# 516. 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.

## Solutions

dp[i][j] is [i,j] longest Palindromic Subsequence in range [i,j]

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

dp[i][j] =

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

• 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 j = 1; j < n; ++j) {
for (int i = j - 1; i >= 0; --i) {
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];
}
}

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

• 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 # pre-set before inner loop
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][-1]

###########

class Solution:
def longestPalindromeSubseq(self, s: str) -> int:
n = len(s)
dp = [[0] * n for _ in range(n)]
for i in range(n):
dp[i][i] = 1 # cannot move it inside 'for j in range(1, n)'
# because below 'for j in range' starting at 1
# so cannot put dp[j][j]=1 inside below for loop
for j in range(1, n):
for i in range(j - 1, -1, -1):
if s[i] == s[j]:
dp[i][j] = dp[i + 1][j - 1] + 2
# "aa" => i=0,j=1 => dp[1][0] is 0, still works
else:
dp[i][j] = max(dp[i + 1][j], dp[i][j - 1])
return dp[0][-1]


• func longestPalindromeSubseq(s string) int {
n := len(s)
dp := make([][]int, n)
for i := 0; i < n; i++ {
dp[i] = make([]int, n)
dp[i][i] = 1
}
for j := 1; j < n; j++ {
for i := j - 1; i >= 0; i-- {
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]
}

• function longestPalindromeSubseq(s: string): number {
const n = s.length;
const f: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
for (let i = 0; i < n; ++i) {
f[i][i] = 1;
}
for (let i = n - 2; ~i; --i) {
for (let j = i + 1; j < n; ++j) {
if (s[i] === s[j]) {
f[i][j] = f[i + 1][j - 1] + 2;
} else {
f[i][j] = Math.max(f[i + 1][j], f[i][j - 1]);
}
}
}
return f[0][n - 1];
}