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.
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Example 1:
\n\n\nInput: a = "x", b = "y"\nOutput: true\nExplaination: If either a or b are palindromes the answer is true since you can split in the following way:\naprefix = "", asuffix = "x"\nbprefix = "", bsuffix = "y"\nThen, aprefix + bsuffix = "" + "y" = "y", which is a palindrome.\n\n\n
Example 2:
\n\n\nInput: a = "xbdef", b = "xecab"\nOutput: false\n\n\n
Example 3:
\n\n\nInput: a = "ulacfd", b = "jizalu"\nOutput: true\nExplaination: Split them at index 3:\naprefix = "ula", asuffix = "cfd"\nbprefix = "jiz", bsuffix = "alu"\nThen, aprefix + bsuffix = "ula" + "alu" = "ulaalu", which is a palindrome.\n\n\n
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Constraints:
\n\n1 <= a.length, b.length <= 105a.length == b.lengtha and b consist of lowercase English lettersCompiled with clang 11 using the latest C++ 20 standard.
Your code is compiled with level two optimization (-O2). AddressSanitizer is also enabled to help detect out-of-bounds and use-after-free bugs.
Most standard library headers are already included automatically for your convenience.
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