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

Intuition

Since the string has no leading zeros (starts with '1'), if there's more than one segment of ones, we must see the pattern "01" a zero followed by a one.

Solution

If "01" appears in the string, it means a segment of ones ended and another begins later.

Simply check if the substring "01" exists:

• If "01" exists → return false (multiple segments)
• Otherwise → return true (at most one segment)

Why This Works

Since s[0] = '1' (no leading zeros):

• All ones at the start form one segment
• If we see "01", a new segment of ones started after some zeros
• If no "01", all ones are contiguous at the beginning

Complexity

• Time: O(n), substring search
• Space: O(1), no extra space

Edge Cases

• Single '1' → true
• All ones "111" → true
• Ones then zeros "110" → true
• Pattern "101" → false (contains "01")

class Solution:
    def checkOnesSegment(self, s: str) -> bool:
        """
        1784. Check if Binary String Has at Most One Segment of Ones
        Time: O(n)
        Space: O(1)
        """
        return "01" not in s

/**
 * 1784. Check if Binary String Has at Most One Segment of Ones
 * Time: O(n)
 * Space: O(1)
 */
class Solution {
    public boolean checkOnesSegment(String s) {
        return !s.contains("01");
    }
}