README.md

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

 

Solutions

Solution: Two Pointers

• Time complexity: O(n)
• Space complexity: O(1)

 

JavaScript

/**
 * @param {string} a
 * @param {string} b
 * @return {boolean}
 */
var checkPalindromeFormation = function(a, b) {
    const checkPalindrome = (a, b) => {
        let left = 0;
        let right = b.length - 1;

        while (left < right && a[left] === b[right]) {
            left += 1;
            right -= 1;
        }
        return isPalindrome(a, left, right) || isPalindrome(b, left, right);
    };
    const isPalindrome = (str, left, right) => {
        while (left < right && str[left] === str[right]) {
            left += 1;
            right -= 1;
        }
        return left >= right;
    };

    return checkPalindrome(a, b) || checkPalindrome(b, a);
};