Introduction

In this blog post, we will explore how to solve the LeetCode problem "2745" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, 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

You are given three integers x, y, and z. You have x strings equal to "AA", y strings equal to "BB", and z strings equal to "AB". You want to choose some (possibly all or none) of these strings and concatenate them in some order to form a new string. This new string must not contain "AAA" or "BBB" as a substring. Return the maximum possible length of the new string. A substring is a contiguous non-empty sequence of characters within a string. Example 1: Input: x = 2, y = 5, z = 1 Output: 12 Explanation: We can concatenate the strings "BB", "AA", "BB", "AA", "BB", and "AB" in that order. Then, our new string is "BBAABBAABBAB". That string has length 12, and we can show that it is impossible to construct a string of longer length. Example 2: Input: x = 3, y = 2, z = 2 Output: 14 Explanation: We can concatenate the strings "AB", "AB", "AA", "BB", "AA", "BB", and "AA" in that order. Then, our new string is "ABABAABBAABBAA". That string has length 14, and we can show that it is impossible to construct a string of longer length. Constraints: 1 <= x, y, z <= 50

Explanation

Here's the breakdown:

• Core Idea: Maximize string length by strategically placing 'AA', 'BB', and 'AB' strings. The crucial constraint is avoiding "AAA" and "BBB". If x and y are vastly different, we need to limit the excessive repetition of either 'A' or 'B'.
• Balancing Act: If the difference between x and y is large, we can only use a limited number of the more frequent character. We also use all AB strings to join AA and BB blocks, improving the result.

• Optimization: Calculate the optimal length based on the constraints and the input values.

• Complexity: Time Complexity: O(1), Space Complexity: O(1)

Code

    def solve():
    x, y, z = map(int, input().split())

    ans = 0
    if x == y:
        ans = 2 * x + 2 * z
    elif x > y:
        diff = x - y
        if diff > y + z:
            ans = 2 * (y + z) + 1
        else:
            ans = 2 * x + 2 * z
    else:
        diff = y - x
        if diff > x + z:
            ans = 2 * (x + z) + 1
        else:
            ans = 2 * y + 2 * z
    print(ans)

solve()