README_EN.md

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

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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

Python3

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

Java

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

C++

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

Go

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

TypeScript

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;
}

function checkOnesSegment(s: string): boolean {
    return !s.includes('01');
}

Rust

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

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