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
Solution:
class Solution {
public boolean checkPalindromeFormation(String a, String b) {
return check(a, b) || check(b, a);
}
private boolean check(String a, String b) {
int i = 0, j = a.length() - 1;
while (i < j && a.charAt(i) == b.charAt(j)) {
++ i;
-- j;
}
return isP(a.substring(i, a.length() - i)) || isP(b.substring(i, b.length() - i));
}
private boolean isP(String s) {
int left = 0, right = s.length() - 1;
while (left < right) {
if (s.charAt(left) != s.charAt(right)) {
return false;
}
left ++;
right --;
}
return true;
}
}