password.py

import string

CONDITIONS = input()
[N, U, L] = [int(x) for x in CONDITIONS.split(" ")]
K_TH = int(input())

# this conversion here is used for easy sorting, 1, 2, 3 = L, U, N
BIGGEST_PERMU = list('1' * L + '2' * U + '3' * N)

def find_kth(current, k):
    if not current:
        return []
    # we begin by finding the significance of the leftest most digit
    tail_numbers, tail_uppers, tail_lowers = current[1:].count('1'), current[1:].count('2'), current[1:].count('3')

    digit_significance = 10**tail_numbers * 26**tail_uppers * 26**tail_lowers

    # what base this digit has
    i = 10 if current[0] == '1' else 26

    success = False
    while i > 0:
        # in decremental order if a digit that fits
        if i * digit_significance <= k:
            success = True
            break
        i = i - 1

    if success:
        # if we manage to actually find such a digit
        sub_k = k - i * digit_significance
        next = current[1:]
        next.sort()

        # we build this digit and recursively call the remaining list
        return [build_digit(current[0], i)] + find_kth(next, sub_k)
    else:
        # if we couldn't find such a digit, either we "shuffle" or "hit base"
        if current[0] == '2': # U
            next_number = current[1:].find('1') + 1 if '1' in current[1:] else 0
            # if we can reduce the group somehow then shuffle down
            if next_number:
                return find_kth(swap(current, 0, next_number), k)

        if current[0] == '3': # L
            next_upper = current[1:].find('2') + 1 if '2' in current[1:] else 0
            if next_upper:
                return find_kth(swap(current, 0, next_upper), k)

            next_number = current[1:].find('1') + 1 if '2' in current[1:] else 0
            if next_number:
                return find_kth(swap(current, 0, next_number), k)

    # if we simply cannot shuffle down, then we "hit base"
    next = current[1:]
    next.sort()

    # and we recursively call the remaining list
    return [build_digit(current[0], i)] + find_kth(next, k)

def swap(l, x, y):
    l[x], l[y] = l[y], l[x]
    return l

def build_digit(type, index):
    if type == '3':
        return string.ascii_lowercase[index]
    if type == '2':
        return string.ascii_uppercase[index]
    if type == '1':
        return str(index)

print(find_kth(BIGGEST_PERMU, K_TH))