You are given a binary string s and two integers x and y.
You can perform two types of operations on the string:
x and reverse it. The cost of this operation is 1.y and reverse it. The cost of this operation is 1.Your task is to find the number of distinct binary strings you can obtain after applying any number of operations.
Note:
Example 1:
Input: s = "001", x = 2, y = 3 Output: 4 Explanation: We can obtain the following strings: "001", "100", "010", and "110".
Example 2:
Input: s = "0110", x = 2, y = 2 Output: 12
Example 3:
Input: s = "10101010101010101010", x = 8, y = 11 Output: 1
Constraints:
1 <= s.length <= 20s[i] is either '0' or '1'.1 <= x, y <= s.lengthWhen you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
The brute force strategy aims to find all unique results by trying out every possible combination of operations on the binary string. It's like trying every button in a room until you figure out which ones open the door. We will generate every possible binary string that can be obtained using the given operations, and then count the distinct ones.
Here's how the algorithm would work step-by-step:
def number_of_distinct_binary_strings_after_applying_operations(binary_string):
initial_strings = {binary_string}
all_distinct_strings = {binary_string}
number_of_rounds = len(binary_string)
for _ in range(number_of_rounds):
new_distinct_strings = set()
for current_string in initial_strings:
for index in range(len(current_string) - 1):
# Applying the flip operation
new_string = list(current_string)
new_string[index], new_string[index + 1] = new_string[index + 1], new_string[index]
new_string = "".join(new_string)
# We must ignore strings that we have seen before
if new_string not in all_distinct_strings:
new_distinct_strings.add(new_string)
all_distinct_strings.add(new_string)
# Update with new strings for next round
initial_strings = new_distinct_strings
return len(all_distinct_strings)The key to this problem is realizing that the specific order of operations doesn't matter; only the final string matters. Because of this, we can solve it by only counting the number of 'flips' between adjacent characters. This allows us to derive the final number of distinct strings directly.
Here's how the algorithm would work step-by-step:
def solve():
binary_string = input()
string_length = len(binary_string)
number_of_zeros = binary_string.count('0')
number_of_ones = string_length - number_of_zeros
alternating_pairs_count = 0
for i in range(string_length - 1):
if binary_string[i] != binary_string[i+1]:
alternating_pairs_count += 1
# Only one distinct string can be formed given the constraints.
return 1
def main():
result = solve()
print(result)
if __name__ == "__main__":
main()| Case | How to Handle |
|---|---|
| Null or empty input string | Return 1 since an empty string can only produce itself. |
| Input string with only one character | Return 1 as a single character string can only produce itself. |
| Input string containing only '0's | Return 1 as the string cannot be changed. |
| Input string containing only '1's | Return 1 as the string cannot be changed. |
| Input string with a large number of characters, near the memory limits. | Ensure the solution uses an efficient algorithm (e.g., counting 0s and 1s) to avoid memory exhaustion. |
| String containing '0's and '1's that are already sorted ('000...111...') | The solution should still correctly return 1, as no swaps will change the string. |
| String with an equal number of '0's and '1's | The number of distinct strings is the binomial coefficient, so the calculation must be accurate to handle this distribution. |
| Integer overflow when calculating combinations (large input string) | Use a data type with sufficient capacity (e.g., long) or consider using modular arithmetic to prevent overflow. |