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feat: add solutions to lc problem: No.1745 #4109

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79 changes: 58 additions & 21 deletions solution/1700-1799/1745.Palindrome Partitioning IV/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -56,11 +56,26 @@ tags:

<!-- solution:start -->

### 方法一：预处理 + 枚举
### 方法一：动态规划

预处理出字符串 `s` 的所有子串是否为回文串，然后枚举 `s` 的所有分割点，判断是否满足条件。
我们定义 $f[i][j]$ 表示字符串 $s$ 的第 $i$ 个字符到第 $j$ 个字符是否为回文串，初始时 $f[i][j] = \textit{true}$。

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。
然后我们可以通过以下的状态转移方程来计算 $f[i][j]$：

$$
f[i][j] = \begin{cases}
\textit{true}, & \text{if } s[i] = s[j] \text{ and } (i + 1 = j \text{ or } f[i + 1][j - 1]) \\
\textit{false}, & \text{otherwise}
\end{cases}
$$

由于 $f[i][j]$ 依赖于 $f[i + 1][j - 1]$，因此，我们需要从大到小的顺序枚举 $i$，从小到大的顺序枚举 $j$，这样才能保证当计算 $f[i][j]$ 时 $f[i + 1][j - 1]$ 已经被计算过。

接下来，我们枚举第一个子串的右端点 $i$，第二个子串的右端点 $j$，那么第三个子串的左端点可以枚举的范围为 $[j + 1, n - 1]$，其中 $n$ 是字符串 $s$ 的长度。如果第一个子串 $s[0..i]$、第二个子串 $s[i+1..j]$ 和第三个子串 $s[j+1..n-1]$ 都是回文串，那么我们就找到了一种可行的分割方案，返回 $\textit{true}$。

枚举完所有的分割方案后，如果没有找到符合要求的分割方案，那么返回 $\textit{false}$。

时间复杂度 $O(n^2)$，空间复杂度 $O(n^2)$。其中 $n$ 是字符串 $s$ 的长度。

<!-- tabs:start -->

Expand All @@ -70,13 +85,13 @@ tags:
class Solution:
def checkPartitioning(self, s: str) -> bool:
n = len(s)
g = [[True] * n for _ in range(n)]
f = [[True] * n for _ in range(n)]
for i in range(n - 1, -1, -1):
for j in range(i + 1, n):
g[i][j] = s[i] == s[j] and (i + 1 == j or g[i + 1][j - 1])
f[i][j] = s[i] == s[j] and (i + 1 == j or f[i + 1][j - 1])
for i in range(n - 2):
for j in range(i + 1, n - 1):
if g[0][i] and g[i + 1][j] and g[j + 1][-1]:
if f[0][i] and f[i + 1][j] and f[j + 1][-1]:
return True
return False
```
Expand All @@ -87,18 +102,18 @@ class Solution:
class Solution {
public boolean checkPartitioning(String s) {
int n = s.length();
boolean[][] g = new boolean[n][n];
for (var e : g) {
Arrays.fill(e, true);
boolean[][] f = new boolean[n][n];
for (var g : f) {
Arrays.fill(g, true);
}
for (int i = n - 1; i >= 0; --i) {
for (int j = i + 1; j < n; ++j) {
g[i][j] = s.charAt(i) == s.charAt(j) && (i + 1 == j || g[i + 1][j - 1]);
f[i][j] = s.charAt(i) == s.charAt(j) && (i + 1 == j || f[i + 1][j - 1]);
}
}
for (int i = 0; i < n - 2; ++i) {
for (int j = i + 1; j < n - 1; ++j) {
if (g[0][i] && g[i + 1][j] && g[j + 1][n - 1]) {
if (f[0][i] && f[i + 1][j] && f[j + 1][n - 1]) {
return true;
}
}
Expand All @@ -115,15 +130,15 @@ class Solution {
public:
bool checkPartitioning(string s) {
int n = s.size();
vector<vector<bool>> g(n, vector<bool>(n, true));
vector<vector<bool>> f(n, vector<bool>(n, true));
for (int i = n - 1; i >= 0; --i) {
for (int j = i + 1; j < n; ++j) {
g[i][j] = s[i] == s[j] && (i + 1 == j || g[i + 1][j - 1]);
f[i][j] = s[i] == s[j] && (i + 1 == j || f[i + 1][j - 1]);
}
}
for (int i = 0; i < n - 2; ++i) {
for (int j = i + 1; j < n - 1; ++j) {
if (g[0][i] && g[i + 1][j] && g[j + 1][n - 1]) {
if (f[0][i] && f[i + 1][j] && f[j + 1][n - 1]) {
return true;
}
}
Expand All @@ -138,21 +153,21 @@ public:
```go
func checkPartitioning(s string) bool {
n := len(s)
g := make([][]bool, n)
for i := range g {
g[i] = make([]bool, n)
for j := range g[i] {
g[i][j] = true
f := make([][]bool, n)
for i := range f {
f[i] = make([]bool, n)
for j := range f[i] {
f[i][j] = true
}
}
for i := n - 1; i >= 0; i-- {
for j := i + 1; j < n; j++ {
g[i][j] = s[i] == s[j] && (i+1 == j || g[i+1][j-1])
f[i][j] = s[i] == s[j] && (i+1 == j || f[i+1][j-1])
}
}
for i := 0; i < n-2; i++ {
for j := i + 1; j < n-1; j++ {
if g[0][i] && g[i+1][j] && g[j+1][n-1] {
if f[0][i] && f[i+1][j] && f[j+1][n-1] {
return true
}
}
Expand All @@ -161,6 +176,28 @@ func checkPartitioning(s string) bool {
}
```

#### TypeScript

