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

split-two-strings-to-make-palindrome.py

class Solution:
    def checkPalindromeFormation(self, a: str, b: str) -> bool:

        def isPalindrome(s, i, j):
            while i < j and s[i] == s[j]:
                i += 1
                j -= 1

            return i >= j

        def check(a, b):
            i, j = 0, len(a) - 1
            while i < j and a[i] == b[j]:
                i += 1
                j -= 1

            return isPalindrome(a,i,j) or isPalindrome(b,i,j)

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

split-two-strings-to-make-palindrome.cpp

class Solution {
public:
    bool isPalindrome(string &s, int i, int j){
        while (i < j && s[i] == s[j])
            ++i, --j;
        return i >= j;
    }

    bool check(string &a, string &b){
        int i = 0, j = a.size() - 1;
        while (i < j && a[i] == b[j])
            ++i, --j;
        return isPalindrome(a, i, j) || isPalindrome(b, i, j);
    }

    bool checkPalindromeFormation(string a, string b) {
        return check(a,b) || check(b,a);   
    }
};