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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 <= 105a.length == b.lengthaandbconsist of lowercase English letters
Solutions
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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) } }