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

/**
 * Problem: LeetCode 1784 - Check if Binary String Has at Most One Segment of Ones
 * Difficulty: Easy
 * Language: Java
 * Author: CodeByTushu
 */
class Solution {
    public boolean checkOnesSegment(String s) {
        // Because s[0] == '1' (no leading zeros),
        // any second segment of ones must be preceded by at least one '0'.
        // If "01" appears anywhere in the string, there is more than one segment.
        return !s.contains("01");
    }
}

"""
Problem: LeetCode 1784 - Check if Binary String Has at Most One Segment of Ones
Difficulty: Easy
Language: Python
Author: CodeByTushu
"""
class Solution:
    def checkOnesSegment(self, s: str) -> bool:
        # Since s has no leading zeros, s starts with '1'.
        # A second segment of '1's would require a '0' followed by a '1' ("01").
        return "01" not in s

/**
 * Problem: LeetCode 1784 - Check if Binary String Has at Most One Segment of Ones
 * Difficulty: Easy
 * Language: C++
 * Author: CodeByTushu
 */
#include <string>

using namespace std;

class Solution {
public:
    bool checkOnesSegment(string s) {
        // Since s has no leading zeros (s[0] == '1'),
        // more than one segment of ones exists if and only if "01" is present.
        return s.find("01") == string::npos;
    }
};

/**
 * Problem: LeetCode 1784 - Check if Binary String Has at Most One Segment of Ones
 * Difficulty: Easy
 * Language: JavaScript
 * Author: CodeByTushu
 */
/**
 * @param {string} s
 * @return {boolean}
 */
var checkOnesSegment = function(s) {
    // Since s starts with '1' (no leading zeros),
    // a second segment of ones exists if and only if "01" appears in the string.
    return !s.includes("01");
};