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‎Problemset/1616-Split Two Strings to Make Palindrome/README.md

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# 1616. Split Two Strings to Make Palindrome
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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: <code>a<sub>prefix</sub></code> and <code>a<sub>suffix</sub></code> where <code>a = a<sub>prefix</sub> + a<sub>suffix</sub></code>, and splitting `b` into two strings: <code>b<sub>prefix</sub></code> and <code>b<sub>suffix</sub></code> where <code>b = b<sub>prefix</sub> + b<sub>suffix</sub></code>. Check if <code>a<sub>prefix</sub> + b<sub>suffix</sub></code> or <code>b<sub>prefix</sub> + a<sub>suffix</sub></code> forms a palindrome.
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When you split a string `s` into <code>s<sub>prefix</sub></code> and <code>s<sub>suffix</sub></code>, either <code>s<sub>suffix</sub></code> or <code>s<sub>prefix</sub></code> is allowed to be empty. For example, if `s = "abc"`, then `"" + "abc"`, `"a" + "bc"`, `"ab" + "c"` , and `"abc" + ""` are valid splits.
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Return `true` *if it is possible to form a palindrome string, otherwise return* `false`.
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**Notice** that `x + y` denotes the concatenation of strings `x` and `y`.
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#### Example 1:
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<pre>
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<strong>Input:</strong> a = "x", b = "y"
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<strong>Output:</strong> true
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<strong>Explanation:</strong> If either a or b are palindromes the answer is true since you can split in the following way:
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aprefix = "", asuffix = "x"
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bprefix = "", bsuffix = "y"
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Then, aprefix + bsuffix = "" + "y" = "y", which is a palindrome.
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</pre>
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#### Example 2:
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<pre>
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<strong>Input:</strong> a = "xbdef", b = "xecab"
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<strong>Output:</strong> false
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</pre>
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#### Example 3:
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<pre>
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<strong>Input:</strong> a = "ulacfd", b = "jizalu"
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<strong>Output:</strong> true
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<strong>Explanation:</strong> Split them at index 3:
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aprefix = "ula", asuffix = "cfd"
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bprefix = "jiz", bsuffix = "alu"
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Then, aprefix + bsuffix = "ula" + "alu" = "ulaalu", which is a palindrome.
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</pre>
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#### Constraints:
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* <code>1 <= a.length, b.length <= 10<sup>5</sup></code>
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* `a.length == b.length`
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* `a` and `b` consist of lowercase English letters
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## Solutions (Rust)
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### 1. Solution
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```Rust
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impl Solution {
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pub fn check_palindrome_formation(a: String, b: String) -> bool {
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let a = a.as_bytes();
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let b = b.as_bytes();
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let n = a.len();
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let mut prefix_i = -1;
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for i in 0..n / 2 {
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if a[i] == b[n - 1 - i] {
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prefix_i = i as i32;
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} else {
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break;
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}
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}
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for i in 0..n / 2 {
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if b[i] == a[n - 1 - i] {
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prefix_i = prefix_i.max(i as i32);
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} else {
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break;
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}
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}
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if prefix_i == n as i32 / 2 - 1 {
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return true;
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}
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for i in (0..=n / 2).rev() {
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if a[i] == a[n - 1 - i] && i as i32 - 1 <= prefix_i {
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return true;
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} else if a[i] != a[n - 1 - i] {
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break;
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}
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}
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for i in (0..=n / 2).rev() {
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if b[i] == b[n - 1 - i] && i as i32 - 1 <= prefix_i {
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return true;
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} else if b[i] != b[n - 1 - i] {
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break;
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}
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}
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false
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}
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}
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```

