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fix: update solutions to lc problem: No.1147 #3388

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Aug 9, 2024
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fix: update solutions to lc problem: No.1147 #3388

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109 changes: 7 additions & 102 deletions solution/1100-1199/1147.Longest Chunked Palindrome Decomposition/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -93,107 +93,6 @@ tags:

#### Python3

```python
class Solution:
def longestDecomposition(self, text: str) -> int:
n = len(text)
if n < 2:
return n
for i in range(n // 2 + 1):
if text[:i] == text[-i:]:
return 2 + self.longestDecomposition(text[i:-i])
return 1
```

#### Java

```java
class Solution {
public int longestDecomposition(String text) {
int n = text.length();
if (n < 2) {
return n;
}
for (int i = 1; i <= n >> 1; ++i) {
if (text.substring(0, i).equals(text.substring(n - i))) {
return 2 + longestDecomposition(text.substring(i, n - i));
}
}
return 1;
}
}
```

#### C++

```cpp
class Solution {
public:
int longestDecomposition(string text) {
int n = text.size();
if (n < 2) return n;
for (int i = 1; i <= n >> 1; ++i) {
if (text.substr(0, i) == text.substr(n - i)) {
return 2 + longestDecomposition(text.substr(i, n - i - i));
}
}
return 1;
}
};
```

#### Go

```go
func longestDecomposition(text string) int {
n := len(text)
if n < 2 {
return n
}
for i := 1; i <= n>>1; i++ {
if text[:i] == text[n-i:] {
return 2 + longestDecomposition(text[i:n-i])
}
}
return 1
}
```

#### TypeScript

```ts
function longestDecomposition(text: string): number {
const n: number = text.length;
if (n < 2) {
return n;
}
for (let i: number = 1; i <= n >> 1; i++) {
if (text.slice(0, i) === text.slice(n - i)) {
return 2 + longestDecomposition(text.slice(i, n - i));
}
}
return 1;
}
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- solution:start -->

### 方法二：字符串哈希

**字符串哈希**是把一个任意长度的字符串映射成一个非负整数，并且其冲突的概率几乎为 $0$。字符串哈希用于计算字符串哈希值，快速判断两个字符串是否相等。

因此，在方法一的基础上，我们可以使用字符串哈希的方法，在 $O(1)$ 时间内比较两个字符串是否相等。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为字符串的长度。

<!-- tabs:start -->

#### Python3

```python
class Solution:
def longestDecomposition(self, text: str) -> int:
Expand Down Expand Up @@ -343,7 +242,13 @@ function longestDecomposition(text: string): number {

<!-- solution:start -->

### 方法三
### 方法二：字符串哈希

**字符串哈希**是把一个任意长度的字符串映射成一个非负整数，并且其冲突的概率几乎为 $0$。字符串哈希用于计算字符串哈希值，快速判断两个字符串是否相等。

因此，在方法一的基础上，我们可以使用字符串哈希的方法，在 $O(1)$ 时间内比较两个字符串是否相等。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为字符串的长度。

<!-- tabs:start -->

Expand Down
109 changes: 7 additions & 102 deletions solution/1100-1199/1147.Longest Chunked Palindrome Decomposition/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -91,107 +91,6 @@ The time complexity is $O(n^2)$, and the space complexity is $O(n)$ or $O(1)$. H

#### Python3

```python
class Solution:
def longestDecomposition(self, text: str) -> int:
n = len(text)
if n < 2:
return n
for i in range(n // 2 + 1):
if text[:i] == text[-i:]:
return 2 + self.longestDecomposition(text[i:-i])
return 1
```

#### Java

```java
class Solution {
public int longestDecomposition(String text) {
int n = text.length();
if (n < 2) {
return n;
}
for (int i = 1; i <= n >> 1; ++i) {
if (text.substring(0, i).equals(text.substring(n - i))) {
return 2 + longestDecomposition(text.substring(i, n - i));
}
}
return 1;
}
}
```

#### C++

```cpp
class Solution {
public:
int longestDecomposition(string text) {
int n = text.size();
if (n < 2) return n;
for (int i = 1; i <= n >> 1; ++i) {
if (text.substr(0, i) == text.substr(n - i)) {
return 2 + longestDecomposition(text.substr(i, n - i - i));
}
}
return 1;
}
};
```

#### Go

```go
func longestDecomposition(text string) int {
n := len(text)
if n < 2 {
return n
}
for i := 1; i <= n>>1; i++ {
if text[:i] == text[n-i:] {
return 2 + longestDecomposition(text[i:n-i])
}
}
return 1
}
```

