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Formatted question description: https://leetcode.ca/all/984.html

# 984. String Without AAA or BBB (Medium)

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"

Note:

1. 0 <= A <= 100
2. 0 <= B <= 100
3. It is guaranteed such an S exists for the given A and B.

Related Topics:
Greedy

## Solution 1.

• class Solution {
public String strWithout3a3b(int A, int B) {
StringBuffer sb = new StringBuffer();
while (A > B && B > 0) {
sb.append("aab");
A -= 2;
B--;
}
while (A < B && A > 0) {
sb.append("bba");
B -= 2;
A--;
}
while (A > 0 && B > 0) {
sb.append("ab");
A--;
B--;
}
while (A > 0) {
sb.append("a");
A--;
}
while (B > 0) {
sb.append("b");
B--;
}
return sb.toString();
}
}

############

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();
}
}

• // OJ: https://leetcode.com/problems/string-without-aaa-or-bbb/
// Time: O(A + B)
// Space: O(1)
class Solution {
public:
string strWithout3a3b(int A, int B) {
char a = 'a', b = 'b';
if (A < B) swap(A, B), swap(a, b);
string ans;
while (A && B) {
for (int i = A > B ? 2 : 1; i > 0; --i, --A) ans += a;
ans += b;
--B;
}
while (A-- > 0) ans += a;
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)

############

class Solution(object):
def strWithout3a3b(self, A, B):
"""
:type A: int
:type B: int
:rtype: str
"""
_len = A + B
# to make A > B
a, b = ('a', 'b') if A > B else ('b', 'a')
A, B = (A, B) if A > B else (B, A)
ab = [a + b] * B
A -= B
res = []
while A:
res.append(a)
if ab:
res.append(ab.pop())
A -= 1
res += ab
return "".join(res)

• 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()
}