real:
                return 'password is {}. found in {} guesses.'.format(guess, attempts)
            loopTime = (time.time() - startTime);
            print(guess, attempts, loopTime)

print("\nIt will take A REALLY LONG TIME to crack a long password. Try this out with a 3 or 4 letter password and see how this program works.\n")
val = input("Enter a password you want to crack that is 9 characters or below: ")
print(guess_password(val.lower()))

""" Binary Search Algorithm 
----------------------------------------
"""
#iterative implementation of binary search in Python
def binary_search(a_list, item):
    """Performs iterative binary search to find the position of an integer in a given, sorted, list.
    a_list -- sorted list of integers
    item -- integer you are searching for the position of
    """
    first = 0
    last = len(a_list) - 1
    while first <= last:
        i = (first + last) / 2
        if a_list[i] == item:
            return ' found at position '.format(item=item, i=i)
        elif a_list[i] > item:
            last = i - 1
        elif a_list[i] < item:
            first = i + 1
        else:
            return ' not found in the list'.format(item=item)
#recursive implementation of binary search in Python
def binary_search_recursive(a_list, item):
    """Performs recursive binary search of an integer in a given, sorted, list.
    a_list -- sorted list of integers
    item -- integer you are searching for the position of
    """
    first = 0
    last = len(a_list) - 1
    if len(a_list) == 0:
        return ' was not found in the list'.format(item=item)
    else:
        i = (first + last) // 2
        if item == a_list[i]:
            return ' found'.format(item=item)
        else:
            if a_list[i] < item:
                return binary_search_recursive(a_list[i+1:], item)
            else:
                return binary_search_recursive(a_list[:i], item)

# Function to check Armstrong number
def is_armstrong_number(number):
    # Convert number to string to iterate over its digits
    num_str = str(number)
    
    # Calculate the sum of the cubes of each digit
    digit_sum = sum(int(digit) ** len(num_str) for digit in num_str)
    
    # Compare the sum with the original number
    if digit_sum == number:
        return True
    else:
        return False

# Prompt user for a number
number = int(input("Enter a number: "))

# Check if the number is an Armstrong number
if is_armstrong_number(number):
    print(number, "is an Armstrong number.")
else:
    print(number, "is not an Armstrong number.")

import itertools

def compute_permutations(string):
    # Generate all permutations of the string
    permutations = itertools.permutations(string)

    # Convert each permutation tuple to a string
    permutations = [''.join(permutation) for permutation in permutations]

    return permutations

# Prompt the user for a string
string = input("Enter a string: ")

# Compute permutations
permutations = compute_permutations(string)

# Display the permutations
print("Permutations:")
for permutation in permutations:
    print(permutation)

# List
lst = [1, 2, 3, 'Alice', 'Alice']

# One-Liner
indices = [i for i in range(len(lst)) if lst[i]=='Alice']

# Result
print(indices)
# [3, 4]