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

## Description

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'.

## Solutions

Solution 1: No ‘1’ After ‘0’

Notice that the string $s$ does not contain leading zeros, which means $s$ starts with ‘1’.

If the string $s$ contains the substring “01”, then $s$ must be a string like “1…01…”, in which case $s$ has at least two consecutive ‘1’ segments, which does not satisfy the problem condition, so we return false.

If the string $s$ does not contain the substring “01”, then $s$ can only be a string like “1..1000…”, in which case $s$ has only one consecutive ‘1’ segment, which satisfies the problem condition, so we return true.

Therefore, we only need to judge whether the string $s$ contains the substring “01”.

The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.

• class Solution {
public boolean checkOnesSegment(String s) {
return !s.contains("01");
}
}

• class Solution {
public:
bool checkOnesSegment(string s) {
return s.find("01") == -1;
}
};

• class Solution:
def checkOnesSegment(self, s: str) -> bool:
return '01' not in s


• func checkOnesSegment(s string) bool {
return !strings.Contains(s, "01")
}

• function checkOnesSegment(s: string): boolean {
let pre = s[0];
for (const c of s) {
if (pre !== c && c === '1') {
return false;
}
pre = c;
}
return true;
}


• impl Solution {
pub fn check_ones_segment(s: String) -> bool {
!s.contains("01")
}
}