What is the time complexity of Maximum Binary String After Change?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Maximum Binary String After Change?

This problem uses String, Greedy. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum Binary String After Change?

Insight 1: The key is to realize that we can move all '0's to the right side of the string except for at most one '0'. All the '1's can be moved to the left side. Insight 2: The operations effectively allow us to swap the positions of '0's and '1's, except that we cannot eliminate '0's. We can transform '00' into '10' and then '10' into '01', effectively swapping two adjacent bits as long as there's a '0' to the left. This suggests we can group all 1s together on the left and all 0s together on the right, except for a single '0'.

Which companies ask LeetCode 1702. Maximum Binary String After Change in interviews?

This problem is frequently asked at Huawei.

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Maximum Binary String After Change - LeetCode 1702 Solution

Maximum Binary String After Change - Complete Solution Guide

Maximum Binary String After Change is LeetCode problem 1702, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a binary string binary consisting of only 0 's or 1 's. You can apply each of the following operations any number of times: Operation 1: If the number contains the substring "00" , you can replace it with "10" . For example, " 00 010" -> " 10 010 " Operation 2: If the number contains the substring "10" , you can replace it with "01" . For example, "000 10 " -> "000 01 " Return the maximum binary string you can obtain after any number of operations. Binary string x is greater than b

Detailed Explanation

The problem asks us to find the maximum possible binary string that can be obtained from a given binary string by applying two operations any number of times. The operations are: 1) Replace "00" with "10", and 2) Replace "10" with "01". The maximum binary string is the one with the largest decimal representation. The input is a binary string consisting of 0s and 1s, and the output is the maximum binary string obtainable after applying the operations.

Solution Approach

The solution counts the number of zeros in the input string and finds the index of the first zero. It then constructs a new string consisting of all '1's, except for a single '0' at the position `first_zero_idx + zeros - 1`. This single '0' represents the combined effect of moving all the other zeros as far to the right as possible by converting pairs of '00' to '10' and then shifting the remaining zeros using operation 2 '10' -> '01'. If the number of zeros is less than or equal to 1, the original string is already the maximum.

Step-by-Step Algorithm

Step 1: Count the total number of zeros in the binary string.
Step 2: If the number of zeros is less than or equal to 1, return the original binary string. No operations are needed.
Step 3: Find the index of the first zero in the binary string.
Step 4: Calculate the position where the final zero should be placed: `first_zero_idx + zeros - 1`.
Step 5: Create a new string of the same length as the original string, filled with '1's.
Step 6: Place a single '0' at the calculated position in the new string.
Step 7: Return the new string.

Key Insights

Insight 1: The key is to realize that we can move all '0's to the right side of the string except for at most one '0'. All the '1's can be moved to the left side.
Insight 2: The operations effectively allow us to swap the positions of '0's and '1's, except that we cannot eliminate '0's. We can transform '00' into '10' and then '10' into '01', effectively swapping two adjacent bits as long as there's a '0' to the left. This suggests we can group all 1s together on the left and all 0s together on the right, except for a single '0'.
Insight 3: We can move any '0' to any position to the right of it. Operation 1 ('00' -> '10') creates a '1' and moves a zero one position to the right. Operation 2 ('10' -> '01') moves a zero one position to the right. By combining these, we can 'shift' zeros towards the right. Therefore, we want to move all zeros except one to the end, leaving one zero in a position such that all positions to its left are ones.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String, Greedy.

Companies

Asked at: Huawei.

Maximum Binary String After Change | LeetCode 1702

Medium

String Greedy

Time: O(n)Space: O(n)

Solution Code

Available in:

class Solution {
    public String maximumBinaryString(String binary) {
        int n = binary.length();
        int zeros = 0;
        for(int i = 0; i < n; i++){
            if(binary.charAt(i) == '0'){
                zeros++;
            }
        }
        
        if (zeros <= 1) {
            return binary;
        }
        
        int first_zero_idx = -1;
        for(int i = 0; i < n; i++){
            if(binary.charAt(i) == '0'){
                first_zero_idx = i;
                break;
            }
        }
        
        int final_zero_pos = first_zero_idx + zeros - 1;
        
        char[] res = new char[n];
        for(int i = 0; i < n; i++){
            res[i] = '1';
        }
        res[final_zero_pos] = '0';
        
        return new String(res);
    }
}

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Problem Info

Description

Problem 1702: Maximum Binary String After Change - Detailed explanation and solution approaches.

