# Solving Leetcode Interviews in Seconds with AI: String Without AAA or BBB


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "984" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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.  

	# Explanation
	Here's the breakdown of the solution:

*   **Greedy Approach:** Prioritize using the character with the higher count, but avoid creating "aaa" or "bbb".
*   **Conditional Logic:** If adding the character with the higher count would result in three consecutive identical characters, switch to the other character.
*   **String Building:** Efficiently construct the string character by character.

*   **Runtime Complexity:** O(a + b), **Storage Complexity:** O(a + b)

	
	# Code
	```python
	def solve():
    a = int(input())
    b = int(input())

    result = ""
    while a > 0 or b > 0:
        if a > b:
            if len(result) >= 2 and result[-1] == 'a' and result[-2] == 'a':
                if b > 0:
                    result += 'b'
                    b -= 1
                else:
                    result += 'a'
                    a -=1
            else:
                result += 'a'
                a -= 1
        else:
            if len(result) >= 2 and result[-1] == 'b' and result[-2] == 'b':
                if a > 0:
                    result += 'a'
                    a -= 1
                else:
                    result += 'b'
                    b -= 1
            else:
                result += 'b'
                b -= 1
    print(result)

solve()
	```