String without aaa or bbb

Time: O(A+B); Space: O(1); easy

Given two integers A and B, return any string S such that: * S has length A + B and contains exactly A ‘a’ letters, and exactly B ‘b’ letters; * The substring ‘aaa’ does not occur in S; * The substring ‘bbb’ does not occur in S.

Example 1:

Input: A = 1, B = 2

Output: “abb”

Explanation:

  • “abb”, “bab” and “bba” are all correct answers.

Example 2:

Input: A = 4, B = 1

Output: “aabaa”

Notes:

  • 0 <= A <= 100

  • 0 <= B <= 100

  • It is guaranteed such an S exists for the given A and B.

[1]:

class Solution1(object):
    """
    Time: O(A+B)
    Space: O(1)
    """
    def strWithout3a3b(self, A, B):
        """
        :type A: int
        :type B: int
        :rtype: str
        """
        result = []
        put_A = None

        while A or B:
            if len(result) >= 2 and result[-1] == result[-2]:
                put_A = result[-1] == 'b'
            else:
                put_A = A >= B

            if put_A:
                A -= 1
                result.append('a')
            else:
                B -= 1
                result.append('b')

        return ''.join(result)

[4]:

s = Solution1()
A = 1
B = 2
assert s.strWithout3a3b(A, B) == "abb" or "abb" or "bab" or "bba"
A = 4
B = 1
assert s.strWithout3a3b(A, B) == "aabaa"