1616. Split Two Strings to Make Palindrome

MediumLeetCode

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

Explanation: 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

Explanation: 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 <= 10^5
• a.length == b.length
• a and b consist of lowercase English letters

How to solve the problem

• Brute Force (Time Limit Exceeded)
• Two Pointers

# Approach 1: Brute Force (Time Limit Exceeded)
class Solution:
    def checkPalindromeFormation(self, a: str, b: str) -> bool:
        for i in range(len(a))+1:
            s1 = a[:i] + b[i:]
            s2 = b[:i] + a[i:]
            if s1 == s1[::-1] or s2 == s2[::-1]:
                return True
        return False

# Approach 2: Two Pointers
class Solution:
    def checkPalindromeFormation(self, a: str, b: str) -> bool:
        def check(a, b):
            i, j = 0, len(a) - 1
            while i < j and a[i] == b[j]:
                i += 1
                j -= 1
            return (a[i : j + 1] == a[i : j + 1][::-1] or b[i : j + 1] == b[i : j + 1][::-1])  
                # Think about this case when you need a counter-exapmle: a = "abcba" b = "axcxa"

        return check(a, b) or check(b, a)
        # Must check both check(a, b) and check(b, a)
        # check(a, b): aprefix + bsuffix
        # check(b, a): bprefix + asuffix
        # Think about this case: a = "efdef", b = "fecab"

Complexity

• Approach 1 (Brute Force): Time complexity: O(n²), Space complexity: O(n)
• Approach 2 (Two Pointers): Time complexity: O(n), Space complexity: O(1)