Check if Binary String Has at Most One Segment of Ones — LeetCode 1784 Python Solution
- Problem
- #1784
- Pattern
- Hash Map
- Reading time
- 2 min
- Source
- leetcode.com
The problem
Given a binary string s without leading zeros, return true if s contains at most one contiguous segment of ones. Otherwise, return false.
Example
- Input
- s = "1001"
- Output
- false
- Explanation
- The ones do not form a contiguous segment.
Python solution
class Solution:
def checkOnesSegment(self, s: str) -> bool:
return '01' not in sComplexity
| Measure | Complexity |
|---|---|
| Time | O(n), where n is the length of the string s |
| Space | O(1) auxiliary |
Pattern: Hash Map
Trade memory for time: remember what you have seen so the second pass never happens. LeetCode 1784. Check if Binary String Has at Most One Segment of Ones is filed here because the reference solution below belongs to the algorithm family this hub collects, even though its LeetCode tags point elsewhere.
The hash map guide has the Python template for the pattern and the 709 LeetCode problems that use it.
Related problems
Frequently asked questions
- How hard is LeetCode 1784. Check if Binary String Has at Most One Segment of Ones?
- LeetCode 1784. Check if Binary String Has at Most One Segment of Ones is rated Easy on LeetCode.
- What is the time complexity of LeetCode 1784. Check if Binary String Has at Most One Segment of Ones?
- The Python solution on this page runs in O(n), where n is the length of the string s.
- What is the space complexity of LeetCode 1784. Check if Binary String Has at Most One Segment of Ones?
- The Python solution on this page uses O(1) auxiliary space.
- What topics does LeetCode 1784. Check if Binary String Has at Most One Segment of Ones cover?
- LeetCode 1784. Check if Binary String Has at Most One Segment of Ones is tagged String on LeetCode.