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feat: add solutions to lc problem: No.1278 (#4122)
No.1278.Palindrome Partitioning III
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‎solution/1200-1299/1278.Palindrome Partitioning III/README.md

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### 方法一：动态规划
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定义 $dp[i][j]$ 表示将字符串 $s$ 的前 $i$ 个字符分割成 $j$ 个回文串所需要的最少修改次数，我们假定 $i$ 下标从 $1$ 开始，答案为 $dp[n][k]$。
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我们定义 $f[i][j]$ 表示将字符串 $s$ 的前 $i$ 个字符分割成 $j$ 个回文串所需要的最少修改次数，我们假定 $i$ 下标从 $1$ 开始，答案为 $f[n][k]$。
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对于 $dp[i][j]$，我们可以枚举第 $j-1$ 个回文串的最后一个字符的位置 $h$，那么 $dp[i][j]$ 就等于 $dp[h][j-1] + g[h][i-1]$ 的较小值，其中 $g[h][i-1]$ 表示将字符串 $s[h..i-1]$ 变成回文串所需要的最少修改次数（这一部分我们可以通过预处理得到，时间复杂度 $O(n^2)$。
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对于 $f[i][j]$，我们可以枚举第 $j-1$ 个回文串的最后一个字符的位置 $h$，那么 $f[i][j]$ 就等于 $f[h][j-1] + g[h][i-1]$ 的较小值，其中 $g[h][i-1]$ 表示将字符串 $s[h..i-1]$ 变成回文串所需要的最少修改次数（这一部分我们可以通过预处理得到，时间复杂度 $O(n^2)$。
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时间复杂度 $O(n^2\times k)$。其中 $n$ 为字符串 $s$ 的长度。
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时间复杂度 $O(n^2 \times k)$，空间复杂度 $O(n \times (n + k))$。其中 $n$ 为字符串 $s$ 的长度。
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<!-- tabs:start -->
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@@ -205,6 +205,76 @@ func palindromePartition(s string, k int) int {
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}
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```
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#### TypeScript
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```ts
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function palindromePartition(s: string, k: number): number {
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const n = s.length;
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const g: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
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for (let i = n - 1; i >= 0; i--) {
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for (let j = i + 1; j < n; j++) {
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g[i][j] = s[i] !== s[j] ? 1 : 0;
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if (i + 1 < j) {
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g[i][j] += g[i + 1][j - 1];
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}
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}
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}
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const f: number[][] = Array.from({ length: n + 1 }, () => Array(k + 1).fill(0));
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for (let i = 1; i <= n; i++) {
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for (let j = 1; j <= Math.min(i, k); j++) {
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if (j === 1) {
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f[i][j] = g[0][i - 1];
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} else {
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f[i][j] = 1 << 30;
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for (let h = j - 1; h < i; h++) {
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f[i][j] = Math.min(f[i][j], f[h][j - 1] + g[h][i - 1]);
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}
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}
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}
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}
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return f[n][k];
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn palindrome_partition(s: String, k: i32) -> i32 {
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let n = s.len();
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let s: Vec<char> = s.chars().collect();
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let mut g = vec![vec![0; n]; n];
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for i in (0..n).rev() {
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for j in i + 1..n {
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g[i][j] = if s[i] != s[j] { 1 } else { 0 };
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if i + 1 < j {
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g[i][j] += g[i + 1][j - 1];
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}
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}
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}
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let mut f = vec![vec![0; (k + 1) as usize]; n + 1];
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let inf = i32::MAX;
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for i in 1..=n {
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for j in 1..=i.min(k as usize) {
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if j == 1 {
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f[i][j] = g[0][i - 1];
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} else {
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f[i][j] = inf;
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for h in (j - 1)..i {
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f[i][j] = f[i][j].min(f[h][j - 1] + g[h][i - 1]);
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}
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}
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}
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}
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f[n][k as usize]
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/1200-1299/1278.Palindrome Partitioning III/README_EN.md

