Problem Description

You are given a binary string binary that contains only the characters '0' and '1'. Two types of operations can be performed on this string an unlimited number of times:

• Operation 1: If the string contains the substring "00", you can replace it with "10".
• Operation 2: If the string contains the substring "10", you can replace it with "01".

Your task is to determine the highest-value binary string possible after any number of these operations. Here, the "value" of a binary string is defined by its decimal equivalent. For example, the binary string "1010" has a decimal equivalent of 10, while "1100" has a decimal equivalent of 12.

The goal is to strategically apply these operations to the initial string to transform it into the highest possible numeric value in binary form.

Intuition

To solve this problem, we should clearly understand the effect of each operation. Let's analyze the operations:

• Operation 1 ("00" -> "10"): This operation increases the value of the binary number. So whenever we see "00", we want to perform this operation.
• Operation 2 ("10" -> "01"): This operation does not increase the value; instead, it shifts the higher value bit ('1') to the right. We may use this to move '1' bits towards the end of the string (right-most side).

Intuitively, we can see that to maximize the binary value, we should have the most number of '1's towards the left side as possible, except possibly one '0' which will be necessary if we started with at least one '0'.

Here are the steps to find the solution:

1. Find the first occurrence of '0'. If there is no '0' in the string, the binary string is already at its maximum value, so return it as is.

2. Count the number of '0's after the first '0'. Every '0' except the last one can be turned into a '1' by using Operation 1 followed by Operation 2 as needed to shift the '1's to the left.

3. Knowing how many '0's there are in total, we can build the resulting string. There will be a sequence of '1's up to the position where the last '0' should be, then one '0', followed by '1's for the rest of the string.

The code achieves this by first looking for the first '0' then counting all subsequent '0's, which tells us where the last '0' will end up. After that, it constructs the string based on the counts.

Learn more about Greedy patterns.

Solution Approach

The solution approach is fairly straightforward with a focus on string manipulation. The key insight here is realizing the final form of the maximum binary string after all possible operations.

The string will be composed of three sections:

• A prefix of all '1's up to the position before the last '0'.
• A single '0'.
• A suffix of all '1's after the last '0'.

Knowing this final form allows us to skip simulating each operation and directly construct the final maximum binary string. Here is how the algorithm does it step by step:

1. The given binary string is first checked for the presence of '0'. This is done using Python's string find method, which returns the index of the first occurrence of a substring—in this case, '0'.

2. If no '0' is found, it means the string is already the maximum binary string possible (since it's composed of only '1's), so the original string is returned.

3. If a '0' is found, we calculate the index k where the last '0' would be after all possible operations. This is done by first setting k to the index of the first '0' found, then increasing k by the count of remaining '0's in the string after the first '0'. This effectively "moves" the last '0' to its final position.

4. Now that we know where the last '0' will be, we construct the final string. This is done by concatenating the following:

   • '1' * k, which creates a string of '1's that has the same length as the number of '1's that should be on the left side of the last '0'.
   • '0', which places the last '0'.
   • '1' * (len(binary) - k - 1), which creates the suffix consisting of the remaining number of '1's after the last '0'.

And that is it. There are no complex data structures involved—only simple string operations. The algorithm is efficient because it calculates the final string in one pass without actually performing the transformations, and the entire logic is implemented within a single Python method.

Here's the key part of the code with comments explaining each operation:

class Solution:
    def maximumBinaryString(self, binary: str) -> str:
        # Find the first '0' in the string.
        k = binary.find('0')
        # If no '0' is found, return the original string as-is.
        if k == -1:
            return binary
        # Increase `k` by the count of '0's following the first '0'.
        # This simulates the position of the last '0' in the string.
        k += binary[k + 1 :].count('0')
        # Construct the final binary string with the calculated numbers 
        # of '1's, the last '0', and then '1's again.
        return '1' * k + '0' + '1' * (len(binary) - k - 1)

Using the count of '0's and string concatenation operations results in a clean, efficient solution that avoids cumbersome loops or conditional operations which would have been necessary if we were to simulate each operation.

Example Walkthrough

Let's use the binary string binary = "00110" as a small example to illustrate the solution approach.

1. We first search for the first occurrence of the character '0'. In this string, it occurs at index 0. The string binary.find('0') would return 0.

2. Next, we count the number of '0's after the first '0'. In the string "00110", there are three '0's in total. After finding the first '0', there are two more '0's to count. Hence, binary[0 + 1:].count('0') would give us 2.

3. We set k to the index of the first '0' found which is 0 and then increase k by the count of remaining '0's, which is 2. Therefore, k becomes 0 + 2 = 2. This is the index where the last '0' will be after performing all the operations possible.

4. Now, we construct the final string. Since k is 2, we know there will be two '1's before the last '0' in the string. And since the original string binary has a length of 5, there will be 5 - 2 - 1 = 2 '1's following the last '0'.

5. The final maximum binary string after all operations have been performed will be '1' * 2 + '0' + '1' * 2 which evaluates to "11011".

Within the given code framework, it would execute as follows:

binary = "00110"
k = binary.find('0')  # k = 0
k += binary[k + 1:].count('0')  # k = 2
maximum_binary = '1' * k + '0' + '1' * (len(binary) - k - 1)  # maximum_binary = "11011"

After these steps, maximum_binary holds the value "11011", which is the highest-value binary string possible from the initial string "00110" by applying the given operations. It has a decimal equivalent of 27, which is greater than the decimal equivalent of the original string (6). This illustrates the effectiveness of the described solution approach.

Solution Implementation

Time and Space Complexity

Time Complexity

The given Python function maximumBinaryString operates on a string binary. The time complexity is analyzed as follows:

1. The find method called on binary to locate the first occurrence of '0' runs in O(n) time, where n is the length of the binary string.
2. Slicing the string after the found index and counting the number of '0's with count('0') also runs in O(n) in the worst case.
3. The multiplication and concatenation of strings return '1' * k + '0' + '1' * (len(binary) - k - 1) comprise constant-time operations for each character in the resulting string, resulting in O(n) complexity.

Thus, by considering each operation, the overall time complexity of the function is O(n).

Space Complexity

The space complexity of this function is considered as follows:

1. The variables k and binary itself, as inputs, do not count towards additional space, since they represent existing memory allocations and not new space that the algorithm requires.
2. The function creates new strings that are returned, which are of length n. Therefore, the space complexity due to the output string is O(n).

Considering the above, the space complexity of the function is O(n) because of the new string that is created and returned.

Overall, the function maximumBinaryString has a time complexity of O(n) and a space complexity of O(n).

Learn more about how to find time and space complexity quickly using problem constraints.