carloscn
commented
1 year ago Analysis
pub fn check_string(s: &str) -> bool
{
if s.len() < 1 {
return false;
}
let s_vec:Vec<char> = s.chars().collect();
let mut a_flag = false;
for i in 0..s_vec.len() {
let e = s_vec[s_vec.len() - i - 1];
if e == 'a' {
a_flag = true;
} else if e == 'b' {
if a_flag == true {
return false;
}
} else {
return false;
}
}
return true;
}
Description
Given a string s consisting of only the characters 'a' and 'b', return true if every 'a' appears before every 'b' in the string. Otherwise, return false.
Example 1:
Input: s = "aaabbb" Output: true Explanation: The 'a's are at indices 0, 1, and 2, while the 'b's are at indices 3, 4, and 5. Hence, every 'a' appears before every 'b' and we return true.
Example 2:
Input: s = "abab" Output: false Explanation: There is an 'a' at index 2 and a 'b' at index 1. Hence, not every 'a' appears before every 'b' and we return false.
Example 3:
Input: s = "bbb" Output: true Explanation: There are no 'a's, hence, every 'a' appears before every 'b' and we return true.
Constraints:
1 <= s.length <= 100 s[i] is either 'a' or 'b'.