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Dynamic Programming
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We define $f[i][j]$ to represent the minimum number of changes needed to partition the first $i$ characters of the string $s$ into $j$ palindromic substrings. We assume the index $i$ starts from 1, and the answer is $f[n][k]$.
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For $f[i][j]$, we can enumerate the position $h$ of the last character of the $(j-1)$-th palindromic substring. Then $f[i][j]$ is equal to the minimum value of $f[h][j-1] + g[h][i-1]$, where $g[h][i-1]$ represents the minimum number of changes needed to turn the substring $s[h..i-1]$ into a palindrome (this part can be preprocessed with a time complexity of $O(n^2)$).
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The time complexity is $O(n^2 \times k)$, and the space complexity is $O(n \times (n + k))$. Where $n$ is the length of the string $s$.
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<!-- tabs:start -->
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}
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```
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#### TypeScript
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```ts
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function palindromePartition(s: string, k: number): number {
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const n = s.length;
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const g: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
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for (let i = n - 1; i >= 0; i--) {
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for (let j = i + 1; j < n; j++) {
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g[i][j] = s[i] !== s[j] ? 1 : 0;
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if (i + 1 < j) {
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g[i][j] += g[i + 1][j - 1];
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}
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}
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}
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const f: number[][] = Array.from({ length: n + 1 }, () => Array(k + 1).fill(0));
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for (let i = 1; i <= n; i++) {
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for (let j = 1; j <= Math.min(i, k); j++) {
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if (j === 1) {
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f[i][j] = g[0][i - 1];
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} else {
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f[i][j] = 1 << 30;
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for (let h = j - 1; h < i; h++) {
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f[i][j] = Math.min(f[i][j], f[h][j - 1] + g[h][i - 1]);
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}
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}
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}
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}
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return f[n][k];
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn palindrome_partition(s: String, k: i32) -> i32 {
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let n = s.len();
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let s: Vec<char> = s.chars().collect();
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let mut g = vec![vec![0; n]; n];
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for i in (0..n).rev() {
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for j in i + 1..n {
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g[i][j] = if s[i] != s[j] { 1 } else { 0 };
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if i + 1 < j {
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g[i][j] += g[i + 1][j - 1];
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}
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}
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}
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let mut f = vec![vec![0; (k + 1) as usize]; n + 1];
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let inf = i32::MAX;
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for i in 1..=n {
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for j in 1..=i.min(k as usize) {
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if j == 1 {
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f[i][j] = g[0][i - 1];
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} else {
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f[i][j] = inf;
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for h in (j - 1)..i {
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f[i][j] = f[i][j].min(f[h][j - 1] + g[h][i - 1]);
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}
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}
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}
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}
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f[n][k as usize]
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/1200-1299/1278.Palindrome Partitioning III/Solution.rs

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impl Solution {
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pub fn palindrome_partition(s: String, k: i32) -> i32 {
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let n = s.len();
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let s: Vec<char> = s.chars().collect();
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let mut g = vec![vec![0; n]; n];
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for i in (0..n).rev() {
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for j in i + 1..n {
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g[i][j] = if s[i] != s[j] { 1 } else { 0 };
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if i + 1 < j {
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g[i][j] += g[i + 1][j - 1];
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}
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}
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}
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let mut f = vec![vec![0; (k + 1) as usize]; n + 1];
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let inf = i32::MAX;
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for i in 1..=n {
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for j in 1..=i.min(k as usize) {
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if j == 1 {
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f[i][j] = g[0][i - 1];
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} else {
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f[i][j] = inf;
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for h in (j - 1)..i {
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f[i][j] = f[i][j].min(f[h][j - 1] + g[h][i - 1]);
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}
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}
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}
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}
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f[n][k as usize]
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}
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}

‎solution/1200-1299/1278.Palindrome Partitioning III/Solution.ts

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function palindromePartition(s: string, k: number): number {
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const n = s.length;
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const g: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
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for (let i = n - 1; i >= 0; i--) {
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for (let j = i + 1; j < n; j++) {
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g[i][j] = s[i] !== s[j] ? 1 : 0;
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if (i + 1 < j) {
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g[i][j] += g[i + 1][j - 1];
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}
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}
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}
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const f: number[][] = Array.from({ length: n + 1 }, () => Array(k + 1).fill(0));
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for (let i = 1; i <= n; i++) {
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for (let j = 1; j <= Math.min(i, k); j++) {
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if (j === 1) {
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f[i][j] = g[0][i - 1];
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} else {
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f[i][j] = 1 << 30;
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for (let h = j - 1; h < i; h++) {
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f[i][j] = Math.min(f[i][j], f[h][j - 1] + g[h][i - 1]);
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}
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}
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}
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}
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return f[n][k];
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}

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