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feat: add solutions to lc problem: No.1216
No.1216.Valid Palindrome III
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‎solution/1200-1299/1216.Valid Palindrome III/README.md

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@@ -41,22 +41,133 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：动态规划**
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题目要求删去最多 $k$ 个字符，使得剩余的字符串是回文串。可以转换为求最长回文子序列的问题。
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我们定义 $f[i][j]$ 表示字符串 $s$ 中下标范围 $[i, j]$ 内的最长回文子序列的长度。初始时 $f[i][i] = 1$，即每个单独的字符都是一个回文子序列。
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当 $s[i] = s[j]$ 时，有 $f[i][j] = f[i + 1][j - 1] + 2$，即去掉 $s[i]$ 和 $s[j]$ 后，剩余的字符串的最长回文子序列长度加 $2$。
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当 $s[i] \neq s[j]$ 时，有 $f[i][j] = \max(f[i + 1][j], f[i][j - 1])$，即去掉 $s[i]$ 或 $s[j]$ 后，剩余的字符串的最长回文子序列长度。
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然后是否存在 $f[i][j] + k \geq n$，如果存在，说明可以通过删去 $k$ 个字符，使得剩余的字符串是回文串。
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时间复杂度 $O(n^2)$，空间复杂度 $O(n^2)$。其中 $n$ 为字符串 $s$ 的长度。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def isValidPalindrome(self, s: str, k: int) -> bool:
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n = len(s)
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f = [[0] * n for _ in range(n)]
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for i in range(n):
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f[i][i] = 1
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for i in range(n - 2, -1, -1):
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for j in range(i + 1, n):
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if s[i] == s[j]:
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f[i][j] = f[i + 1][j - 1] + 2
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else:
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f[i][j] = max(f[i + 1][j], f[i][j - 1])
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if f[i][j] + k >= n:
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return True
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return False
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public boolean isValidPalindrome(String s, int k) {
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int n = s.length();
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int[][] f = new int[n][n];
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for (int i = 0; i < n; ++i) {
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f[i][i] = 1;
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}
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for (int i = n - 2; i >= 0; --i) {
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for (int j = i + 1; j < n; ++j) {
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if (s.charAt(i) == s.charAt(j)) {
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f[i][j] = f[i + 1][j - 1] + 2;
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} else {
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f[i][j] = Math.max(f[i + 1][j], f[i][j - 1]);
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}
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if (f[i][j] + k >= n) {
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return true;
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}
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}
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}
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return false;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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bool isValidPalindrome(string s, int k) {
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int n = s.length();
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int f[n][n];
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memset(f, 0, sizeof f);
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for (int i = 0; i < n; ++i) {
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f[i][i] = 1;
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}
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for (int i = n - 2; i >= 0; --i) {
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for (int j = i + 1; j < n; ++j) {
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if (s[i] == s[j]) {
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f[i][j] = f[i + 1][j - 1] + 2;
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} else {
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f[i][j] = max(f[i + 1][j], f[i][j - 1]);
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}
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if (f[i][j] + k >= n) {
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return true;
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}
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}
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}
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return false;
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}
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};
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```
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### **Go**
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```go
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func isValidPalindrome(s string, k int) bool {
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n := len(s)
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f := make([][]int, n)
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for i := range f {
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f[i] = make([]int, n)
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f[i][i] = 1
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}
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for i := n - 2; i >= 0; i-- {
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for j := i + 1; j < n; j++ {
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if s[i] == s[j] {
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f[i][j] = f[i+1][j-1] + 2
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} else {
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f[i][j] = max(f[i+1][j], f[i][j-1])
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}
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if f[i][j]+k >= n {
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return true
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}
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}
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}
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return false
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/1200-1299/1216.Valid Palindrome III/README_EN.md

