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Time Needed to Rearrange a Binary String

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Strings

You are given a binary string s. In one second, all occurrences of "01" are simultaneously replaced with "10". This process repeats until no occurrences of "01" exist.

Return the number of seconds needed to complete this process.

Example 1:

Input: s = "0110101"
Output: 4
Explanation: 
After one second, s becomes "1011010".
After another second, s becomes "1101100".
After the third second, s becomes "1110100".
After the fourth second, s becomes "1111000".
No occurrence of "01" exists any longer, and the process needed 4 seconds to complete,
so we return 4.

Example 2:

Input: s = "11100"
Output: 0
Explanation:
No occurrence of "01" exists in s, and the processes needed 0 seconds to complete,
so we return 0.

Constraints:

1 <= s.length <= 1000
s[i] is either '0' or '1'.

Follow up:

Can you solve this problem in O(n) time complexity?

Time Needed to Rearrange a Binary String

Medium

PayPal

7 views

Topics:

Strings

You are given a binary string s. In one second, all occurrences of "01" are simultaneously replaced with "10". This process repeats until no occurrences of "01" exist.

Return the number of seconds needed to complete this process.

Example 1:

Input: s = "0110101"
Output: 4
Explanation: 
After one second, s becomes "1011010".
After another second, s becomes "1101100".
After the third second, s becomes "1110100".
After the fourth second, s becomes "1111000".
No occurrence of "01" exists any longer, and the process needed 4 seconds to complete,
so we return 4.

Example 2:

Input: s = "11100"
Output: 0
Explanation:
No occurrence of "01" exists in s, and the processes needed 0 seconds to complete,
so we return 0.

Constraints:

1 <= s.length <= 1000
s[i] is either '0' or '1'.

Follow up:

Can you solve this problem in O(n) time complexity?

Solution

Clarifying Questions

When 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:

What is the maximum length of the binary string `s`?
Can the input string `s` be empty or null?
If the string initially contains no occurrences of '01', should I return 0?
Are there any specific limitations on memory usage I should be aware of?
Is the replacement of '01' with '10' truly simultaneous? Meaning, if I have '0101', does it become '1010' in one step, or do intermediate replacements affect subsequent ones in the same second?

Brute Force Solution

Approach

The brute force method for rearranging a binary string involves repeatedly swapping adjacent characters. We keep doing this until the string is sorted according to the desired pattern, calculating the time taken for each swap.

Here's how the algorithm would work step-by-step:

Start with the original binary string.
Look at all possible pairs of adjacent characters that are out of order (a '1' before a '0').
For each such pair, imagine swapping them.
Check if this swap makes the string closer to the sorted version.
Keep track of how much time it takes to do each swap.
Repeat this process of finding out-of-order pairs, swapping them, and recording the time, until the string is completely sorted with all '0's before all '1's.
The total time taken for all the swaps is the answer.

Code Implementation

def time_needed_to_rearrange(
    binary_string):

    total_time = 0

    while True:
        swapped = False
        string_list = list(binary_string)

        for index in range(
            len(string_list) - 1):

            # Check for out-of-order pairs ('1' before '0')
            if (
                string_list[index] == '1'
                and string_list[index + 1] == '0'):

                # Swap the adjacent characters
                string_list[index], \
                    string_list[index + 1] = \
                    string_list[index + 1], \
                    string_list[index]


                binary_string = "".join(
                    string_list)
                total_time += 1
                swapped = True
                break #Only perform one swap per pass


        # If no swaps were made, the string is sorted
        if not swapped:
            break

    return total_time

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates until the binary string of length n is sorted, which means all '0's are before all '1's. In the worst-case scenario, each '1' might need to move from the end to the beginning of the string. In each iteration, we scan the string to find adjacent '10' pairs, taking O(n) time. The maximum number of swaps needed in the worst case is proportional to the number of inversions, which can be n * (n-1) / 2 in the worst case (e.g., "111...1000...0"). Therefore, the overall time complexity is O(n * number of swaps) which can be O(n * n) in the worst case, thus O(n²).

Space Complexity

O(1) – The provided algorithm operates primarily by swapping adjacent characters in the input string. It does not mention the creation of any auxiliary data structures such as lists, arrays, or hash maps. The space complexity depends solely on the number of extra variables the algorithm requires, which appears to be a constant number of variables like loop indices or temporary variables for swapping. Therefore, the space complexity remains constant, irrespective of the input string's length N.

Optimal Solution

Approach

The key idea is to analyze the string from right to left. For each '0', we need to find the earliest '1' to its right to swap with. By counting these swaps, we effectively determine the maximum time needed for the rearrangement.

Here's how the algorithm would work step-by-step:

Start by looking at the very end of the string.
Move backwards through the string, one position at a time.
Whenever you find a '0', start searching for a '1' to its right.
If you find a '1' to the right of the '0', it means that '0' needs to move.
The time it takes for that '0' to move depends on how many '1's are still waiting to move to the left of it.
Keep a running total of the number of '1's that are waiting to move.
Whenever you find a '1', increase the waiting '1's count.
When you find a '0', add the number of waiting '1's to a 'total time' count, because that '0' will eventually need to move past all of them.
As you move through the string, keep track of the maximum value for the total time count. This will give the final answer.

