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feat: add solutions to lc problem: No.2002
No.2002.Maximum Product of the Length of Two Palindromic Subsequences
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‎solution/2000-2099/2002.Maximum Product of the Length of Two Palindromic Subsequences/README.md

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@@ -53,22 +53,177 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：二进制枚举**
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我们注意到，字符串 $s$ 的长度不超过 $12$，因此我们可以使用二进制枚举的方法来枚举 $s$ 的所有子序列。不妨设 $s$ 的长度为 $n$，我们可以使用 $2^n$ 个长度为 $n$ 的二进制数来表示 $s$ 的所有子序列。对于每个二进制数，第 $i$ 位为 $1$ 表示 $s$ 的第 $i$ 个字符在子序列中，为 $0$ 表示不在子序列中。我们对于每个二进制数，判断其是否为回文子序列，并且记录在数组 $p$ 中。
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接下来，我们枚举 $p$ 中的每个数 $i$，如果 $i$ 是回文子序列，那么我们可以从 $i$ 的补集 $mx = (2^n - 1) \& \overline{i}$ 中枚举一个数 $j$，如果 $j$ 也是回文子序列，那么 $i$ 和 $j$ 就是我们要找的两个回文子序列，它们的长度分别为 $i$ 和 $j$ 的二进制表示中的 $1$ 的个数，我们记为 $a$ 和 $b$，那么它们的乘积就是 $a \times b$，我们取所有可能的 $a \times b$ 中的最大值即可。
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时间复杂度 $(2^n \times n + 3^n)$，空间复杂度 $O(2^n)$。其中 $n$ 为字符串 $s$ 的长度。
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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 maxProduct(self, s: str) -> int:
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n = len(s)
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p = [True] * (1 << n)
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for k in range(1, 1 << n):
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i, j = 0, n - 1
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while i < j:
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while i < j and (k >> i & 1) == 0:
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i += 1
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while i < j and (k >> j & 1) == 0:
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j -= 1
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if i < j and s[i] != s[j]:
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p[k] = False
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break
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i, j = i + 1, j - 1
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ans = 0
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for i in range(1, 1 << n):
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if p[i]:
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mx = ((1 << n) - 1) ^ i
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j = mx
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a = i.bit_count()
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while j:
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if p[j]:
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b = j.bit_count()
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ans = max(ans, a * b)
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j = (j - 1) & mx
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return ans
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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 maxProduct(String s) {
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int n = s.length();
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boolean[] p = new boolean[1 << n];
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Arrays.fill(p, true);
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for (int k = 1; k < 1 << n; ++k) {
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for (int i = 0, j = n - 1; i < n; ++i, --j) {
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while (i < j && (k >> i & 1) == 0) {
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++i;
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}
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while (i < j && (k >> j & 1) == 0) {
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--j;
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}
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if (i < j && s.charAt(i) != s.charAt(j)) {
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p[k] = false;
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break;
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}
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}
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}
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int ans = 0;
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for (int i = 1; i < 1 << n; ++i) {
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if (p[i]) {
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int a = Integer.bitCount(i);
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int mx = ((1 << n) - 1) ^ i;
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for (int j = mx; j > 0; j = (j - 1) & mx) {
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if (p[j]) {
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int b = Integer.bitCount(j);
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ans = Math.max(ans, a * b);
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}
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}
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}
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}
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return ans;
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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 maxProduct(string s) {
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int n = s.size();
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vector<bool> p(1 << n, true);
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for (int k = 1; k < 1 << n; ++k) {
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for (int i = 0, j = n - 1; i < j; ++i, --j) {
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while (i < j && !(k >> i & 1)) {
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++i;
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}
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while (i < j && !(k >> j & 1)) {
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--j;
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}
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if (i < j && s[i] != s[j]) {
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p[k] = false;
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break;
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}
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}
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}
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int ans = 0;
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for (int i = 1; i < 1 << n; ++i) {
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if (p[i]) {
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int a = __builtin_popcount(i);
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int mx = ((1 << n) - 1) ^ i;
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for (int j = mx; j; j = (j - 1) & mx) {
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if (p[j]) {
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int b = __builtin_popcount(j);
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ans = max(ans, a * b);
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}
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}
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}
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}
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return ans;
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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 maxProduct(s string) (ans int) {
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n := len(s)
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p := make([]bool, 1<<n)
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for i := range p {
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p[i] = true
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}
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for k := 1; k < 1<<n; k++ {
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for i, j := 0, n-1; i < j; i, j = i+1, j-1 {
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for i < j && (k>>i&1) == 0 {
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i++
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}
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for i < j && (k>>j&1) == 0 {
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j--
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}
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if i < j && s[i] != s[j] {
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p[k] = false
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break
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}
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}
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}
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for i := 1; i < 1<<n; i++ {
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if p[i] {
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a := bits.OnesCount(uint(i))
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mx := (1<<n - 1) ^ i
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for j := mx; j > 0; j = (j - 1) & mx {
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if p[j] {
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b := bits.OnesCount(uint(j))
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ans = max(ans, a*b)
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}
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}
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}
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}
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return
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}
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func max(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/2000-2099/2002.Maximum Product of the Length of Two Palindromic Subsequences/README_EN.md

Lines changed: 148 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -53,13 +53,160 @@ The product of their lengths is: 5 * 5 = 25.
