Given a binary string s ​​​​​without leading zeros, return true​​​ if s contains at most one contiguous segment of ones. Otherwise, return false.

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Example 1:

Input: s = "1001"
Output: false
Explanation: The ones do not form a contiguous segment.


Example 2:

Input: s = "110"
Output: true

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Constraints:


	1 &lt;= s.length &lt;= 100
	s[i]​​​​ is either '0' or '1'.
	s[0] is&nbsp;'1'.


class Solution:
    def checkOnesSegment(self, s: str) -> bool:
        return "01" not in s
class Solution {
    public boolean checkOnesSegment(String s) {
        return !s.contains("01");
    }
}
class Solution {
public:
	bool checkOnesSegment(string s) {
		for(int i = 1; i < s.size(); i++){
			if(s[i - 1] == '0' and s[i] == '1'){
				return false;
			}
		}
		return true;
	}
};
var checkOnesSegment = function(s) {
   return s.indexOf("01") == -1
};
Check if Binary String Has at Most One Segment of Ones