1616. Split Two Strings to Make Palindrome: AC

Author: tan-wei

src/solution/mod.rs

@@ -1218,3 +1218,4 @@ mod s1610_maximum_number_of_visible_points;
 mod s1611_minimum_one_bit_operations_to_make_integers_zero;
 mod s1614_maximum_nesting_depth_of_the_parentheses;
 mod s1615_maximal_network_rank;
+mod s1616_split_two_strings_to_make_palindrome;

src/solution/s1616_split_two_strings_to_make_palindrome.rs

@@ -0,0 +1,110 @@
+/**
+ * [1616] Split Two Strings to Make Palindrome
+ *
+ * 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 <= 10^5
+ * 	a.length == b.length
+ * 	a and b consist of lowercase English letters
+ *
+ */
+pub struct Solution {}
+
+// problem: https://leetcode.com/problems/split-two-strings-to-make-palindrome/
+// discuss: https://leetcode.com/problems/split-two-strings-to-make-palindrome/discuss/?currentPage=1&orderBy=most_votes&query=
+
+// submission codes start here
+
+impl Solution {
+    pub fn check_palindrome_formation(a: String, b: String) -> bool {
+        let (a, b) = (a.as_bytes(), b.as_bytes());
+
+        Self::check(a, b) || Self::check(b, a)
+    }
+
+    fn is_palindrome(s: &[u8], i: usize, j: usize) -> bool {
+        let (mut i, mut j) = (i, j);
+        while i < j && s[i] == s[j] {
+            i += 1;
+            j -= 1;
+        }
+        i >= j
+    }
+
+    fn check(a: &[u8], b: &[u8]) -> bool {
+        let (mut i, mut j) = (0, a.len() - 1);
+
+        while i < j && a[i] == b[j] {
+            i += 1;
+            j -= 1;
+        }
+
+        Self::is_palindrome(a, i, j) || Self::is_palindrome(b, i, j)
+    }
+}
+
+// submission codes end
+
+#[cfg(test)]
+mod tests {
+    use super::*;
+
+    #[test]
+    fn test_1616_example_1() {
+        let a = "x".to_string();
+        let b = "y".to_string();
+
+        let result = true;
+
+        assert_eq!(Solution::check_palindrome_formation(a, b), result);
+    }
+
+    #[test]
+    fn test_1616_example_2() {
+        let a = "xbdef".to_string();
+        let b = "xecab".to_string();
+
+        let result = false;
+
+        assert_eq!(Solution::check_palindrome_formation(a, b), result);
+    }
+
+    #[test]
+    fn test_1616_example_3() {
+        let a = "ulacfd".to_string();
+        let b = "jizalu".to_string();
+
+        let result = true;
+
+        assert_eq!(Solution::check_palindrome_formation(a, b), result);
+    }
+}