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516. Longest Palindromic Subsequence

Description

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.

Solutions

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])

• Java
• C++
• Python
• Go
• TypeScript

• class Solution {
      public int longestPalindromeSubseq(String s) {
          int n = s.length();
          int[][] dp = new int[n][n];
          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.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
                      # "aa" => i=0,j=1 => dp[1][0] is 0, still works
                  else:
                      dp[i][j] = max(dp[i + 1][j], dp[i][j - 1])
          return dp[0][-1]
  
  

• 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]
  }
  

• function longestPalindromeSubseq(s: string): number {
      const n = s.length;
      const f: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
      for (let i = 0; i < n; ++i) {
          f[i][i] = 1;
      }
      for (let i = n - 2; ~i; --i) {
          for (let j = i + 1; j < n; ++j) {
              if (s[i] === s[j]) {
                  f[i][j] = f[i + 1][j - 1] + 2;
              } else {
                  f[i][j] = Math.max(f[i + 1][j], f[i][j - 1]);
              }
          }
      }
      return f[0][n - 1];
  }
  
  

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