Question

2002. Maximum Product of the Length of Two Palindromic Subsequences

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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 $s$.

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

}