LeetCode in Kotlin

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

@Suppress("NAME_SHADOWING")
class Solution {
    fun checkPalindromeFormation(a: String, b: String): Boolean {
        return check(a, b) || check(b, a)
    }

    private fun check(a: String, b: String): Boolean {
        var i = 0
        var j = b.length - 1
        while (j > i && a[i] == b[j]) {
            ++i
            --j
        }
        return isPalindrome(a, i, j) || isPalindrome(b, i, j)
    }

    private fun isPalindrome(s: String, i: Int, j: Int): Boolean {
        var i = i
        var j = j
        while (j > i && s[i] == s[j]) {
            ++i
            --j
        }
        return i >= j
    }
}

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