#1702

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Maximum Binary String After Change

The problem asks to maximize a binary string using specific operations to get the highest possible value.

Problem Statement

You are given a binary string binary consisting only of '0's and '1's. You can repeatedly apply the following operation to the string: If there is a '01' substring, you can change it to '10'. Perform this operation any number of times.

Return the maximum binary string you can obtain after performing any number of operations. A binary string x is greater than binary string y if x's decimal value is greater than y's.

Examples

Example 1

Input: binary = "000110"

Output: "111011"

A valid transformation sequence can be: "000110" -> "000101" "000101" -> "100101" "100101" -> "110101" "110101" -> "110011" "110011" -> "111011"

Example 2

Input: binary = "01"

Output: "01"

"01" cannot be transformed any further.

Constraints

1 <= binary.length <= 105
binary consist of '0' and '1'.

Solution Approach

Greedy Transformation

The solution uses a greedy approach by iterating through the string and changing '01' to '10' until no further changes can be made. This helps maximize the number of '1's and push zeros towards the end, which increases the overall value of the binary string.

Maximizing Zeros

One key observation is that at most one zero can remain in the string. By iterating and swapping '01' pairs, the string will eventually have at most one '0', making the string as large as possible.

Invariant Validation

After each operation, the string's structure must be validated to ensure that no more '01' pairs are present. Once the string cannot be transformed further, the current configuration is the largest possible binary string.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the final approach and may range from O(n) to O(n^2), depending on the method used to track and transform '01' pairs. The space complexity also depends on the final approach, but it generally remains O(1) as the operations only modify the original string in place.

What Interviewers Usually Probe

Evaluate understanding of greedy algorithms in string manipulation.
Look for the candidate's ability to handle edge cases, such as no valid operations or already maximized strings.
Assess their ability to optimize time complexity and avoid unnecessary operations.

Common Pitfalls or Variants

Common pitfalls

Failing to recognize that only one zero can remain in the string after transformations.
Not validating the string structure after each operation and missing potential '01' pairs to change.
Overcomplicating the solution by applying unnecessary operations or using suboptimal approaches.

Follow-up variants

What if you were allowed to apply the operation multiple times in a row for the same '01' pair?
How would the solution change if you had to ensure the resulting string contains at least one '1'?
Can this approach be adapted to other string manipulation problems with different operations?

FAQ

What is the maximum binary string after change?

The maximum binary string is the result of repeatedly transforming '01' pairs into '10', until no more transformations can be made, resulting in a string with at most one '0'.

How does the greedy approach work in this problem?

The greedy approach focuses on iterating through the string and applying the '01' to '10' transformation as often as possible to push zeros toward the end, maximizing the binary string's value.

What is the time complexity of solving the problem?

The time complexity depends on the approach used. In the optimal case, it can be O(n), but in some implementations, it might reach O(n^2).

Can this problem be solved using dynamic programming?

Although dynamic programming could be used for some problems in string manipulation, this problem is best solved with a greedy approach due to the simplicity of the operations and the fact that each transformation can be applied directly to maximize the string value.

What happens if the binary string is already maximized?

If the binary string is already maximized, no operations will be needed, and the string remains unchanged. The algorithm should handle this efficiently without unnecessary steps.

terminal

Solution

Solution 1: Quick Thinking

We observe that operation $2$ can move all $1$s to the end of the string, and operation $1$ can change all `0000..000` strings to `111..110`.

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class Solution:
    def maximumBinaryString(self, binary: str) -> str:
        k = binary.find('0')
        if k == -1:
            return binary
        k += binary[k + 1 :].count('0')
        return '1' * k + '0' + '1' * (len(binary) - k - 1)

Continue Practicing

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