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Longest Palindromic Subsequence

Problem​

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.

Examples​

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.

Approach​

To solve this problem, we can use dynamic programming. We will create a 2D array dp where dp[i][j] represents the length of the longest palindromic subsequence in the substring s[i:j+1].

Steps:​

  1. Initialize a 2D array dp of size n x n where n is the length of the string s. Set all elements to 0.
  2. For each single character, set dp[i][i] = 1 because a single character is a palindrome of length 1.
  3. Fill the dp array in a bottom-up manner:
    • For each substring length l from 2 to n:
      • For each starting index i from 0 to n-l:
        • Set j = i + l - 1.
        • If s[i] == s[j], then dp[i][j] = dp[i+1][j-1] + 2.
        • Otherwise, dp[i][j] = max(dp[i+1][j], dp[i][j-1]).
  4. The result will be dp[0][n-1], the length of the longest palindromic subsequence in the entire string.

Solution​

Java Solution​

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];
    }
}

C++ Solution​

class Solution {
public:
    int longestPalindromeSubseq(string s) {
        int n = s.length();
        vector<vector<int>> dp(n, vector<int>(n, 0));
        
        for (int i = n - 1; i >= 0; i--) {
            dp[i][i] = 1;
            for (int j = i + 1; j < n; j++) {
                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];
    }
};

Python Solution​

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
            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][n - 1]

Complexity Analysis​

Time Complexity: O(n^2)

Reason: We are filling an n x n table, and each cell takes constant time to compute.

Space Complexity: O(n^2)

Reason: We use a 2D array dp of size n x n to store the intermediate results.

References​

LeetCode Problem: Longest Palindromic Subsequence