# Longest Palindromic Subsequence LeetCode Solution

Here, We see Longest Palindromic Subsequence LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

# List of all LeetCode Solution

## Problem Statement

Given a string `s`, find the longest palindromic subsequence‘s length in `s`.

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”.

## Longest Palindromic Subsequence LeetCode SolutionC++

``````class Solution {
public:
int longestPalindromeSubseqTabular(string& s) {
if(s.empty())
return 0;
const 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 l = 1; l < N; l++) {
for(int i = 0; i < N - l; i++) {
int j = i + l;
if((j - i + 1) == 2) {
dp[i][j] = 1 + (s[i] == s[j]);
}
else {
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];
}

int longestPalindSubseqRec(vector<vector<int> >& dp, string& s, int i, int j) {
if(i > j)
return 0;
if(i == j)
return dp[i][j] = 1;
if(dp[i][j] == -1) {
if(s[i] == s[j])
dp[i][j] = 2 + longestPalindSubseqRec(dp, s, i+1, j-1);
else
dp[i][j] = max(longestPalindSubseqRec(dp, s, i+1, j),
longestPalindSubseqRec(dp, s, i, j-1));
}
return dp[i][j];
}

int longestPalindSubseqRecDriver(string& s) {
if(s.empty())
return 0;
const int N = s.size();
vector<vector<int> > dp(N, vector<int>(N, -1));
return longestPalindSubseqRec(dp, s, 0, N-1);
}

int longestPalindromeSubseq(string s) {
return longestPalindSubseqRecDriver(s);
}
};```Code language: PHP (php)```

## Longest Palindromic Subsequence LeetCode SolutionJava

``````class Solution {
public int longestPalindromeSubseq(String s) {
int n = s.length();
int[][] dp = new int[n][n];
for (int i = n-1; i >= 0; i--) {
dp[i][i] = 1;
for (int j = i+1; j < n; j++) {
if (s.charAt(i) == s.charAt(j)) {
dp[i][j] = 2 + dp[i+1][j-1];
} else {
dp[i][j] = Math.max(dp[i+1][j], dp[i][j-1]);
}
}
}
return dp[0][n-1];
}
}```Code language: JavaScript (javascript)```

## Longest Palindromic Subsequence SolutionJavaScript

``````var longestPalindromeSubseq = function(s) {
const n = s.length;
const dp = Array(n).fill().map(() => Array(n).fill(0));
for (let i = n-1; i >= 0; i--) {
dp[i][i] = 1;
for (let j = i+1; j < n; j++) {
if (s[i] === s[j]) {
dp[i][j] = 2 + dp[i+1][j-1];
} else {
dp[i][j] = Math.max(dp[i+1][j], dp[i][j-1]);
}
}
}
return dp[0][n-1];
};```Code language: JavaScript (javascript)```

## Longest Palindromic Subsequence SolutionPython

``````class Solution(object):
def longestPalindromeSubseq(self, s):
n = len(s)
dp = [[0]*n for _ in range(n)]
for i in range(n-1, -1, -1):
dp[i][i] = 1
for j in range(i+1, n):
if s[i] == s[j]:
dp[i][j] = 2 + dp[i+1][j-1]
else:
dp[i][j] = max(dp[i+1][j], dp[i][j-1])
return dp[0][n-1]``````
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