```ts
function checkPartitioning(s: string): boolean {
const n = s.length;
const f: boolean[][] = Array.from({ length: n }, () => Array(n).fill(true));
for (let i = n - 1; i >= 0; --i) {
for (let j = i + 1; j < n; ++j) {
f[i][j] = s[i] === s[j] && f[i + 1][j - 1];
}
}
for (let i = 0; i < n - 2; ++i) {
for (let j = i + 1; j < n - 1; ++j) {
if (f[0][i] && f[i + 1][j] && f[j + 1][n - 1]) {
return true;
}
}
}
return false;
}
```

<!-- tabs:end -->

<!-- solution:end -->
Expand Down
79 changes: 60 additions & 19 deletions solution/1700-1799/1745.Palindrome Partitioning IV/README_EN.md
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Expand Up @@ -54,7 +54,26 @@ tags:

<!-- solution:start -->

### Solution 1
### Solution 1: Dynamic Programming

We define $f[i][j]$ to indicate whether the substring of $s$ from the $i$-th character to the $j$-th character is a palindrome, initially $f[i][j] = \textit{true}$.

Then we can calculate $f[i][j]$ using the following state transition equation:

$$
f[i][j] = \begin{cases}
\textit{true}, & \text{if } s[i] = s[j] \text{ and } (i + 1 = j \text{ or } f[i + 1][j - 1]) \\
\textit{false}, & \text{otherwise}
\end{cases}
$$

Since $f[i][j]$ depends on $f[i + 1][j - 1]$, we need to enumerate $i$ from large to small and $j$ from small to large, so that when calculating $f[i][j]$, $f[i + 1][j - 1]$ has already been calculated.

Next, we enumerate the right endpoint $i$ of the first substring and the right endpoint $j$ of the second substring. The left endpoint of the third substring can be enumerated in the range $[j + 1, n - 1]$, where $n$ is the length of the string $s$. If the first substring $s[0..i]$, the second substring $s[i+1..j]$, and the third substring $s[j+1..n-1]$ are all palindromes, then we have found a feasible partitioning scheme and return $\textit{true}$.

After enumerating all partitioning schemes, if no valid partitioning scheme is found, return $\textit{false}$.

Time complexity is $O(n^2)$, and space complexity is $O(n^2)$. Where $n$ is the length of the string $s$.

<!-- tabs:start -->

Expand All @@ -64,13 +83,13 @@ tags:
class Solution:
def checkPartitioning(self, s: str) -> bool:
n = len(s)
g = [[True] * n for _ in range(n)]
f = [[True] * n for _ in range(n)]
for i in range(n - 1, -1, -1):
