LeetCode 1616. Split Two Strings to Make Palindrome

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Question

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.

Constraints:

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

Example

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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.

Input: a = "abdef", b = "fecab"
Output: true

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.

Input: a = "xbdef", b = "xecab"
Output: false

Solution

题目意思比较明确，即我们找一个相同的位置分割a,b字符串，然后看a的前部分和b的后半部分或者b的前半部分和a的后半部分组成的字符串是否为回文字符串。所以，我们可以简单的遍历每一个位置然后分别分割然后组成字符串，之后再验证是否符合回文的特性，但是这样子就是N^2的算法。这样的算法会TLE，所以我们需要更加高效的方式，由于我们都是用前半部分或者是后半部分合成，我们可以看例子3，a = "ulacfd", b = "jizalu"，我们其实可以得到最长的a前缀子串以及b最长的后缀子串，即ula和alu，然后再验证中间剩下的部分，如果符合条件，即符合。具体可以看看Huahua的视频

Code

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package com.leetcode.string.Palindrome;

/**
 * @Date: 10/14/2020
 * @Description: greedy
 **/
public class _1616_Split_Two_Strings_to_Make_Palindrome {
    public boolean checkPalindromeFormation(String a, String b) {
        // 找最长的前后缀匹配。
        // 相当于可以看a，b两个字符串prefix 以及suffix匹配最长的palindrome
        return check(a, b) || check(b, a);
    }


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

    public boolean isValid(String s, int i, int j) {
        while (i < j) {
            if (s.charAt(i) != s.charAt(j)) {
                return false;
            }
            i++;
            j--;
        }
        return true;
    }
}

Reference