# 1616. Split Two Strings to Make Palindrome

## Description

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

## Solutions

• class Solution {
public boolean checkPalindromeFormation(String a, String b) {
return check1(a, b) || check1(b, a);
}

private boolean check1(String a, String b) {
int i = 0;
int j = b.length() - 1;
while (i < j && a.charAt(i) == b.charAt(j)) {
i++;
j--;
}
return i >= j || check2(a, i, j) || check2(b, i, j);
}

private boolean check2(String a, int i, int j) {
while (i < j && a.charAt(i) == a.charAt(j)) {
i++;
j--;
}
return i >= j;
}
}

• class Solution {
public:
bool checkPalindromeFormation(string a, string b) {
return check1(a, b) || check1(b, a);
}

private:
bool check1(string& a, string& b) {
int i = 0, j = b.size() - 1;
while (i < j && a[i] == b[j]) {
++i;
--j;
}
return i >= j || check2(a, i, j) || check2(b, i, j);
}

bool check2(string& a, int i, int j) {
while (i <= j && a[i] == a[j]) {
++i;
--j;
}
return i >= j;
}
};

• class Solution:
def checkPalindromeFormation(self, a: str, b: str) -> bool:
def check1(a: str, b: str) -> bool:
i, j = 0, len(b) - 1
while i < j and a[i] == b[j]:
i, j = i + 1, j - 1
return i >= j or check2(a, i, j) or check2(b, i, j)

def check2(a: str, i: int, j: int) -> bool:
return a[i : j + 1] == a[i : j + 1][::-1]

return check1(a, b) or check1(b, a)


• func checkPalindromeFormation(a string, b string) bool {
return check1(a, b) || check1(b, a)
}

func check1(a, b string) bool {
i, j := 0, len(b)-1
for i < j && a[i] == b[j] {
i++
j--
}
return i >= j || check2(a, i, j) || check2(b, i, j)
}

func check2(a string, i, j int) bool {
for i < j && a[i] == a[j] {
i++
j--
}
return i >= j
}

• function checkPalindromeFormation(a: string, b: string): boolean {
const check1 = (a: string, b: string) => {
let i = 0;
let j = b.length - 1;
while (i < j && a.charAt(i) === b.charAt(j)) {
i++;
j--;
}
return i >= j || check2(a, i, j) || check2(b, i, j);
};

const check2 = (a: string, i: number, j: number) => {
while (i < j && a.charAt(i) === a.charAt(j)) {
i++;
j--;
}
return i >= j;
};
return check1(a, b) || check1(b, a);
}


• impl Solution {
pub fn check_palindrome_formation(a: String, b: String) -> bool {
fn check1(a: &[u8], b: &[u8]) -> bool {
let (mut i, mut j) = (0, b.len() - 1);
while i < j && a[i] == b[j] {
i += 1;
j -= 1;
}
if i >= j {
return true;
}
check2(a, i, j) || check2(b, i, j)
}

fn check2(a: &[u8], mut i: usize, mut j: usize) -> bool {
while i < j && a[i] == a[j] {
i += 1;
j -= 1;
}
i >= j
}

let a = a.as_bytes();
let b = b.as_bytes();
check1(a, b) || check1(b, a)
}
}