Code Implementation

def time_needed_to_rearrange(binary_string):
    waiting_ones_count = 0
    total_time = 0
    maximum_time = 0

    # Iterate backwards through the string.
    for i in range(len(binary_string) - 1, -1, -1):
        if binary_string[i] == '1':
            waiting_ones_count += 1

        # Count time based on waiting ones.
        elif binary_string[i] == '0':
            total_time += waiting_ones_count

            # Keep track of the maximum time.
            maximum_time = max(maximum_time, total_time)

    return maximum_time

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through the binary string of length n from right to left using a single loop. Inside this loop, when a '0' is encountered, it potentially searches for a '1' to its right. In the worst-case scenario, for each '0', the algorithm might iterate through the remaining portion of the string to find a '1'. This nested behavior, where an inner loop (searching for '1') depends on the outer loop (iterating through the string), leads to a quadratic time complexity. The number of operations can approximate n * n/2, which simplifies to O(n²).

Space Complexity

O(1) – The algorithm uses a fixed number of integer variables: one to track the count of waiting '1's, and one to store the maximum time. The amount of memory used by these variables does not depend on the length of the input string. Therefore, the space complexity is constant, O(1).

Edge Cases

Case	How to Handle
Empty string	Return 0, as no rearrangement is needed for an empty string.
String with length 1	Return 0, as a single character string has no '01' sequence.
String with all '0's	Return 0, as a string with only '0's has no '01' sequence.
String with all '1's	Return 0, as a string with only '1's has no '01' sequence.
String with a large number of '01' sequences	The algorithm should efficiently handle a large number of swaps without exceeding time limits.
String alternates '01' many times like '01010101'	Ensure that the swapping logic correctly handles alternating sequences.
String with a single '0' followed by many '1's ('011111')	The algorithm should efficiently handle the rearrangement in one step.
String with many '0's followed by a single '1' ('000001')	The algorithm should identify no '01' sequence exists and return 0

Solution

Clarifying Questions

When 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:

What is the maximum length of the binary string `s`?
Can the input string `s` be empty or null?
If the string initially contains no occurrences of '01', should I return 0?
Are there any specific limitations on memory usage I should be aware of?
Is the replacement of '01' with '10' truly simultaneous? Meaning, if I have '0101', does it become '1010' in one step, or do intermediate replacements affect subsequent ones in the same second?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start with the original binary string.
Look at all possible pairs of adjacent characters that are out of order (a '1' before a '0').
For each such pair, imagine swapping them.
Check if this swap makes the string closer to the sorted version.
Keep track of how much time it takes to do each swap.
Repeat this process of finding out-of-order pairs, swapping them, and recording the time, until the string is completely sorted with all '0's before all '1's.
The total time taken for all the swaps is the answer.

Code Implementation

def time_needed_to_rearrange(
    binary_string):

    total_time = 0

    while True:
        swapped = False
        string_list = list(binary_string)

        for index in range(
            len(string_list) - 1):

            # Check for out-of-order pairs ('1' before '0')
            if (
                string_list[index] == '1'
                and string_list[index + 1] == '0'):

                # Swap the adjacent characters
                string_list[index], \
                    string_list[index + 1] = \
                    string_list[index + 1], \
                    string_list[index]


                binary_string = "".join(
                    string_list)
                total_time += 1
                swapped = True
                break #Only perform one swap per pass


        # If no swaps were made, the string is sorted
        if not swapped:
            break

    return total_time

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Start by looking at the very end of the string.
Move backwards through the string, one position at a time.
Whenever you find a '0', start searching for a '1' to its right.
If you find a '1' to the right of the '0', it means that '0' needs to move.
The time it takes for that '0' to move depends on how many '1's are still waiting to move to the left of it.
Keep a running total of the number of '1's that are waiting to move.
Whenever you find a '1', increase the waiting '1's count.
When you find a '0', add the number of waiting '1's to a 'total time' count, because that '0' will eventually need to move past all of them.
As you move through the string, keep track of the maximum value for the total time count. This will give the final answer.

Code Implementation

def time_needed_to_rearrange(binary_string):
    waiting_ones_count = 0
    total_time = 0
    maximum_time = 0

    # Iterate backwards through the string.
    for i in range(len(binary_string) - 1, -1, -1):
        if binary_string[i] == '1':
            waiting_ones_count += 1

        # Count time based on waiting ones.
        elif binary_string[i] == '0':
            total_time += waiting_ones_count

            # Keep track of the maximum time.
            maximum_time = max(maximum_time, total_time)

    return maximum_time

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty string	Return 0, as no rearrangement is needed for an empty string.
String with length 1	Return 0, as a single character string has no '01' sequence.
String with all '0's	Return 0, as a string with only '0's has no '01' sequence.
String with all '1's	Return 0, as a string with only '1's has no '01' sequence.
String with a large number of '01' sequences	The algorithm should efficiently handle a large number of swaps without exceeding time limits.
String alternates '01' many times like '01010101'	Ensure that the swapping logic correctly handles alternating sequences.
String with a single '0' followed by many '1's ('011111')	The algorithm should efficiently handle the rearrangement in one step.
String with many '0's followed by a single '1' ('000001')	The algorithm should identify no '01' sequence exists and return 0