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### **Python3**
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```python
56-
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class Solution:
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def maxProduct(self, s: str) -> int:
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n = len(s)
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p = [True] * (1 << n)
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for k in range(1, 1 << n):
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i, j = 0, n - 1
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while i < j:
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while i < j and (k >> i & 1) == 0:
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i += 1
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while i < j and (k >> j & 1) == 0:
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j -= 1
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if i < j and s[i] != s[j]:
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p[k] = False
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break
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i, j = i + 1, j - 1
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ans = 0
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for i in range(1, 1 << n):
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if p[i]:
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mx = ((1 << n) - 1) ^ i
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j = mx
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a = i.bit_count()
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while j:
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if p[j]:
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b = j.bit_count()
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ans = max(ans, a * b)
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j = (j - 1) & mx
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return ans
5783
```
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### **Java**
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```java
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class Solution {
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public int maxProduct(String s) {
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int n = s.length();
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boolean[] p = new boolean[1 << n];
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Arrays.fill(p, true);
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for (int k = 1; k < 1 << n; ++k) {
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for (int i = 0, j = n - 1; i < n; ++i, --j) {
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while (i < j && (k >> i & 1) == 0) {
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++i;
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}
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while (i < j && (k >> j & 1) == 0) {
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--j;
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}
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if (i < j && s.charAt(i) != s.charAt(j)) {
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p[k] = false;
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break;
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}
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}
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}
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int ans = 0;
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for (int i = 1; i < 1 << n; ++i) {
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if (p[i]) {
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int a = Integer.bitCount(i);
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int mx = ((1 << n) - 1) ^ i;
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for (int j = mx; j > 0; j = (j - 1) & mx) {
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if (p[j]) {
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int b = Integer.bitCount(j);
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ans = Math.max(ans, a * b);
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}
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}
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}
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}
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return ans;
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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:
130+
int maxProduct(string s) {
131+
int n = s.size();
132+
vector<bool> p(1 << n, true);
133+
for (int k = 1; k < 1 << n; ++k) {
134+
for (int i = 0, j = n - 1; i < j; ++i, --j) {
135+
while (i < j && !(k >> i & 1)) {
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++i;
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}
138+
while (i < j && !(k >> j & 1)) {
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--j;
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}
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if (i < j && s[i] != s[j]) {
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p[k] = false;
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break;
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}
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}
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}
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int ans = 0;
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for (int i = 1; i < 1 << n; ++i) {
149+
if (p[i]) {
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int a = __builtin_popcount(i);
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int mx = ((1 << n) - 1) ^ i;
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for (int j = mx; j; j = (j - 1) & mx) {
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if (p[j]) {
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int b = __builtin_popcount(j);
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ans = max(ans, a * b);
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}
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}
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
168+
func maxProduct(s string) (ans int) {
169+
n := len(s)
170+
p := make([]bool, 1<<n)
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for i := range p {
172+
p[i] = true
173+
}
174+
for k := 1; k < 1<<n; k++ {
175+
for i, j := 0, n-1; i < j; i, j = i+1, j-1 {
176+
for i < j && (k>>i&1) == 0 {
177+
i++
178+
}
179+
for i < j && (k>>j&1) == 0 {
180+
j--
181+
}
182+
if i < j && s[i] != s[j] {
183+
p[k] = false
184+
break
185+
}
186+
}
187+
}
188+
for i := 1; i < 1<<n; i++ {
189+
if p[i] {
190+
a := bits.OnesCount(uint(i))
191+
mx := (1<<n - 1) ^ i
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for j := mx; j > 0; j = (j - 1) & mx {
193+
if p[j] {
194+
b := bits.OnesCount(uint(j))
195+
ans = max(ans, a*b)
196+
}
197+
}
198+
}
199+
}
200+
return
201+
202+
}
203+
204+
func max(a, b int) int {
205+
if a > b {
206+
return a
207+
}
208+
return b
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}
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```
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### **...**

‎solution/2000-2099/2002.Maximum Product of the Length of Two Palindromic Subsequences/Solution.cpp

Lines changed: 35 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,35 @@
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class Solution {
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public:
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int maxProduct(string s) {
4+
int n = s.size();
5+
vector<bool> p(1 << n, true);
6+
for (int k = 1; k < 1 << n; ++k) {
7+
for (int i = 0, j = n - 1; i < j; ++i, --j) {
8+
while (i < j && !(k >> i & 1)) {
9+
++i;
10+
}
11+
while (i < j && !(k >> j & 1)) {
12+
--j;
13+
}
14+
if (i < j && s[i] != s[j]) {
15+
p[k] = false;
16+
break;
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}
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}
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}
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int ans = 0;
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for (int i = 1; i < 1 << n; ++i) {
22+
if (p[i]) {
23+
int a = __builtin_popcount(i);
24+
int mx = ((1 << n) - 1) ^ i;
25+
for (int j = mx; j; j = (j - 1) & mx) {
26+
if (p[j]) {
27+
int b = __builtin_popcount(j);
28+
ans = max(ans, a * b);
29+
}
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}
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}
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}
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return ans;
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}
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};

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