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 1st subsequence and "cdc" for the 2nd 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 1st subsequence and "b" (the second character) for the 2nd 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 1st subsequence and "xxcxx" for the 2nd 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 $s$.

Python3JavaC++Go

class Solution:
    def maxProduct(self, s: str) -> int:
        n = len(s)
        p = [True] * (1 << n)
        for k in range(1, 1 << n):
            i, j = 0, n - 1
            while i < j:
                while i < j and (k >> i & 1) == 0:
                    i += 1
                while i < j and (k >> j & 1) == 0:
                    j -= 1
                if i < j and s[i] != s[j]:
                    p[k] = False
                    break
                i, j = i + 1, j - 1
        ans = 0
        for i in range(1, 1 << n):
            if p[i]:
                mx = ((1 << n) - 1) ^ i
                j = mx
                a = i.bit_count()
                while j:
                    if p[j]:
                        b = j.bit_count()
                        ans = max(ans, a * b)
                    j = (j - 1) & mx
        return ans

class Solution {
    public int maxProduct(String s) {
        int n = s.length();
        boolean[] p = new boolean[1 << n];
        Arrays.fill(p, true);
        for (int k = 1; k < 1 << n; ++k) {
            for (int i = 0, j = n - 1; i < n; ++i, --j) {
                while (i < j && (k >> i & 1) == 0) {
                    ++i;
                }
                while (i < j && (k >> j & 1) == 0) {
                    --j;
                }
                if (i < j && s.charAt(i) != s.charAt(j)) {
                    p[k] = false;
                    break;
                }
            }
        }
        int ans = 0;
        for (int i = 1; i < 1 << n; ++i) {
            if (p[i]) {
                int a = Integer.bitCount(i);
                int mx = ((1 << n) - 1) ^ i;
                for (int j = mx; j > 0; j = (j - 1) & mx) {
                    if (p[j]) {
                        int b = Integer.bitCount(j);
                        ans = Math.max(ans, a * b);
                    }
                }
            }
        }
        return ans;
    }
}

class Solution {
public:
    int maxProduct(string s) {
        int n = s.size();
        vector<bool> p(1 << n, true);
        for (int k = 1; k < 1 << n; ++k) {
            for (int i = 0, j = n - 1; i < j; ++i, --j) {
                while (i < j && !(k >> i & 1)) {
                    ++i;
                }
                while (i < j && !(k >> j & 1)) {
                    --j;
                }
                if (i < j && s[i] != s[j]) {
                    p[k] = false;
                    break;
                }
            }
        }
        int ans = 0;
        for (int i = 1; i < 1 << n; ++i) {
            if (p[i]) {
                int a = __builtin_popcount(i);
                int mx = ((1 << n) - 1) ^ i;
                for (int j = mx; j; j = (j - 1) & mx) {
                    if (p[j]) {
                        int b = __builtin_popcount(j);
                        ans = max(ans, a * b);
                    }
                }
            }
        }
        return ans;
    }
};

func maxProduct(s string) (ans int) {
    n := len(s)
    p := make([]bool, 1<<n)
    for i := range p {
        p[i] = true
    }
    for k := 1; k < 1<<n; k++ {
        for i, j := 0, n-1; i < j; i, j = i+1, j-1 {
            for i < j && (k>>i&1) == 0 {
                i++
            }
            for i < j && (k>>j&1) == 0 {
                j--
            }
            if i < j && s[i] != s[j] {
                p[k] = false
                break
            }
        }
    }
    for i := 1; i < 1<<n; i++ {
        if p[i] {
            a := bits.OnesCount(uint(i))
            mx := (1<<n - 1) ^ i
            for j := mx; j > 0; j = (j - 1) & mx {
                if p[j] {
                    b := bits.OnesCount(uint(j))
                    ans = max(ans, a*b)
                }
            }
        }
    }
    return

}

Comments