LeetCode-in-Java

1616. Split Two Strings to Make Palindrome

Medium

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:

a_prefix = “”, a_suffix = “x”

b_prefix = “”, b_suffix = “y”

Then, a_prefix + b_suffix = “” + “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:

a_prefix = “ula”, a_suffix = “cfd”

b_prefix = “jiz”, b_suffix = “alu”

Then, a_prefix + b_suffix = “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

@SuppressWarnings("java:S2234")
public class Solution {
    public boolean checkPalindromeFormation(String a, String b) {
        return check(a, b) || check(b, a);
    }

    private boolean check(String a, String b) {
        int i = 0;
        int j = b.length() - 1;
        while (j > i && a.charAt(i) == b.charAt(j)) {
            ++i;
            --j;
        }
        return isPalindrome(a, i, j) || isPalindrome(b, i, j);
    }

    private boolean isPalindrome(String s, int i, int j) {
        while (j > i && s.charAt(i) == s.charAt(j)) {
            ++i;
            --j;
        }
        return i >= j;
    }
}

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