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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])
  • 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]
    }
    

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