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Commit 37fb2dc

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feat: add solutions to lc problem: No.1312
No.1312.Minimum Insertion Steps to Make a String Palindrome
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  • solution
    • 1000-1099/1039.Minimum Score Triangulation of Polygon
    • 1300-1399/1312.Minimum Insertion Steps to Make a String Palindrome

3 files changed

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‎solution/1000-1099/1039.Minimum Score Triangulation of Polygon/README.md‎

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时间复杂度 $O(n^3),ドル空间复杂度 $O(n^2)$。其中 $n$ 为多边形的顶点数。
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相似题目:
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- [1312. 让字符串成为回文串的最少插入次数](/solution/1300-1399/1312.Minimum%20Insertion%20Steps%20to%20Make%20a%20String%20Palindrome/README.md)
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<!-- tabs:start -->
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### **Python3**

‎solution/1300-1399/1312.Minimum Insertion Steps to Make a String Palindrome/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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**方法一:记忆化搜索**
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我们设计一个函数 $dfs(i, j),ドル表示将字符串 $s[i..j]$ 变成回文串所需要的最少操作次数。那么答案就是 $dfs(0, n - 1)$。
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函数 $dfs(i, j)$ 的计算过程如下:
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如果 $i \geq j,ドル此时无需插入任何字符,我们直接返回 0ドル$。
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否则,我们判断 $s[i]$ 与 $s[j]$ 是否相等,如果 $s[i]=s[j],ドル那么我们只需要将 $s[i+1..j-1]$ 变成回文串,那么我们返回 $dfs(i + 1, j - 1)$。否则,我们可以在 $s[i]$ 的左侧或者 $s[j]$ 的右侧插入一个与另一侧相同的字符,那么 $dfs(i, j) = \min(dfs(i + 1, j), dfs(i, j - 1)) + 1$。
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为了避免重复计算,我们可以使用记忆化搜索,即使用哈希表或者数组来存储已经计算过的函数值。
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最后,我们返回 $dfs(0, n - 1)$ 即可。
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时间复杂度 $O(n^2),ドル空间复杂度 $O(n^2)$。其中 $n$ 为字符串 $s$ 的长度。
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**方法二:动态规划(区间 DP)**
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我们定义 $f[i][j]$ 表示将字符串 $s[i..j]$ 变成回文串所需要的最少操作次数。初始时 $f[i][j]=0,ドル答案即为 $f[0][n-1]$。
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时间复杂度 $O(n^2),ドル空间复杂度 $O(n^2)$。其中 $n$ 为字符串 $s$ 的长度。
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相似题目:
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- [1039. 多边形三角剖分的最低得分](/solution/1000-1099/1039.Minimum%20Score%20Triangulation%20of%20Polygon/README.md)
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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 minInsertions(self, s: str) -> int:
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@cache
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def dfs(i: int, j: int) -> int:
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if i >= j:
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return 0
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if s[i] == s[j]:
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return dfs(i + 1, j - 1)
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return 1 + min(dfs(i + 1, j), dfs(i, j - 1))
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return dfs(0, len(s) - 1)
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```
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```python
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class Solution:
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def minInsertions(self, s: str) -> int:
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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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private Integer[][] f;
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private String s;
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public int minInsertions(String s) {
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this.s = s;
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int n = s.length();
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f = new Integer[n][n];
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return dfs(0, n - 1);
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}
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private int dfs(int i, int j) {
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if (i >= j) {
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return 0;
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}
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if (f[i][j] != null) {
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return f[i][j];
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}
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int ans = 1 << 30;
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if (s.charAt(i) == s.charAt(j)) {
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ans = dfs(i + 1, j - 1);
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} else {
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ans = Math.min(dfs(i + 1, j), dfs(i, j - 1)) + 1;
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}
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return f[i][j] = ans;
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}
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}
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```
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```java
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class Solution {
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public int minInsertions(String s) {
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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 minInsertions(string s) {
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int n = s.size();
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int f[n][n];
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memset(f, -1, sizeof(f));
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function<int(int, int)> dfs = [&](int i, int j) -> int {
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if (i >= j) {
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return 0;
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}
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if (f[i][j] != -1) {
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return f[i][j];
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}
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int ans = 1 << 30;
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if (s[i] == s[j]) {
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ans = dfs(i + 1, j - 1);
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} else {
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ans = min(dfs(i + 1, j), dfs(i, j - 1)) + 1;
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}
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return f[i][j] = ans;
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};
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return dfs(0, n - 1);
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}
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};
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```
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```cpp
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class Solution {
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public:
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### **Go**
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```go
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func minInsertions(s string) int {
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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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for j := range f[i] {
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f[i][j] = -1
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}
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}
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var dfs func(i, j int) int
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dfs = func(i, j int) int {
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if i >= j {
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return 0
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}
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if f[i][j] != -1 {
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return f[i][j]
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}
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ans := 1 << 30
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if s[i] == s[j] {
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ans = dfs(i+1, j-1)
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} else {
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ans = min(dfs(i+1, j), dfs(i, j-1)) + 1
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}
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f[i][j] = ans
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return ans
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}
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return dfs(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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```go
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func minInsertions(s string) int {
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n := len(s)

