2002. Maximum Product of the Length of Two Palindromic Subsequences
Description
Given a string s
, find two disjoint palindromic subsequences of s
such that the product of their lengths is maximized. The two subsequences are disjoint if they do not both pick a character at the same index.
Return the maximum possible product of the lengths of the two palindromic subsequences.
A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters. A string is palindromic if it reads the same forward and backward.
Example 1:
Input: s = "leetcodecom" Output: 9 Explanation: An optimal solution is to choose "ete" for the 1^{st} subsequence and "cdc" for the 2^{nd} subsequence. The product of their lengths is: 3 * 3 = 9.
Example 2:
Input: s = "bb" Output: 1 Explanation: An optimal solution is to choose "b" (the first character) for the 1^{st} subsequence and "b" (the second character) for the 2^{nd} subsequence. The product of their lengths is: 1 * 1 = 1.
Example 3:
Input: s = "accbcaxxcxx" Output: 25 Explanation: An optimal solution is to choose "accca" for the 1^{st} subsequence and "xxcxx" for the 2^{nd} subsequence. The product of their lengths is: 5 * 5 = 25.
Constraints:
2 <= s.length <= 12
s
consists of lowercase English letters only.
Solutions
Solution 1: Binary Enumeration
We notice that the length of the string $s$ does not exceed $12$, so we can use the method of binary enumeration to enumerate all subsequences of $s$. Suppose the length of $s$ is $n$, we can use $2^n$ binary numbers of length $n$ to represent all subsequences of $s$. For each binary number, the $i$th bit being $1$ means the $i$th character of $s$ is in the subsequence, and $0$ means it is not in the subsequence. For each binary number, we judge whether it is a palindrome subsequence and record it in the array $p$.
Next, we enumerate each number $i$ in $p$. If $i$ is a palindrome subsequence, then we can enumerate a number $j$ from the complement of $i$, $mx = (2^n  1) \oplus i$. If $j$ is also a palindrome subsequence, then $i$ and $j$ are the two palindrome subsequences we are looking for. Their lengths are the number of $1$s in the binary representation of $i$ and $j$, denoted as $a$ and $b$, respectively. Then their product is $a \times b$. We take the maximum of all possible $a \times b$.
The time complexity is $(2^n \times n + 3^n)$, and the space complexity is $O(2^n)$. Here, $n$ is the length of the string $s$.







