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 = "xbdef", b = "xecab"
Output: false
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.
Constraints:
1 <= a.length, b.length <= 105
a.length == b.length
a and b consist of lowercase English letters
Answer:
class Solution {
public int maxPalindromeFromCenter(String str, int s, int e) {
int len = 0;
while (s >= 0 && str.charAt(s) == str.charAt(e)) {
len++;
s--;
e++;
}
return len;
}
// Find max common prefix & suffix between the 2 strings
public int maxCommon(String a, String b, int n) {
int s = 0, e = n - 1;
int len1 = 0;
while (s <= e) {
if (a.charAt(s) == b.charAt(e))
len1++;
else
break;
s++;
e--;
}
int len2 = 0;
s = 0;
e = n - 1;
while (s <= e) {
if (b.charAt(s) == a.charAt(e))
len2++;
else
break;
s++;
e--;
}
return Math.max(len1, len2);
}
public boolean checkPalindromeFormation(String a, String b) {
int n = a.length();
int P1 = maxPalindromeFromCenter(a, (int) Math.floor((n - 1) / 2.0), (int) Math.ceil((n - 1) / 2.0));
int P2 = maxPalindromeFromCenter(b, (int) Math.floor((n - 1) / 2.0), (int) Math.ceil((n - 1) / 2.0));
int common = maxCommon(a, b, n);
return common + Math.max(P1, P2) >= (n + 1) / 2;
}
}