Split Two Strings to Make Palindrome

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

Answer:

class Solution {
    public int maxPalindromeFromCenter(String str, int s, int e) {
        int len = 0;
        while (s >= 0 && str.charAt(s) == str.charAt(e)) {
            len++;
            s--;
            e++;
        }
        return len;
    }

    // Find max common prefix & suffix between the 2 strings
    public int maxCommon(String a, String b, int n) {
        int s = 0, e = n - 1;
        int len1 = 0;
        while (s <= e) {
            if (a.charAt(s) == b.charAt(e))
                len1++;
            else
                break;
            s++;
            e--;
        }
        int len2 = 0;
        s = 0;
        e = n - 1;
        while (s <= e) {
            if (b.charAt(s) == a.charAt(e))
                len2++;
            else
                break;
            s++;
            e--;
        }
        return Math.max(len1, len2);
    }

    public boolean checkPalindromeFormation(String a, String b) {
        int n = a.length();
        int P1 = maxPalindromeFromCenter(a, (int) Math.floor((n - 1) / 2.0), (int) Math.ceil((n - 1) / 2.0));
        int P2 = maxPalindromeFromCenter(b, (int) Math.floor((n - 1) / 2.0), (int) Math.ceil((n - 1) / 2.0));
        int common = maxCommon(a, b, n);
        return common + Math.max(P1, P2) >= (n + 1) / 2;
    }
}