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### **Python3**
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```python
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class Solution:
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def isValidPalindrome(self, s: str, k: int) -> bool:
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n = len(s)
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f = [[0] * n for _ in range(n)]
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for i in range(n):
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f[i][i] = 1
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for i in range(n - 2, -1, -1):
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for j in range(i + 1, n):
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if s[i] == s[j]:
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f[i][j] = f[i + 1][j - 1] + 2
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else:
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f[i][j] = max(f[i + 1][j], f[i][j - 1])
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if f[i][j] + k >= n:
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return True
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return False
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```
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### **Java**
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```java
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class Solution {
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public boolean isValidPalindrome(String s, int k) {
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int n = s.length();
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int[][] f = new int[n][n];
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for (int i = 0; i < n; ++i) {
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f[i][i] = 1;
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}
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for (int i = n - 2; i >= 0; --i) {
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for (int j = i + 1; j < n; ++j) {
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if (s.charAt(i) == s.charAt(j)) {
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f[i][j] = f[i + 1][j - 1] + 2;
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} else {
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f[i][j] = Math.max(f[i + 1][j], f[i][j - 1]);
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}
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if (f[i][j] + k >= n) {
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return true;
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}
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}
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}
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return false;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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bool isValidPalindrome(string s, int k) {
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int n = s.length();
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int f[n][n];
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memset(f, 0, sizeof f);
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for (int i = 0; i < n; ++i) {
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f[i][i] = 1;
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}
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for (int i = n - 2; i >= 0; --i) {
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for (int j = i + 1; j < n; ++j) {
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if (s[i] == s[j]) {
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f[i][j] = f[i + 1][j - 1] + 2;
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} else {
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f[i][j] = max(f[i + 1][j], f[i][j - 1]);
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}
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if (f[i][j] + k >= n) {
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return true;
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}
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}
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}
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return false;
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}
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};
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```
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### **Go**
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```go
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func isValidPalindrome(s string, k int) bool {
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n := len(s)
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f := make([][]int, n)
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for i := range f {
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f[i] = make([]int, n)
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f[i][i] = 1
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}
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for i := n - 2; i >= 0; i-- {
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for j := i + 1; j < n; j++ {
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if s[i] == s[j] {
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f[i][j] = f[i+1][j-1] + 2
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} else {
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f[i][j] = max(f[i+1][j], f[i][j-1])
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}
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if f[i][j]+k >= n {
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return true
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}
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}
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}
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return false
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/1200-1299/1216.Valid Palindrome III/Solution.cpp

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class Solution {
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public:
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bool isValidPalindrome(string s, int k) {
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int n = s.length();
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int f[n][n];
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memset(f, 0, sizeof f);
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for (int i = 0; i < n; ++i) {
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f[i][i] = 1;
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}
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for (int i = n - 2; i >= 0; --i) {
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for (int j = i + 1; j < n; ++j) {
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if (s[i] == s[j]) {
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f[i][j] = f[i + 1][j - 1] + 2;
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} else {
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f[i][j] = max(f[i + 1][j], f[i][j - 1]);
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}
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if (f[i][j] + k >= n) {
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return true;
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}
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}
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}
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return false;
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}
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};

‎solution/1200-1299/1216.Valid Palindrome III/Solution.go

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func isValidPalindrome(s string, k int) bool {
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n := len(s)
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f := make([][]int, n)
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for i := range f {
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f[i] = make([]int, n)
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f[i][i] = 1
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}
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for i := n - 2; i >= 0; i-- {
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for j := i + 1; j < n; j++ {
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if s[i] == s[j] {
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f[i][j] = f[i+1][j-1] + 2
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} else {
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f[i][j] = max(f[i+1][j], f[i][j-1])
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}
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if f[i][j]+k >= n {
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return true
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}
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}
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}
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return false
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}

‎solution/1200-1299/1216.Valid Palindrome III/Solution.java

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class Solution {
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public boolean isValidPalindrome(String s, int k) {
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int n = s.length();
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int[][] f = new int[n][n];
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for (int i = 0; i < n; ++i) {
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f[i][i] = 1;
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}
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for (int i = n - 2; i >= 0; --i) {
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for (int j = i + 1; j < n; ++j) {
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if (s.charAt(i) == s.charAt(j)) {
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f[i][j] = f[i + 1][j - 1] + 2;
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} else {
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f[i][j] = Math.max(f[i + 1][j], f[i][j - 1]);
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}
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if (f[i][j] + k >= n) {
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return true;
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}
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}
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}
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return false;
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}
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}

‎solution/1200-1299/1216.Valid Palindrome III/Solution.py

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class Solution:
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def isValidPalindrome(self, s: str, k: int) -> bool:
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n = len(s)
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f = [[0] * n for _ in range(n)]
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for i in range(n):
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f[i][i] = 1
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for i in range(n - 2, -1, -1):
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for j in range(i + 1, n):
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if s[i] == s[j]:
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f[i][j] = f[i + 1][j - 1] + 2
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else:
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f[i][j] = max(f[i + 1][j], f[i][j - 1])
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if f[i][j] + k >= n:
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return True
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return False

‎solution/1200-1299/1217.Minimum Cost to Move Chips to The Same Position/README.md

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将所有偶数下标的芯片移动到 0 号位置，所有奇数下标的芯片移动到 1 号位置，所有的代价为 0，接下来只需要在 0/1 号位置中选择其中一个较小数量的芯片，移动到另一个位置。所需的最小代价就是那个较小的数量。
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时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为芯片的数量。
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<!-- tabs:start -->
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### **Python3**

‎solution/1200-1299/1221.Split a String in Balanced Strings/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：贪心**
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我们用变量 $l$ 维护当前字符串的平衡度，即 $l$ 的值为当前字符串中 $L$ 的数量减去 $R$ 的数量。当 $l$ 的值为 0 时，我们就找到了一个平衡字符串。
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遍历字符串 $s$，当遍历到第 $i$ 个字符时，如果 $s[i] = L$，则 $l$ 的值加 1，否则 $l$ 的值减 1。当 $l$ 的值为 0 时，我们将答案加 1。
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时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为字符串 $s$ 的长度。
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<!-- tabs:start -->
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### **Python3**

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