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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 $s$.

• Java
• C++
• Python
• Go

• 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;
      }
  };
  

• 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
  
  

• 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
  
  }
  

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