LeetCode [1616] Split Two Strings to Make Palindrome

 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:
aprefix = "", asuffix = "x"
bprefix = "", bsuffix = "y"
Then, aprefix + bsuffix = "" + "y" = "y", which is a palindrome.

Example 2:

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

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.

Example 4:

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

 

Constraints:

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

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class Solution { public boolean checkPalindromeFormation(String a, String b) { if(helper(a, b, 0, b.length()-1, false)) return true; return helper(b, a, 0, a.length()-1, false); } boolean helper(String a, String b, int l, int r, boolean oneString){ if(l>=r) return true; if(a.charAt(l)!=b.charAt(r)) return false; if(helper(a, b, l+1, r-1, oneString)) return true; if(!oneString){ if(helper(a, a, l+1, r-1, true)) return true; if(helper(b, b, l+1, r-1, true)) return true; } return false; } }