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Use a 2D table to keep track of the longest palindromic subsequence. The idea is to fill out this table by solving smaller subproblems and using their results to build up solutions to bigger ones.
The table, dp[i][j], will store the length of the longest palindromic subsequence in s[i...j]. Start by assuming that each character is a palindrome of length 1, and fill the table based on whether the characters at the current indices are the same.
Time Complexity: O(n^2), where n is the length of the string, because we need to fill an n x n table.
Space Complexity: O(n^2), for the same reason due to the dp table.
1class Solution {
2 public int longestPalindromeSubseq(String s) {
3 int n = s.length();
4 int[][] dp = new int[n][n];
5 for (int i = n - 1; i >= 0; i--) {
6 dp[i][i] = 1;
7 for (int j = i + 1; j < n; j++) {
8 if (s.charAt(i) == s.charAt(j))
9 dp[i][j] = dp[i + 1][j - 1] + 2;
10 else
11 dp[i][j] = Math.max(dp[i + 1][j], dp[i][j - 1]);
12 }
13 }
14 return dp[0][n - 1];
15 }
16}
17
18public class Main {
19 public static void main(String[] args) {
20 Solution solution = new Solution();
21 System.out.println(solution.longestPalindromeSubseq("bbbab"));
22 }
23}Uses a Java 2D array for the dp table, and systematically calculates the length of the longest palindromic subsequence by checking character equalities and using previously calculated results.
This approach defines a recursive function that explores all potential palindromic subsequences, using memoization to store results of overlapping subproblems, thereby avoiding redundant calculations. The function checks if the characters at current indices are equal and proceeds based on that, with memoization avoiding recalculations.
Time Complexity: O(n^2), where n is the length of the string due to memoization.
Space Complexity: O(n^2) for the memoization table.
1class
The Java solution also uses a helper function along with a memoization array, providing a recursive methodology to solve the problem efficiently by avoiding redundant computations.