for j in range(i + 1, n):
g[i][j] = s[i] == s[j] and (i + 1 == j or g[i + 1][j - 1])
f[i][j] = s[i] == s[j] and (i + 1 == j or f[i + 1][j - 1])
for i in range(n - 2):
for j in range(i + 1, n - 1):
if g[0][i] and g[i + 1][j] and g[j + 1][-1]:
if f[0][i] and f[i + 1][j] and f[j + 1][-1]:
return True
return False
```
Expand All @@ -81,18 +100,18 @@ class Solution:
class Solution {
public boolean checkPartitioning(String s) {
int n = s.length();
boolean[][] g = new boolean[n][n];
for (var e : g) {
Arrays.fill(e, true);
boolean[][] f = new boolean[n][n];
for (var g : f) {
Arrays.fill(g, true);
}
for (int i = n - 1; i >= 0; --i) {
for (int j = i + 1; j < n; ++j) {
g[i][j] = s.charAt(i) == s.charAt(j) && (i + 1 == j || g[i + 1][j - 1]);
f[i][j] = s.charAt(i) == s.charAt(j) && (i + 1 == j || f[i + 1][j - 1]);
}
}
for (int i = 0; i < n - 2; ++i) {
for (int j = i + 1; j < n - 1; ++j) {
if (g[0][i] && g[i + 1][j] && g[j + 1][n - 1]) {
if (f[0][i] && f[i + 1][j] && f[j + 1][n - 1]) {
return true;
}
}
Expand All @@ -109,15 +128,15 @@ class Solution {
public:
bool checkPartitioning(string s) {
int n = s.size();
vector<vector<bool>> g(n, vector<bool>(n, true));
vector<vector<bool>> f(n, vector<bool>(n, true));
for (int i = n - 1; i >= 0; --i) {
for (int j = i + 1; j < n; ++j) {
g[i][j] = s[i] == s[j] && (i + 1 == j || g[i + 1][j - 1]);
f[i][j] = s[i] == s[j] && (i + 1 == j || f[i + 1][j - 1]);
}
}
for (int i = 0; i < n - 2; ++i) {
for (int j = i + 1; j < n - 1; ++j) {
if (g[0][i] && g[i + 1][j] && g[j + 1][n - 1]) {
if (f[0][i] && f[i + 1][j] && f[j + 1][n - 1]) {
return true;
}
}
Expand All @@ -132,21 +151,21 @@ public:
```go
func checkPartitioning(s string) bool {
n := len(s)
g := make([][]bool, n)
for i := range g {
g[i] = make([]bool, n)
for j := range g[i] {
g[i][j] = true
f := make([][]bool, n)
for i := range f {
f[i] = make([]bool, n)
for j := range f[i] {
f[i][j] = true
}
}
for i := n - 1; i >= 0; i-- {
for j := i + 1; j < n; j++ {
g[i][j] = s[i] == s[j] && (i+1 == j || g[i+1][j-1])
f[i][j] = s[i] == s[j] && (i+1 == j || f[i+1][j-1])
}
}
for i := 0; i < n-2; i++ {
for j := i + 1; j < n-1; j++ {
if g[0][i] && g[i+1][j] && g[j+1][n-1] {
if f[0][i] && f[i+1][j] && f[j+1][n-1] {
return true
}
}
Expand All @@ -155,6 +174,28 @@ func checkPartitioning(s string) bool {
}
```

