What is the time complexity of Time Needed to Rearrange a Binary String?

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

What data structures are used to solve Time Needed to Rearrange a Binary String?

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

What are the key insights for solving Time Needed to Rearrange a Binary String?

Insight 1: We don't need to actually modify the string in each step. We only need to track the number of zeros encountered so far. Insight 2: The number of seconds needed is determined by how far each '1' has to move to the left, which is influenced by the number of '0's to its left.

Which companies ask LeetCode 2380. Time Needed to Rearrange a Binary String in interviews?

This problem is frequently asked at PayPal, ServiceNow.

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Time Needed to Rearrange a Binary String - LeetCode 2380 Solution

Time Needed to Rearrange a Binary String - Complete Solution Guide

Time Needed to Rearrange a Binary String is LeetCode problem 2380, 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 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,

Detailed Explanation

The problem asks us to simulate the rearrangement of a binary string. In each second, all occurrences of the substring "01" are simultaneously replaced with "10". We need to determine how many seconds it takes for the string to reach a stable state where no more "01" substrings exist. The input is a binary string 's' consisting of '0's and '1's, and the output is an integer representing the number of seconds needed.

Solution Approach

The provided code implements an efficient O(n) solution. It iterates through the string, keeping track of the number of zeros encountered (`zeros_count`). When a '1' is encountered, it checks if there are any zeros to its left. If so, it calculates the time needed for that '1' to move to its final position. The overall time is the maximum time needed for any '1' to reach its final position.

Step-by-Step Algorithm

Step 1: Initialize `seconds` to 0 and `zeros_count` to 0.
Step 2: Iterate through the input string `s` character by character.
Step 3: If the current character is '0', increment `zeros_count`.
Step 4: If the current character is '1' and `zeros_count` is greater than 0 (meaning there are zeros to the left of the current one), calculate the time needed for this '1' to move. This time is the maximum between the current `seconds + 1` (the next second if the 1 moves) and the number of `zeros_count` (how many positions the 1 must cross). Update the `seconds` with the max of the two values.
Step 5: After iterating through the entire string, return the final `seconds`.

Key Insights

Insight 1: We don't need to actually modify the string in each step. We only need to track the number of zeros encountered so far.
Insight 2: The number of seconds needed is determined by how far each '1' has to move to the left, which is influenced by the number of '0's to its left.
Insight 3: The critical point is when the current '1' can't move further left in the next second because the previously swapped '1' is blocking its path.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String, Dynamic Programming, Simulation.

Companies

Asked at: PayPal, ServiceNow.

Time Needed to Rearrange a Binary String | LeetCode 2380

Medium

String Dynamic Programming Simulation

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int secondsToRemoveOccurrences(String s) {
        int seconds = 0;
        int zeros_count = 0;
        for (char charr : s.toCharArray()) {
            if (charr == '0') {
                zeros_count += 1;
            } else if (zeros_count > 0) {
                seconds = Math.max(seconds + 1, zeros_count);
            }
        }
        return seconds;
    }
}

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Problem Info

Description

Problem 2380: Time Needed to Rearrange a Binary String - Detailed explanation and solution approaches.

Topics

String Dynamic Programming Simulation

Companies

PayPalServiceNow

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2380

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

5 steps

1
Step 1: Initialize `seconds` to 0 and `zeros_count` to 0.
2
Step 2: Iterate through the input string `s` character by character.
3
Step 3: If the current character is '0', increment `zeros_count`.
4
Step 4: If the current character is '1' and `zeros_count` is greater than 0 (meaning there are zeros to the left of the current one), calculate the time needed for this '1' to move. This time is the maximum between the current `seconds + 1` (the next second if the 1 moves) and the number of `zeros_count` (how many positions the 1 must cross). Update the `seconds` with the max of the two values.
5
Step 5: After iterating through the entire string, return the final `seconds`.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is linear because the algorithm performs a single pass through the input string. The space complexity is constant because it uses only a few integer variables (seconds, zeros_count) which do not depend on the input size.

📖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

Simulation

Simulate the process step by step by repeatedly finding and replacing "01" with "10" until no such occurrences exist. This approach directly follows the problem description.

Time: O(n^2) in the worst case, where n is the length of the string. Each step might take O(n) to find and replace the occurrences, and the number of steps can be up to O(n).Space: O(n) because the string needs to be modified.

Trade-offs: The simulation approach is more intuitive but less efficient than the optimal solution. It's easier to understand but can be slow for larger inputs.

✨

Key Insights

1Insight 1: We don't need to actually modify the string in each step. We only need to track the number of zeros encountered so far.
2Insight 2: The number of seconds needed is determined by how far each '1' has to move to the left, which is influenced by the number of '0's to its left.
3Insight 3: The critical point is when the current '1' can't move further left in the next second because the previously swapped '1' is blocking its path.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

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

Edge Cases

Edge case 1: Input string contains only '1's (e.g., "111"). The solution correctly returns 0 because `zeros_count` will always be 0.
Edge case 2: Input string contains only '0's (e.g., "000"). The solution correctly returns 0 because the `else if` condition will never be met.
Edge case 3: Input string is empty (this is technically disallowed by constraints but good to consider). The for loop will not execute and the initial value of 'seconds' will be returned which is 0.
Edge case 4: Input string starts with '1's (e.g., "11001"). Only the zeros after these 1s are relevant to calculating the seconds.
Edge case 5: A string such as "01010101". In this case the answer must be calculated according to which 1 must shift the furthest.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Modifying the string directly in each iteration. This is inefficient and can lead to incorrect results due to simultaneous replacements.

✗Common mistake 2: Not considering the 'blocking' effect of previous swaps. The time for a '1' to move is not simply the number of zeros to its left; it's limited by the time it took for the previous '1' to move.

✗Common mistake 3: Assuming all '1's move simultaneously without considering the sequential dependence.

✗Common mistake 4: Incorrectly handling edge cases like strings with only 0s or only 1s.

✅

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: Task scheduling where tasks need to be reordered based on dependencies. A '0' might represent an available slot and a '1' represents a task that needs to be moved to an available slot.

🏦Real-world scenario 2: Data packet reordering in network communication, where packets arrive out of order and need to be rearranged based on sequence numbers. Similar logic could apply if sequence 0s/1s needed sorting

💡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.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: July 11, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

❓

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 identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

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 "Time Needed to Rearrange a Binary String"?

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 dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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