Topics

String Greedy

Companies

Huawei

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1702

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Count the total number of zeros in the binary string.
2
Step 2: If the number of zeros is less than or equal to 1, return the original binary string. No operations are needed.
3
Step 3: Find the index of the first zero in the binary string.
4
Step 4: Calculate the position where the final zero should be placed: `first_zero_idx + zeros - 1`.
5
Step 5: Create a new string of the same length as the original string, filled with '1's.
6
Step 6: Place a single '0' at the calculated position in the new string.
7
Step 7: Return the new string.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The loops to count zeros and find the first zero iterate through the string once, resulting in O(n) time complexity. Creating a new string of length n contributes O(n) space complexity. The other operations (addition, assignment) take constant time, O(1). Therefore, the overall time complexity is O(n) and the overall space complexity is O(n).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

In-place Modification

Instead of creating a new string, we could attempt to modify the original string in-place. However, this may be more complex to implement correctly in languages like Java and Python due to the immutability of strings.

Time: O(n)Space: O(1) if in-place modification is truly possible in the chosen language, otherwise O(n).

Trade-offs: While potentially saving space (if truly in-place is possible in the chosen language which is generally not the case for most mainstream languages), this approach can increase code complexity and the risk of errors.

✨

Key Insights

1Insight 1: The key is to realize that we can move all '0's to the right side of the string except for at most one '0'. All the '1's can be moved to the left side.
2Insight 2: The operations effectively allow us to swap the positions of '0's and '1's, except that we cannot eliminate '0's. We can transform '00' into '10' and then '10' into '01', effectively swapping two adjacent bits as long as there's a '0' to the left. This suggests we can group all 1s together on the left and all 0s together on the right, except for a single '0'.
3Insight 3: We can move any '0' to any position to the right of it. Operation 1 ('00' -> '10') creates a '1' and moves a zero one position to the right. Operation 2 ('10' -> '01') moves a zero one position to the right. By combining these, we can 'shift' zeros towards the right. Therefore, we want to move all zeros except one to the end, leaving one zero in a position such that all positions to its left are ones.

📚

Key Terms Explained

For Students

Greedy Algorithm

Making the best choice at each step without looking ahead.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: The input string contains no zeros. In this case, `binary.count('0')` returns 0, which is less than or equal to 1, so the original string is returned, which is correct.
Edge case 2: The input string contains only one zero. In this case, `binary.count('0')` returns 1, which is less than or equal to 1, so the original string is returned, which is correct.
Edge case 3: The input string contains only zeros (e.g., "0000"). The algorithm correctly transforms it to "1110".
Edge case 4: Input string starts with all ones. The algorithm correctly handle and computes correct index of the final zero

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the edge cases where the string contains 0 or 1 zero. This will lead to incorrect results as the algorithm's core logic isn't needed in such cases.

✗Common mistake 2: Incorrectly calculating the final zero position, which will result in the '0' being placed in the wrong location or out of bounds.

✗Common mistake 3: Modifying the input string directly without considering its immutability in some languages, which will lead to errors or unexpected behavior.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data compression. Rearranging bits based on patterns to achieve better compression ratios uses principles of bit manipulation and optimization, similar to the logic here.

🏦Real-world scenario 2: Optimizing memory allocation. Algorithms to arrange data in memory for efficient access sometimes involve rearranging bits to minimize fragmentation or improve cache performance. This requires understanding bit manipulation and tradeoffs similar to what's explored in this problem.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 22, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How do I verify if a greedy approach is correct?

Prove the greedy choice property: the locally optimal choice leads to globally optimal solution. Use exchange argument: show that swapping any element with the greedy choice doesn't improve the solution. If you can't prove it, test extensively or consider DP.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Maximum Binary String After Change"?

For this medium-level String problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether greedy techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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