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Question

Formatted question description: https://leetcode.ca/all/516.html

Given a string s, find the longest palindromic subsequence's length in s.

A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.

 

Example 1:

Input: s = "bbbab"
Output: 4
Explanation: One possible longest palindromic subsequence is "bbbb".

Example 2:

Input: s = "cbbd"
Output: 2
Explanation: One possible longest palindromic subsequence is "bb".

 

Constraints:

• 1 <= s.length <= 1000
• s consists only of lowercase English letters.

Algorithm

dp[i][j] is [i,j] longest Palindromic Subsequence in range [i,j]

              /  dp[i + 1][j - 1] + 2                       if (s[i] == s[j])

dp[i][j] =

              \  max(dp[i + 1][j], dp[i][j - 1])        if (s[i] != s[j])

Code

• Java
• C++
• Python
• Go

• public class Longest_Palindromic_Subsequence {
  
      public class Solution {
          public int longestPalindromeSubseq(String s) {
              int n = s.length();
              int[][] dp = new int[n][n];
              for (int i = n - 1; i >= 0; --i) {
                  dp[i][i] = 1;
                  for (int j = i + 1; j < n; ++j) {
                      if (s.charAt(i) == s.charAt(j)) {
                          dp[i][j] = dp[i + 1][j - 1] + 2;
                      } else {
                          dp[i][j] = Math.max(dp[i + 1][j], dp[i][j - 1]);
                      }
                  }
              }
              return dp[0][n - 1];
          }
      }
  
  }
  

• class Solution {
  public:
      int longestPalindromeSubseq(string s) {
          int n = s.size();
          vector<vector<int>> dp(n, vector<int>(n, 0));
          for (int i = 0; i < n; ++i) {
              dp[i][i] = 1;
          }
          for (int j = 1; j < n; ++j) {
              for (int i = j - 1; i >= 0; --i) {
                  if (s[i] == s[j]) {
                      dp[i][j] = dp[i + 1][j - 1] + 2;
                  } else {
                      dp[i][j] = max(dp[i + 1][j], dp[i][j - 1]);
                  }
              }
          }
          return dp[0][n - 1];
      }
  };
  

• class Solution:
      def longestPalindromeSubseq(self, s: str) -> int:
          n = len(s)
          dp = [[0] * n for _ in range(n)]
          for i in range(n - 1, -1, -1):
              dp[i][i] = 1 # pre-set before inner loop
              for j in range(i + 1, n):
                  if s[i] == s[j]:
                      dp[i][j] = dp[i + 1][j - 1] + 2
                  else:
                      dp[i][j] = max(dp[i + 1][j], dp[i][j - 1])
          return dp[0][-1]
  
  
  class Solution:
      def longestPalindromeSubseq(self, s: str) -> int:
          n = len(s)
          dp = [[0] * n for _ in range(n)]
          for i in range(n):
              dp[i][i] = 1 # cannot move it inside 'for j in range(1, n)'
              # because below 'for j in range' starting at 1
              # so cannot put dp[j][j]=1 inside below for loop
          for j in range(1, n):
              for i in range(j - 1, -1, -1):
                  if s[i] == s[j]:
                      dp[i][j] = dp[i + 1][j - 1] + 2
                  else:
                      dp[i][j] = max(dp[i + 1][j], dp[i][j - 1])
          return dp[0][-1]
  
  ############
  
  class Solution(object):
    def longestPalindromeSubseq(self, s):
      """
      :type s: str
      :rtype: int
      """
      n = len(s)
      dp = [1] * n
      for j in range(1, len(s)):
        pre = dp[j]
        for i in reversed(range(0, j)):
          tmp = dp[i]
          if s[i] == s[j]:
            dp[i] = 2 + pre if i + 1 <= j - 1 else 2
          else:
            dp[i] = max(dp[i + 1], dp[i])
          pre = tmp
      return dp[0]
  
  

• func longestPalindromeSubseq(s string) int {
  	n := len(s)
  	dp := make([][]int, n)
  	for i := 0; i < n; i++ {
  		dp[i] = make([]int, n)
  		dp[i][i] = 1
  	}
  	for j := 1; j < n; j++ {
  		for i := j - 1; i >= 0; i-- {
  			if s[i] == s[j] {
  				dp[i][j] = dp[i+1][j-1] + 2
  			} else {
  				dp[i][j] = max(dp[i+1][j], dp[i][j-1])
  			}
  		}
  	}
  	return dp[0][n-1]
  }
  
  func max(a, b int) int {
  	if a > b {
  		return a
  	}
  	return b
  }
  

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