Is Check if Binary String Has at Most One Segment of Ones easy or hard?

The problem is classified as Easy. It focuses on careful string traversal and state tracking rather than complex algorithms or data structures.

Check if Binary String Has at Most One Segment of Ones Python/Java solution

In Python or Java, iterate through the characters and track whether a zero has appeared after the first 1. If another 1 appears afterward, return false. The implementation takes O(n) time and O(1) extra space.

How to solve Check if Binary String Has at Most One Segment of Ones in O(n)?

Iterate through the string while tracking whether a zero has appeared after the first block of ones. If you encounter a 1 after such a zero, immediately return false. Completing the scan without this condition means there is at most one segment of ones.

What is the best approach for Check if Binary String Has at Most One Segment of Ones?

The most efficient approach is a single-pass traversal using a flag to track whether the sequence of ones has ended. Once a zero appears after the first 1, any later 1 indicates a second segment. This method runs in O(n) time with O(1) extra space.

Is Check if Binary String Has at Most One Segment of Ones asked at Google/Amazon/Meta?

This exact problem is commonly used for practice on coding platforms and reflects the type of basic string reasoning questions asked in screening rounds. Similar problems appear in interviews at companies like Amazon and Google when testing simple traversal and state tracking skills.

What data structure is used in Check if Binary String Has at Most One Segment of Ones?

No complex data structures are required. The optimal approach only uses simple variables while iterating through the string. Some alternative solutions temporarily create arrays of substrings after splitting the string by zeros.

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

The optimal solution runs in O(n) time because the string is scanned once from left to right. Only constant extra variables are used to track state, so the space complexity is O(1).

1784. Check if Binary String Has at Most One Segment of Ones - Practice Online

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'.