‎Problemset/1616-Split Two Strings to Make Palindrome/README_CN.md

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# 1616. 分割两个字符串得到回文串
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给你两个字符串 `a` 和 `b` ，它们长度相同。请你选择一个下标，将两个字符串都在 **相同的下标** 分割开。由 `a` 可以得到两个字符串： <code>a<sub>prefix</sub></code> 和 <code>a<sub>suffix</sub></code> ，满足 <code>a = a<sub>prefix</sub> + a<sub>suffix</sub></code> ，同理，由 `b` 可以得到两个字符串 <code>b<sub>prefix</sub></code> 和 <code>b<sub>suffix</sub></code> ，满足 <code>b = b<sub>prefix</sub> + b<sub>suffix</sub></code> 。请你判断 <code>a<sub>prefix</sub> + b<sub>suffix</sub></code> 或者 <code>b<sub>prefix</sub> + a<sub>suffix</sub></code> 能否构成回文串。
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当你将一个字符串 `s` 分割成 <code>s<sub>prefix</sub></code> 和 <code>s<sub>suffix</sub></code> 时， <code>s<sub>suffix</sub></code> 或者 <code>s<sub>prefix</sub></code> 可以为空。比方说， `s = "abc"` 那么 `"" + "abc"` ， `"a" + "bc"` ， `"ab" + "c"` 和 `"abc" + ""` 都是合法分割。
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如果 **能构成回文字符串** ，那么请返回 `true`，否则返回 `false` 。
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**注意**， `x + y` 表示连接字符串 `x` 和 `y` 。
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#### 示例 1:
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<pre>
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<strong>输入:</strong> a = "x", b = "y"
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<strong>输出:</strong> true
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<strong>解释:</strong> 如果 a 或者 b 是回文串，那么答案一定为 true ，因为你可以如下分割：
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aprefix = "", asuffix = "x"
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bprefix = "", bsuffix = "y"
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那么 aprefix + bsuffix = "" + "y" = "y" 是回文串。
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</pre>
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#### 示例 2:
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<pre>
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<strong>输入:</strong> a = "xbdef", b = "xecab"
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<strong>输出:</strong> false
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</pre>
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#### 示例 3:
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<pre>
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<strong>输入:</strong> a = "ulacfd", b = "jizalu"
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<strong>输出:</strong> true
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<strong>解释:</strong> 在下标为 3 处分割：
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aprefix = "ula", asuffix = "cfd"
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bprefix = "jiz", bsuffix = "alu"
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那么 aprefix + bsuffix = "ula" + "alu" = "ulaalu" 是回文串。
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</pre>
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#### 提示:
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* <code>1 <= a.length, b.length <= 10<sup>5</sup></code>
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* `a.length == b.length`
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* `a` 和 `b` 都只包含小写英文字母
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## 题解 (Rust)
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### 1. 题解
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```Rust
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impl Solution {
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pub fn check_palindrome_formation(a: String, b: String) -> bool {
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let a = a.as_bytes();
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let b = b.as_bytes();
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let n = a.len();
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let mut prefix_i = -1;
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for i in 0..n / 2 {
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if a[i] == b[n - 1 - i] {
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prefix_i = i as i32;
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} else {
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break;
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}
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}
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for i in 0..n / 2 {
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if b[i] == a[n - 1 - i] {
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prefix_i = prefix_i.max(i as i32);
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} else {
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break;
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}
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}
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if prefix_i == n as i32 / 2 - 1 {
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return true;
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}
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for i in (0..=n / 2).rev() {
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if a[i] == a[n - 1 - i] && i as i32 - 1 <= prefix_i {
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return true;
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} else if a[i] != a[n - 1 - i] {
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break;
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}
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}
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for i in (0..=n / 2).rev() {
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if b[i] == b[n - 1 - i] && i as i32 - 1 <= prefix_i {
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return true;
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} else if b[i] != b[n - 1 - i] {
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break;
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}
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}
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false
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}
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}
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```

‎Problemset/1616-Split Two Strings to Make Palindrome/Solution.rs

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impl Solution {
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pub fn check_palindrome_formation(a: String, b: String) -> bool {
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let a = a.as_bytes();
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let b = b.as_bytes();
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let n = a.len();
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let mut prefix_i = -1;
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for i in 0..n / 2 {
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if a[i] == b[n - 1 - i] {
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prefix_i = i as i32;
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} else {
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break;
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}
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}
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for i in 0..n / 2 {
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if b[i] == a[n - 1 - i] {
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prefix_i = prefix_i.max(i as i32);
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} else {
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break;
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}
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}
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if prefix_i == n as i32 / 2 - 1 {
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return true;
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}
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for i in (0..=n / 2).rev() {
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if a[i] == a[n - 1 - i] && i as i32 - 1 <= prefix_i {
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return true;
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} else if a[i] != a[n - 1 - i] {
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break;
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}
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}
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for i in (0..=n / 2).rev() {
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if b[i] == b[n - 1 - i] && i as i32 - 1 <= prefix_i {
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return true;
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} else if b[i] != b[n - 1 - i] {
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break;
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}
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}
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false
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}
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}

