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

The optimal solution for Check if Binary String Has at Most One Segment of Ones runs in O(n) time and uses O(1) space. The complexity is O(n) because we only make a single pass through the string, checking each character once.

What is the brute force solution for Check if Binary String Has at Most One Segment of Ones?

The brute force approach for Check if Binary String Has at Most One Segment of Ones is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every possible segment of the string to count the number of contiguous segments of '1's. If we find more than one segment, we return false. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Check if Binary String Has at Most One Segment of Ones?

Check if Binary String Has at Most One Segment of Ones uses the following concepts: String. The recommended patterns to study are: Two Pointers, Sliding Window.

What are the interview tips for Check if Binary String Has at Most One Segment of Ones?

Always clarify the problem statement and constraints before diving into coding. Think about edge cases, such as strings with only '1's or '0's. Practice writing clean and efficient code, focusing on readability.

What are common mistakes when solving Check if Binary String Has at Most One Segment of Ones?

Not checking the transition correctly, leading to incorrect segment counts. Overcomplicating the problem with unnecessary checks or loops.

#1784

Check if Binary String Has at Most One Segment of Ones

Easy

StringTwo PointersSliding Window

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal solution involves a single pass through the string to count the segments of '1's. This is efficient because we only need to check transitions from '0' to '1'.

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Algorithm

5 steps

1Step 1: Initialize a counter for segments of '1's.
2Step 2: Traverse the string character by character.
3Step 3: Increment the counter when a '1' is found that follows a '0' or is the first character.
4Step 4: If the counter exceeds 1 at any point, return false.
5Step 5: If the loop completes and the counter is 1 or less, return true.

solution.py8 lines

1def checkOnesSegment(s):
2    count = 0
3    for i in range(len(s)):
4        if s[i] == '1' and (i == 0 or s[i-1] == '0'):
5            count += 1
6        if count > 1:
7            return False
8    return True

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Complexity note: The complexity is O(n) because we only make a single pass through the string, checking each character once.

1A segment of '1's is defined by transitions from '0' to '1'.
2The problem can be solved efficiently by counting these transitions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.