‎solution/1300-1399/1312.Minimum Insertion Steps to Make a String Palindrome/README_EN.md‎

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### **Python3**
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```python
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class Solution:
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def minInsertions(self, s: str) -> int:
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@cache
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def dfs(i: int, j: int) -> int:
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if i >= j:
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return 0
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if s[i] == s[j]:
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return dfs(i + 1, j - 1)
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return 1 + min(dfs(i + 1, j), dfs(i, j - 1))
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return dfs(0, len(s) - 1)
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```
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```python
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class Solution:
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def minInsertions(self, s: str) -> int:
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### **Java**
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```java
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class Solution {
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private Integer[][] f;
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private String s;
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public int minInsertions(String s) {
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this.s = s;
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int n = s.length();
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f = new Integer[n][n];
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return dfs(0, n - 1);
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}
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private int dfs(int i, int j) {
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if (i >= j) {
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return 0;
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}
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if (f[i][j] != null) {
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return f[i][j];
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}
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int ans = 1 << 30;
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if (s.charAt(i) == s.charAt(j)) {
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ans = dfs(i + 1, j - 1);
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} else {
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ans = Math.min(dfs(i + 1, j), dfs(i, j - 1)) + 1;
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}
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return f[i][j] = ans;
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}
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}
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```
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```java
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class Solution {
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public int minInsertions(String s) {
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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 minInsertions(string s) {
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int n = s.size();
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int f[n][n];
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memset(f, -1, sizeof(f));
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function<int(int, int)> dfs = [&](int i, int j) -> int {
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if (i >= j) {
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return 0;
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}
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if (f[i][j] != -1) {
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return f[i][j];
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}
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int ans = 1 << 30;
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if (s[i] == s[j]) {
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ans = dfs(i + 1, j - 1);
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} else {
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ans = min(dfs(i + 1, j), dfs(i, j - 1)) + 1;
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}
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return f[i][j] = ans;
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};
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return dfs(0, n - 1);
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}
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};
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```
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```cpp
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class Solution {
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public:
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### **Go**
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```go
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func minInsertions(s string) int {
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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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for j := range f[i] {
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f[i][j] = -1
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}
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}
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var dfs func(i, j int) int
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dfs = func(i, j int) int {
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if i >= j {
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return 0
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}
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if f[i][j] != -1 {
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return f[i][j]
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}
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ans := 1 << 30
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if s[i] == s[j] {
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ans = dfs(i+1, j-1)
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} else {
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ans = min(dfs(i+1, j), dfs(i, j-1)) + 1
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}
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f[i][j] = ans
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return ans
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}
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return dfs(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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```go
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func minInsertions(s string) int {
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n := len(s)

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