# 984. String Without AAA or BBB

## Description

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, and
• 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"


Constraints:

• 0 <= a, b <= 100
• It is guaranteed such an s exists for the given a and b.

## Solutions

• class Solution {
public String strWithout3a3b(int a, int b) {
StringBuilder ans = new StringBuilder();
while (a > 0 && b > 0) {
if (a > b) {
ans.append("aab");
a -= 2;
b -= 1;
} else if (a < b) {
ans.append("bba");
a -= 1;
b -= 2;
} else {
ans.append("ab");
--a;
--b;
}
}
if (a > 0) {
ans.append("a".repeat(a));
}
if (b > 0) {
ans.append("b".repeat(b));
}
return ans.toString();
}
}

• class Solution {
public:
string strWithout3a3b(int a, int b) {
string ans;
while (a && b) {
if (a > b) {
ans += "aab";
a -= 2;
b -= 1;
} else if (a < b) {
ans += "bba";
a -= 1;
b -= 2;
} else {
ans += "ab";
--a;
--b;
}
}
if (a) ans += string(a, 'a');
if (b) ans += string(b, 'b');
return ans;
}
};

• class Solution:
def strWithout3a3b(self, a: int, b: int) -> str:
ans = []
while a and b:
if a > b:
ans.append('aab')
a, b = a - 2, b - 1
elif a < b:
ans.append('bba')
a, b = a - 1, b - 2
else:
ans.append('ab')
a, b = a - 1, b - 1
if a:
ans.append('a' * a)
if b:
ans.append('b' * b)
return ''.join(ans)


• func strWithout3a3b(a int, b int) string {
var ans strings.Builder
for a > 0 && b > 0 {
if a > b {
ans.WriteString("aab")
a -= 2
b -= 1
} else if a < b {
ans.WriteString("bba")
a -= 1
b -= 2
} else {
ans.WriteString("ab")
a--
b--
}
}
if a > 0 {
ans.WriteString(strings.Repeat("a", a))
}
if b > 0 {
ans.WriteString(strings.Repeat("b", b))
}
return ans.String()
}