README.md

1616. 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
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 <= 105
• a.length == b.length
• a and b consist of lowercase English letters

Solutions (Rust)

1. Solution

impl Solution {
    pub fn check_palindrome_formation(a: String, b: String) -> bool {
        let a = a.as_bytes();
        let b = b.as_bytes();
        let n = a.len();
        let mut prefix_i = -1;

        for i in 0..n / 2 {
            if a[i] == b[n - 1 - i] {
                prefix_i = i as i32;
            } else {
                break;
            }
        }
        for i in 0..n / 2 {
            if b[i] == a[n - 1 - i] {
                prefix_i = prefix_i.max(i as i32);
            } else {
                break;
            }
        }

        if prefix_i == n as i32 / 2 - 1 {
            return true;
        }

        for i in (0..=n / 2).rev() {
            if a[i] == a[n - 1 - i] && i as i32 - 1 <= prefix_i {
                return true;
            } else if a[i] != a[n - 1 - i] {
                break;
            }
        }
        for i in (0..=n / 2).rev() {
            if b[i] == b[n - 1 - i] && i as i32 - 1 <= prefix_i {
                return true;
            } else if b[i] != b[n - 1 - i] {
                break;
            }
        }

        false
    }
}