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feat: add solutions to lc problem: No.1246
No.1246.Palindrome Removal
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‎solution/1200-1299/1246.Palindrome Removal/README.md

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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（区间 DP）**
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我们定义 $f[i][j]$ 表示删除下标区间 $[i,..j]$ 内的所有数字所需的最少操作次数。初始时 $f[i][i] = 1$，表示只有一个数字时，需要执行一次删除操作。
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对于 $f[i][j]$，如果 $i + 1 = j$，即只有两个数字时，如果 $arr[i]=arr[j]$，则 $f[i][j] = 1$，否则 $f[i][j] = 2$。
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对于超过两个数字的情况，如果 $arr[i]=arr[j]$，那么 $f[i][j]$ 可以取 $f[i + 1][j - 1]$，或者我们可以在下标范围 $[i,..j-1]$ 范围内枚举 $k$，取 $f[i][k] + f[k + 1][j]$ 的最小值。将最小值赋给 $f[i][j]$。
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答案即为 $f[0][n - 1]$。
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时间复杂度 $O(n^3)$，空间复杂度 $O(n^2)$。其中 $n$ 为数组长度。
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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 minimumMoves(self, arr: List[int]) -> int:
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n = len(arr)
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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 i + 1 == j:
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f[i][j] = 1 if arr[i] == arr[j] else 2
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else:
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t = f[i + 1][j - 1] if arr[i] == arr[j] else inf
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for k in range(i, j):
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t = min(t, f[i][k] + f[k + 1][j])
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f[i][j] = t
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return f[0][n - 1]
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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 int minimumMoves(int[] arr) {
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int n = arr.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 (i + 1 == j) {
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f[i][j] = arr[i] == arr[j] ? 1 : 2;
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} else {
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int t = arr[i] == arr[j] ? f[i + 1][j - 1] : 1 << 30;
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for (int k = i; k < j; ++k) {
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t = Math.min(t, f[i][k] + f[k + 1][j]);
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}
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f[i][j] = t;
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}
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}
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}
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return f[0][n - 1];
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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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int minimumMoves(vector<int>& arr) {
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int n = arr.size();
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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 (i + 1 == j) {
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f[i][j] = arr[i] == arr[j] ? 1 : 2;
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} else {
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int t = arr[i] == arr[j] ? f[i + 1][j - 1] : 1 << 30;
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for (int k = i; k < j; ++k) {
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t = min(t, f[i][k] + f[k + 1][j]);
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}
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f[i][j] = t;
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}
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}
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}
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return f[0][n - 1];
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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 minimumMoves(arr []int) int {
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n := len(arr)
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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 i+1 == j {
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f[i][j] = 2
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if arr[i] == arr[j] {
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f[i][j] = 1
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}
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} else {
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t := 1 << 30
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if arr[i] == arr[j] {
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t = f[i+1][j-1]
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}
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for k := i; k < j; k++ {
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t = min(t, f[i][k]+f[k+1][j])
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}
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f[i][j] = t
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}
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}
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}
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return f[0][n-1]
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}
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func min(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/1200-1299/1246.Palindrome Removal/README_EN.md

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### **Python3**
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```python
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class Solution:
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def minimumMoves(self, arr: List[int]) -> int:
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n = len(arr)
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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 i + 1 == j:
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f[i][j] = 1 if arr[i] == arr[j] else 2
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else:
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t = f[i + 1][j - 1] if arr[i] == arr[j] else inf
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for k in range(i, j):
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t = min(t, f[i][k] + f[k + 1][j])
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f[i][j] = t
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return f[0][n - 1]
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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 int minimumMoves(int[] arr) {
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int n = arr.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 (i + 1 == j) {
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f[i][j] = arr[i] == arr[j] ? 1 : 2;
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} else {
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int t = arr[i] == arr[j] ? f[i + 1][j - 1] : 1 << 30;
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for (int k = i; k < j; ++k) {
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t = Math.min(t, f[i][k] + f[k + 1][j]);
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}
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f[i][j] = t;
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}
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}
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}
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return f[0][n - 1];
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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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int minimumMoves(vector<int>& arr) {
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int n = arr.size();
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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 (i + 1 == j) {
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f[i][j] = arr[i] == arr[j] ? 1 : 2;
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} else {
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int t = arr[i] == arr[j] ? f[i + 1][j - 1] : 1 << 30;
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for (int k = i; k < j; ++k) {
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t = min(t, f[i][k] + f[k + 1][j]);
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}
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f[i][j] = t;
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}
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}
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}
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return f[0][n - 1];
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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 minimumMoves(arr []int) int {
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n := len(arr)
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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 i+1 == j {
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f[i][j] = 2
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if arr[i] == arr[j] {
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f[i][j] = 1
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}
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} else {
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t := 1 << 30
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if arr[i] == arr[j] {
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t = f[i+1][j-1]
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}
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for k := i; k < j; k++ {
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t = min(t, f[i][k]+f[k+1][j])
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}
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f[i][j] = t
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}
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}
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}
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return f[0][n-1]
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}
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func min(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/1200-1299/1246.Palindrome Removal/Solution.cpp

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class Solution {
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public:
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int minimumMoves(vector<int>& arr) {
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int n = arr.size();
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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 (i + 1 == j) {
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f[i][j] = arr[i] == arr[j] ? 1 : 2;
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} else {
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int t = arr[i] == arr[j] ? f[i + 1][j - 1] : 1 << 30;
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for (int k = i; k < j; ++k) {
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t = min(t, f[i][k] + f[k + 1][j]);
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}
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f[i][j] = t;
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}
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}
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}
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return f[0][n - 1];
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}
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};

‎solution/1200-1299/1246.Palindrome Removal/Solution.go

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func minimumMoves(arr []int) int {
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n := len(arr)
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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 i+1 == j {
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f[i][j] = 2
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if arr[i] == arr[j] {
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f[i][j] = 1
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}
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} else {
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t := 1 << 30
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if arr[i] == arr[j] {
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t = f[i+1][j-1]
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}
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for k := i; k < j; k++ {
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t = min(t, f[i][k]+f[k+1][j])
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}
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f[i][j] = t
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}
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}
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}
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return f[0][n-1]
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}
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func min(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/1200-1299/1246.Palindrome Removal/Solution.java

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class Solution {
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public int minimumMoves(int[] arr) {
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int n = arr.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 (i + 1 == j) {
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f[i][j] = arr[i] == arr[j] ? 1 : 2;
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} else {
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int t = arr[i] == arr[j] ? f[i + 1][j - 1] : 1 << 30;
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for (int k = i; k < j; ++k) {
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t = Math.min(t, f[i][k] + f[k + 1][j]);
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}
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f[i][j] = t;
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}
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}
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}
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return f[0][n - 1];
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}
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}

‎solution/1200-1299/1246.Palindrome Removal/Solution.py

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class Solution:
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def minimumMoves(self, arr: List[int]) -> int:
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n = len(arr)
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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 i + 1 == j:
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f[i][j] = 1 if arr[i] == arr[j] else 2
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else:
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t = f[i + 1][j - 1] if arr[i] == arr[j] else inf
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for k in range(i, j):
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t = min(t, f[i][k] + f[k + 1][j])
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f[i][j] = t
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return f[0][n - 1]

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