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
andb
consist of lowercase English letters
Related Topics:
String
// OJ: https://leetcode.com/problems/split-two-strings-to-make-palindrome/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
bool check(string a, string b) {
int N = a.size(), i = N / 2;
while (i - 1 >= 0 && a[i - 1] == a[N - i]) --i; // a[i..(N-i-1)] is palindrome
reverse(begin(b), end(b));
return a.substr(0, i) == b.substr(0, i) || a.substr(N - i) == b.substr(N - i);
}
public:
bool checkPalindromeFormation(string a, string b) {
return check(a, b) || check(b, a);
}
};
Or don't create substring.
// OJ: https://leetcode.com/problems/split-two-strings-to-make-palindrome/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
bool equal(string &a, string &b, int i) {
while (i - 1 >= 0 && a[i - 1] == b[a.size() - i]) --i;
return i == 0;
}
bool check(string &a, string &b) {
int N = a.size(), i = N / 2;
while (i - 1 >= 0 && a[i - 1] == a[N - i]) --i;
return equal(a, b, i) || equal(b, a, i);
}
public:
bool checkPalindromeFormation(string a, string b) {
return check(a, b) || check(b, a);
}
};