‎README.md

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10101010
[1610][1610l]|[Maximum Number of Visible Points][1610] |![rs]
10111011
[1614][1614l]|[Maximum Nesting Depth of the Parentheses][1614] |![rb]&nbsp;&nbsp;![rs]
10121012
[1615][1615l]|[Maximal Network Rank][1615] |![rs]
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[1616][1616l]|[Split Two Strings to Make Palindrome][1616] |![rs]
10131014
[1619][1619l]|[Mean of Array After Removing Some Elements][1619] |![rb]&nbsp;&nbsp;![rs]
10141015
[1620][1620l]|[Coordinate With Maximum Network Quality][1620] |![rs]
10151016
[1621][1621l]|[Number of Sets of K Non-Overlapping Line Segments][1621] |![py]
@@ -2565,6 +2566,7 @@
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[1610]:Problemset/1610-Maximum%20Number%20of%20Visible%20Points/README.md#1610-maximum-number-of-visible-points
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[1614]:Problemset/1614-Maximum%20Nesting%20Depth%20of%20the%20Parentheses/README.md#1614-maximum-nesting-depth-of-the-parentheses
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[1615]:Problemset/1615-Maximal%20Network%20Rank/README.md#1615-maximal-network-rank
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[1616]:Problemset/1616-Split%20Two%20Strings%20to%20Make%20Palindrome/README.md#1616-split-two-strings-to-make-palindrome
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[1619]:Problemset/1619-Mean%20of%20Array%20After%20Removing%20Some%20Elements/README.md#1619-mean-of-array-after-removing-some-elements
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[1620]:Problemset/1620-Coordinate%20With%20Maximum%20Network%20Quality/README.md#1620-coordinate-with-maximum-network-quality
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[1621]:Problemset/1621-Number%20of%20Sets%20of%20K%20Non-Overlapping%20Line%20Segments/README.md#1621-number-of-sets-of-k-non-overlapping-line-segments
@@ -4119,6 +4121,7 @@
41194121
[1610l]:https://leetcode.com/problems/maximum-number-of-visible-points/
41204122
[1614l]:https://leetcode.com/problems/maximum-nesting-depth-of-the-parentheses/
41214123
[1615l]:https://leetcode.com/problems/maximal-network-rank/
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[1616l]:https://leetcode.com/problems/split-two-strings-to-make-palindrome/
41224125
[1619l]:https://leetcode.com/problems/mean-of-array-after-removing-some-elements/
41234126
[1620l]:https://leetcode.com/problems/coordinate-with-maximum-network-quality/
41244127
[1621l]:https://leetcode.com/problems/number-of-sets-of-k-non-overlapping-line-segments/

‎README_CN.md

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10101010
[1610][1610l]|[可见点的最大数目][1610] |![rs]
10111011
[1614][1614l]|[括号的最大嵌套深度][1614] |![rb]&nbsp;&nbsp;![rs]
10121012
[1615][1615l]|[最大网络秩][1615] |![rs]
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[1616][1616l]|[分割两个字符串得到回文串][1616] |![rs]
10131014
[1619][1619l]|[删除某些元素后的数组均值][1619] |![rb]&nbsp;&nbsp;![rs]
10141015
[1620][1620l]|[网络信号最好的坐标][1620] |![rs]
10151016
[1621][1621l]|[大小为 K 的不重叠线段的数目][1621] |![py]
@@ -2565,6 +2566,7 @@
25652566
[1610]:Problemset/1610-Maximum%20Number%20of%20Visible%20Points/README_CN.md#1610-可见点的最大数目
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[1614]:Problemset/1614-Maximum%20Nesting%20Depth%20of%20the%20Parentheses/README_CN.md#1614-括号的最大嵌套深度
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[1615]:Problemset/1615-Maximal%20Network%20Rank/README_CN.md#1615-最大网络秩
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[1616]:Problemset/1616-Split%20Two%20Strings%20to%20Make%20Palindrome/README_CN.md#1616-分割两个字符串得到回文串
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[1619]:Problemset/1619-Mean%20of%20Array%20After%20Removing%20Some%20Elements/README_CN.md#1619-删除某些元素后的数组均值
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[1620]:Problemset/1620-Coordinate%20With%20Maximum%20Network%20Quality/README_CN.md#1620-网络信号最好的坐标
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[1621]:Problemset/1621-Number%20of%20Sets%20of%20K%20Non-Overlapping%20Line%20Segments/README_CN.md#1621-大小为-k-的不重叠线段的数目
@@ -4119,6 +4121,7 @@
41194121
[1610l]:https://leetcode.cn/problems/maximum-number-of-visible-points/
41204122
[1614l]:https://leetcode.cn/problems/maximum-nesting-depth-of-the-parentheses/
41214123
[1615l]:https://leetcode.cn/problems/maximal-network-rank/
4124+
[1616l]:https://leetcode.cn/problems/split-two-strings-to-make-palindrome/
41224125
[1619l]:https://leetcode.cn/problems/mean-of-array-after-removing-some-elements/
41234126
[1620l]:https://leetcode.cn/problems/coordinate-with-maximum-network-quality/
41244127
[1621l]:https://leetcode.cn/problems/number-of-sets-of-k-non-overlapping-line-segments/

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