2124. Check if All A's Appears Before All B's
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 <= 100s[i]is either'a'or'b'.
Solutions
Solution 1: Brain Teaser
Thinking
The string contains only a and b. All as precede all bs iff no a appears after a b, i.e., the substring ba is absent.
A linear scan or a substring test decides this in \(O(n)\).
Return whether "ba" does not occur in \(s\).
According to the problem statement, the string \(s\) consists only of characters a and b.
To ensure that all as appear before all bs, the condition that must be met is that b should not appear before a. In other words, the substring "ba" should not be present in the string \(s\).
The time complexity is \(O(n)\), where \(n\) is the length of the string \(s\). The space complexity is \(O(1)\).
1 2 3 | |
1 2 3 4 5 | |
1 2 3 4 5 6 | |
1 2 3 | |
1 2 3 | |
1 2 3 4 5 | |