Introduction

In this blog post, we will explore how to solve the LeetCode problem "1616" 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 two strings a and b of the same length. Choose an index and split both strings at the same index, splitting a into two strings: aprefix and asuffix where a = aprefix + asuffix, and splitting b into two strings: bprefix and bsuffix where b = bprefix + bsuffix. Check if aprefix + bsuffix or bprefix + asuffix forms a palindrome. When you split a string s into sprefix and ssuffix, either ssuffix or sprefix is allowed to be empty. For example, if s = "abc", then "" + "abc", "a" + "bc", "ab" + "c" , and "abc" + "" are valid splits. Return true if it is possible to form a palindrome string, otherwise return false. Notice that x + y denotes the concatenation of strings x and y. Example 1: Input: a = "x", b = "y" Output: true Explaination: If either a or b are palindromes the answer is true since you can split in the following way: aprefix = "", asuffix = "x" bprefix = "", bsuffix = "y" Then, aprefix + bsuffix = "" + "y" = "y", which is a palindrome. Example 2: Input: a = "xbdef", b = "xecab" Output: false Example 3: Input: a = "ulacfd", b = "jizalu" Output: true Explaination: Split them at index 3: aprefix = "ula", asuffix = "cfd" bprefix = "jiz", bsuffix = "alu" Then, aprefix + bsuffix = "ula" + "alu" = "ulaalu", which is a palindrome. Constraints: 1 <= a.length, b.length <= 105 a.length == b.length a and b consist of lowercase English letters

Explanation

Here's the breakdown of the solution:

• Palindrome Check: Implement a helper function to efficiently check if a string is a palindrome.
• Split and Combine: Iterate to split the strings a and b at every possible index. For each split, create the combined strings aprefix + bsuffix and bprefix + asuffix.

• Early Exit: If either of the combined strings is a palindrome, return True immediately. If the loop completes without finding a palindrome, return False.

• Time Complexity: O(n), where n is the length of the strings.

• Space Complexity: O(1).

Code

    def checkPalindromeFormation(a, b):
    """
    Checks if it's possible to form a palindrome string by splitting a and b at the same index
    and combining the prefix of one with the suffix of the other.

    Args:
        a (str): The first string.
        b (str): The second string.

    Returns:
        bool: True if it is possible to form a palindrome, False otherwise.
    """

    def is_palindrome(s):
        """
        Helper function to check if a string is a palindrome.
        """
        left, right = 0, len(s) - 1
        while left < right:
            if s[left] != s[right]:
                return False
            left += 1
            right -= 1
        return True

    def check(a, b):
        """
        Checks if a prefix of a and suffix of b can form a palindrome or vice-versa
        """
        l, r = 0, len(a) - 1
        while l < r and a[l] == b[r]:
            l += 1
            r -= 1

        if l >= r:
            return True

        return is_palindrome(a[l:r+1]) or is_palindrome(b[l:r+1])

    return check(a,b) or check(b,a)