#### TypeScript

```ts
function checkPartitioning(s: string): boolean {
const n = s.length;
const f: boolean[][] = Array.from({ length: n }, () => Array(n).fill(true));
for (let i = n - 1; i >= 0; --i) {
for (let j = i + 1; j < n; ++j) {
f[i][j] = s[i] === s[j] && f[i + 1][j - 1];
}
}
for (let i = 0; i < n - 2; ++i) {
for (let j = i + 1; j < n - 1; ++j) {
if (f[0][i] && f[i + 1][j] && f[j + 1][n - 1]) {
return true;
}
}
}
return false;
}
```

<!-- tabs:end -->

<!-- solution:end -->
Expand Down
8 changes: 4 additions & 4 deletions solution/1700-1799/1745.Palindrome Partitioning IV/Solution.cpp
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Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -2,19 +2,19 @@ class Solution {
public:
bool checkPartitioning(string s) {
int n = s.size();
vector<vector<bool>> g(n, vector<bool>(n, true));
vector<vector<bool>> f(n, vector<bool>(n, true));
for (int i = n - 1; i >= 0; --i) {
for (int j = i + 1; j < n; ++j) {
g[i][j] = s[i] == s[j] && (i + 1 == j || g[i + 1][j - 1]);
f[i][j] = s[i] == s[j] && (i + 1 == j || f[i + 1][j - 1]);
}
}
for (int i = 0; i < n - 2; ++i) {
for (int j = i + 1; j < n - 1; ++j) {
if (g[0][i] && g[i + 1][j] && g[j + 1][n - 1]) {
if (f[0][i] && f[i + 1][j] && f[j + 1][n - 1]) {
return true;
}
}
}
return false;
}
};
};
16 changes: 8 additions & 8 deletions solution/1700-1799/1745.Palindrome Partitioning IV/Solution.go
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Original file line number Diff line number Diff line change
@@ -1,23 +1,23 @@
func checkPartitioning(s string) bool {
n := len(s)
g := make([][]bool, n)
for i := range g {
g[i] = make([]bool, n)
for j := range g[i] {
g[i][j] = true
f := make([][]bool, n)
for i := range f {
f[i] = make([]bool, n)
for j := range f[i] {
f[i][j] = true
}
}
for i := n - 1; i >= 0; i-- {
for j := i + 1; j < n; j++ {
g[i][j] = s[i] == s[j] && (i+1 == j || g[i+1][j-1])
f[i][j] = s[i] == s[j] && (i+1 == j || f[i+1][j-1])
}
}
for i := 0; i < n-2; i++ {
for j := i + 1; j < n-1; j++ {
if g[0][i] && g[i+1][j] && g[j+1][n-1] {
if f[0][i] && f[i+1][j] && f[j+1][n-1] {
return true
}
}
}
return false
}
}
12 changes: 6 additions & 6 deletions solution/1700-1799/1745.Palindrome Partitioning IV/Solution.java
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Original file line number Diff line number Diff line change
@@ -1,22 +1,22 @@
class Solution {
public boolean checkPartitioning(String s) {
int n = s.length();
boolean[][] g = new boolean[n][n];
for (var e : g) {
Arrays.fill(e, true);
boolean[][] f = new boolean[n][n];
for (var g : f) {
Arrays.fill(g, true);
}
for (int i = n - 1; i >= 0; --i) {
for (int j = i + 1; j < n; ++j) {
g[i][j] = s.charAt(i) == s.charAt(j) && (i + 1 == j || g[i + 1][j - 1]);
f[i][j] = s.charAt(i) == s.charAt(j) && (i + 1 == j || f[i + 1][j - 1]);
}
}
for (int i = 0; i < n - 2; ++i) {
for (int j = i + 1; j < n - 1; ++j) {
if (g[0][i] && g[i + 1][j] && g[j + 1][n - 1]) {
if (f[0][i] && f[i + 1][j] && f[j + 1][n - 1]) {
return true;
}
}
}
return false;
}
}
}
6 changes: 3 additions & 3 deletions solution/1700-1799/1745.Palindrome Partitioning IV/Solution.py
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Original file line number Diff line number Diff line change
@@ -1,12 +1,12 @@
class Solution:
def checkPartitioning(self, s: str) -> bool:
n = len(s)
g = [[True] * n for _ in range(n)]
f = [[True] * n for _ in range(n)]
for i in range(n - 1, -1, -1):
for j in range(i + 1, n):
g[i][j] = s[i] == s[j] and (i + 1 == j or g[i + 1][j - 1])
f[i][j] = s[i] == s[j] and (i + 1 == j or f[i + 1][j - 1])
for i in range(n - 2):
for j in range(i + 1, n - 1):
if g[0][i] and g[i + 1][j] and g[j + 1][-1]:
if f[0][i] and f[i + 1][j] and f[j + 1][-1]:
return True
return False
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