516. Longest Palindromic Subsequence

TC: O(n^2)
SC: O(n^2)
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 i = n-1; i >= 0; i--) {
            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][j-1], dp[i+1][j]);
                }
            }
        }
        return dp[0][n-1];
    }
}
/*

Longest Palindromic Subsequence


dp[i][j]: Longest Palindromic Subsequence for s[i:j]

i xxxxxx j

if (s[i] == s[j])
dp[i][j] = dp[i+1][j-1] + 2

else 
dp[i][j] = max(dp[i][j-1], dp[i+1][j])


s[i] != s[j] so we can't have i, j at the same time
[       ]
i xxxxxx j
  [      ]
  
base case, if i == j , value is 1 (only one character)
if (i < j ) value is all 0

[i, j-1]   [i,j]
[i+1, j-1] [i+1, j]

      j
      -> ->
   1  x. x  ^
i  0  1  x  |
   0  0  1   for right buttom to exe

TC: O(n^2)
SC: O(n^2)

*/

DFS + MEMO

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