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.

Code

Brute Force Recursion

$O(2^n)$, Time Limit Exceed.

class Solution {
public:
    int longestPalindromeSubseq(string s) {
        int l = 0, r = s.length() - 1;
        return lps(l, r, s);
    }
 
    int lps(int left, int right, string& s) {
        if(left == right) return 1;
        if(left > right) return 0;
        if(s[left] == s[right]) {
            return lps(left + 1, right - 1, s) + 2;
        } else {
            return max(lps(left+1, right, s), lps(left, right - 1, s));
        }
    }
};

With Memo

$O(n^2)$, pass all tests

class Solution {
public:
    int longestPalindromeSubseq(string s) {
        int l = 0, r = s.length() - 1, n = s.length();
        vector<vector<int>> dp(n, vector<int>(n, NULL));
        return lps(l, r, s, dp);
    }
 
    int lps(int left, int right, string& s, vector<vector<int>>& dp) {
        if(dp[left][right] != NULL) return dp[left][right];
 
        if(left == right) {
            return dp[left][right] = 1;
        }
        if(left > right) {
            return dp[left][right] = 0;
        }
 
        if(s[left] == s[right]) {
            return dp[left][right] = lps(left + 1, right - 1, s, dp) + 2;
        } else {
            return dp[left][right] = max(lps(left+1, right, s, dp), lps(left, right - 1, s, dp));
        }
    }
};

DP

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

Source

• Longest Palindromic Subsequence - LeetCode