answer 4

def reduce_ticket_price(tickets, k):
    n = len(tickets)
    if k == n:
        return "0"  # If we have to remove all digits, the result will be "0"

    result = []  # This will store the final digits

    for digit in tickets:
        # Remove digits from result if they are greater than current digit, and we still have k removals left
        while k > 0 and result and result[-1] > digit:
            result.pop()
            k -= 1
        result.append(digit)

    # If we still have removals left, remove from the end
    while k > 0:
        result.pop()
        k -= 1

    # Join the result and remove leading zeros
    final_result = ''.join(result).lstrip('0')

    # If the final result is empty, return '0'
    return final_result if final_result else '0'

# Test case 1
tickets = "203"
k = 2
print(reduce_ticket_price(tickets, k))  # Output: "0"

# Test case 2
tickets = "1432219"
k = 3
print(reduce_ticket_price(tickets, k))  # Output: "1219"