What is the time complexity of Check if Binary String Has at Most One Segment of Ones?

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

What data structures are used to solve Check if Binary String Has at Most One Segment of Ones?

This problem uses String. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Check if Binary String Has at Most One Segment of Ones?

Insight 1: The problem can be efficiently solved by iterating through the string only once, keeping track of whether a segment of ones has ended. Insight 2: The presence of a '0' followed by a '1' indicates more than one segment of ones. This is a simple condition to check.

Which companies ask LeetCode 1784. Check if Binary String Has at Most One Segment of Ones in interviews?

This problem is frequently asked at Cisco.

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Check if Binary String Has at Most One Segment of Ones - LeetCode 1784 Solution

Q: How to solve Check if Binary String Has at Most One Segment of Ones?

The most efficient solutions leverage the observation that if a '0' is encountered and later a '1' is encountered, it signifies more than one segment of ones. The provided solutions iterate through the string, using a flag or a simple check to detect this condition.

Check if Binary String Has at Most One Segment of Ones - Complete Solution Guide

Check if Binary String Has at Most One Segment of Ones is LeetCode problem 1784, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given a binary string s without leading zeros , return true if s contains at most one contiguous segment of ones . Otherwise, return false . Example 1: Input: s = "1001" Output: false Explanation: The ones do not form a contiguous segment. Example 2: Input: s = "110" Output: true Constraints: 1 <= s.length <= 100 s[i] is either '0' or '1' . s[0] is '1' .

Detailed Explanation

The problem asks you to determine if a binary string (a string containing only '0's and '1's) has at most one contiguous segment of ones. A contiguous segment of ones means a sequence of one or more consecutive '1's. The input string is guaranteed to start with '1' and not contain leading zeros. The output is a boolean value: `true` if there's at most one contiguous segment of ones, and `false` otherwise.

Solution Approach

The most efficient solutions leverage the observation that if a '0' is encountered and later a '1' is encountered, it signifies more than one segment of ones. The provided solutions iterate through the string, using a flag or a simple check to detect this condition.

Step-by-Step Algorithm

Step 1: Initialize a flag (e.g., `foundZero`) to `false`. This flag indicates whether a '0' has been encountered.
Step 2: Iterate through the binary string.
Step 3: If a '0' is encountered, set the `foundZero` flag to `true`.
Step 4: If a '1' is encountered and the `foundZero` flag is already `true`, it means there's more than one segment of ones. Return `false`.
Step 5: If the loop completes without finding a '0' followed by a '1', return `true` (at most one segment of ones).

Key Insights

Insight 1: The problem can be efficiently solved by iterating through the string only once, keeping track of whether a segment of ones has ended.
Insight 2: The presence of a '0' followed by a '1' indicates more than one segment of ones. This is a simple condition to check.
Insight 3: The constraint that the string starts with '1' simplifies the problem, eliminating the need for special handling of leading zeros.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String.

Companies

Asked at: Cisco.

Check if Binary String Has at Most One Segment of Ones | LeetCode 1784

Easy

String

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

Solution Code

Available in:

class Solution {
    public boolean checkOnesSegment(String s) {
        int n = s.length();
        int count = 0;
        for (int i = 0; i < n; i++) {
            if (s.charAt(i) == '1') {
                count++;
                while (i < n && s.charAt(i) == '1') {
                    i++;
                }
                if (i < n && s.charAt(i) == '0') {
                    while (i < n && s.charAt(i) == '0') {
                        i++;
                    }
                    if (i < n && s.charAt(i) == '1') {
                        return false;
                    } else {
                        break;
                    }
                } else {
                    break;
                }
            }
        }
        return true;
    }
}

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Problem Info

Description

Problem 1784: Check if Binary String Has at Most One Segment of Ones - Detailed explanation and solution approaches.

Topics

String

Companies

Cisco

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1784

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 a flag (e.g., `foundZero`) to `false`. This flag indicates whether a '0' has been encountered.
2
Step 2: Iterate through the binary string.
3
Step 3: If a '0' is encountered, set the `foundZero` flag to `true`.
4
Step 4: If a '1' is encountered and the `foundZero` flag is already `true`, it means there's more than one segment of ones. Return `false`.
5
Step 5: If the loop completes without finding a '0' followed by a '1', return `true` (at most one segment of ones).

💡

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 O(n) because the algorithm iterates through the string once. The space complexity is O(1) because only a constant amount of extra space (the `foundZero` flag) is used, regardless of the input string's length.

📖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

Using String Methods (Python)

The Python solution cleverly uses the `'01'` substring check. If '01' is a substring, it means there's more than one segment of ones.

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

Trade-offs: This approach is concise and readable in Python, but it might not be as directly translatable to other languages as the iterative approach.

Regular Expressions

A regular expression could be used to check if the string matches a pattern representing at most one contiguous segment of ones.

Time: O(n)Space: Depending on regex engine, could be O(1) or O(n)

Trade-offs: While potentially concise, regular expressions can be less readable and might have higher overhead compared to the simple iterative approach, especially for beginners.

✨

Key Insights

1Insight 1: The problem can be efficiently solved by iterating through the string only once, keeping track of whether a segment of ones has ended.
2Insight 2: The presence of a '0' followed by a '1' indicates more than one segment of ones. This is a simple condition to check.
3Insight 3: The constraint that the string starts with '1' simplifies the problem, eliminating the need for special handling of leading zeros.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: The input string is '1'. The function should correctly return `true` as there is only one segment of ones.
Edge case 2: The input string is '10000'. The function should correctly return `true` as there is only one segment of ones, even if it's followed by zeros.
Edge case 3: The input string is '11101'. The function should correctly return `false` because there are two segments of ones.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where the input string contains only '1's. Some inefficient solutions might mistakenly process this edge case incorrectly.

✗Common mistake 2: Incorrectly handling multiple '0's between segments of '1's. A robust solution needs to account for sequences like '10001'.

✅

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: Network packet analysis – checking for consecutive bits set in a status flag.

🏦Real-world scenario 2: Data validation – ensuring a specific format constraint in binary data representing flags or states.

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ 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: November 12, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

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

@DevExpert92Jul 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 should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "Check if Binary String Has at Most One Segment of Ones"?

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

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