516. Longest Palindromic Subsequence

Problem statement: 

Given a string s, find the longest palindromic subsequence's length in s. You may assume that the maximum length of s is 1000.

Example 1:
Input:

"bbbab"

Output:

4

One possible longest palindromic subsequence is "bbbb".

Example 2:
Input:

"cbbd"

Output:

2

One possible longest palindromic subsequence is "bb".

Solution:

This is a dynamic programming problem. The key point also is to find the dynamic formula. Although there is only one string, but we need a two dimension array. The value in diagonal is 1 and we only need right up part above diagonal.

dp[i][j] means the max subsequence from position i to j. The formula is:

dp[i][j] = dp[i + 1][j - 1]   if s[i] == s[j]

　　　　　= max(dp[i + 1][j], dp[i][j + 1]) if s[i] != s[j]

Since there is i + 1 and j - 1 is essential for current dp[i][j], we should do DP from bottom to up, left to right.

The time complexity is O(n*n), n is the size of string.

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