Leetcode 1784

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

The problem description is:

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

This is kind of hard to understand, so I am going to explain it. A contiguous segment of ones looks like:

And some of the ones that are not contiguous are:

You should be able to see that the non-consecutive strings have a "01". We have to check whether there is a "01" in the string. If so, return false.

Note: We don’t have to worry about strings with leading zeros. That is why we can check whether there is a "01" in a string and return false if so. If we had leading zeros, there would be "01"’s in the string even if there are consecutive ones, such as "001110". We should return true, but there is a "01".

First Solution:

func checkOnesSegment(s string) bool {

    for i := 0; i < len(s)-1; i++ {
        if s[i:i+2] == "01" { return false }
    }
    return true
}

Second Solution:

func checkOnesSegment(s string) bool {

    for i := 0; i < len(s)-1; i++ {
        if s[i] == '0' && s[i+1] == '1' { return false }
    }
    return true
}