leetcode - 1616. Split Two Strings to Make Palindrome

Description

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

Solution

The palindrome would only be like:

So get two pointers, one working on A, one working at B reversely, if the char that are pointed are not the same, then either change pointer A to work on B or change pointer B to work on A.

Time complexity: $o(n)$
Space complexity: $o(1)$

Code

class Solution:
    def checkPalindromeFormation(self, a: str, b: str) -> bool:
        def checker(a: str, b: str) -> bool:
            a_index, b_index = 0, -1
            cmp_a, cmp_b = a, b
            # cmp_a -> b
            flag_a_to_b, flag_b_to_a = True, True
            while a_index - b_index < len(b):
                if cmp_a[a_index] == cmp_b[b_index]:
                    a_index += 1
                    b_index -= 1
                else:
                    if cmp_a == a:
                        cmp_a = b
                    else:
                        flag_a_to_b = False
                        break
            # cmp_b -> a
            cmp_a, cmp_b = a, b
            a_index, b_index = 0, -1
            while a_index - b_index < len(b):
                if cmp_a[a_index] == cmp_b[b_index]:
                    a_index += 1
                    b_index -= 1
                else:
                    if cmp_b == b:
                        cmp_b = a
                    else:
                        flag_b_to_a = False
                        break
            return flag_a_to_b or flag_b_to_a
        return checker(a, b) or checker(b, a)