#### TypeScript

```ts
function longestDecomposition(text: string): number {
const n: number = text.length;
if (n < 2) {
return n;
}
for (let i: number = 1; i <= n >> 1; i++) {
if (text.slice(0, i) === text.slice(n - i)) {
return 2 + longestDecomposition(text.slice(i, n - i));
}
}
return 1;
}
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- solution:start -->

### Solution 2: String Hash

**String hash** is to map a string of any length to a non-negative integer, and its collision probability is almost $0$. String hash is used to calculate the hash value of a string and quickly determine whether two strings are equal.

Therefore, based on Solution 1, we can use the method of string hash to compare whether two strings are equal in $O(1)$ time.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string.

<!-- tabs:start -->

#### Python3

```python
class Solution:
def longestDecomposition(self, text: str) -> int:
Expand Down Expand Up @@ -341,7 +240,13 @@ function longestDecomposition(text: string): number {

<!-- solution:start -->

### Solution 3
### Solution 2: String Hash

**String hash** is to map a string of any length to a non-negative integer, and its collision probability is almost $0$. String hash is used to calculate the hash value of a string and quickly determine whether two strings are equal.

Therefore, based on Solution 1, we can use the method of string hash to compare whether two strings are equal in $O(1)$ time.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string.

<!-- tabs:start -->

Expand Down
30 changes: 24 additions & 6 deletions solution/1100-1199/1147.Longest Chunked Palindrome Decomposition/Solution.cpp
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Original file line number Diff line number Diff line change
@@ -1,13 +1,31 @@
class Solution {
public:
int longestDecomposition(string text) {
int n = text.size();
if (n < 2) return n;
for (int i = 1; i <= n >> 1; ++i) {
if (text.substr(0, i) == text.substr(n - i)) {
return 2 + longestDecomposition(text.substr(i, n - i - i));
int ans = 0;
auto check = [&](int i, int j, int k) -> bool {
while (k--) {
if (text[i++] != text[j++]) {
return false;
}
}
return true;
};
for (int i = 0, j = text.size() - 1; i <= j;) {
bool ok = false;
for (int k = 1; i + k - 1 < j - k + 1; ++k) {
if (check(i, j - k + 1, k)) {
ans += 2;
i += k;
j -= k;
ok = true;
break;
}
}
if (!ok) {
ans += 1;
break;
}
}
return 1;
return ans;
}
};
25 changes: 16 additions & 9 deletions solution/1100-1199/1147.Longest Chunked Palindrome Decomposition/Solution.go
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Original file line number Diff line number Diff line change
@@ -1,12 +1,19 @@
func longestDecomposition(text string) int {
n := len(text)
if n < 2 {
return n
}
for i := 1; i <= n>>1; i++ {
if text[:i] == text[n-i:] {
return 2 + longestDecomposition(text[i:n-i])
func longestDecomposition(text string) (ans int) {
for i, j := 0, len(text)-1; i <= j; {
ok := false
for k := 1; i+k-1 < j-k+1; k++ {
if text[i:i+k] == text[j-k+1:j+1] {
ans += 2
i += k
j -= k
ok = true
break
}
}
if !ok {
ans++
break
}
}
return 1
return
}
31 changes: 24 additions & 7 deletions solution/1100-1199/1147.Longest Chunked Palindrome Decomposition/Solution.java
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Original file line number Diff line number Diff line change
@@ -1,14 +1,31 @@
class Solution {
public int longestDecomposition(String text) {
int n = text.length();
if (n < 2) {
return n;
int ans = 0;
for (int i = 0, j = text.length() - 1; i <= j;) {
boolean ok = false;
for (int k = 1; i + k - 1 < j - k + 1; ++k) {
if (check(text, i, j - k + 1, k)) {
ans += 2;
i += k;
j -= k;
ok = true;
break;
}
}
if (!ok) {
++ans;
break;
}
}
for (int i = 1; i <= n >> 1; ++i) {
if (text.substring(0, i).equals(text.substring(n - i))) {
return 2 + longestDecomposition(text.substring(i, n - i));
return ans;
}

private boolean check(String s, int i, int j, int k) {
while (k-- > 0) {
if (s.charAt(i++) != s.charAt(j++)) {
return false;
}
}
return 1;
return true;
}
}
24 changes: 17 additions & 7 deletions solution/1100-1199/1147.Longest Chunked Palindrome Decomposition/Solution.py
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Original file line number Diff line number Diff line change
@@ -1,9 +1,19 @@
class Solution:
def longestDecomposition(self, text: str) -> int:
n = len(text)
if n < 2:
return n
for i in range(n // 2 + 1):
if text[:i] == text[-i:]:
return 2 + self.longestDecomposition(text[i:-i])
return 1
ans = 0
i, j = 0, len(text) - 1
while i <= j:
k = 1
ok = False
while i + k - 1 < j - k + 1:
if text[i : i + k] == text[j - k + 1 : j + 1]:
ans += 2
i += k
j -= k
ok = True
break
k += 1
if not ok:
ans += 1
break
return ans
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