Longest Palindromic Subsequence

shriyanshi24jindal

Mar 28th, 2026

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Java 8.77 KB | None | 0 0

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class Solution {
// idea: longest palindromic subsequence will be length of LCS
public int longestPalindromeSubseq(String s) {
int n = s.length();
// reverse the string and stroe it in new string
String s2 = new StringBuilder(s).reverse().toString();
int m = s2.length();
// step-1 create a matrix
// t[i][j] = LCS of first i chars of s and first j chars of reversed string
int[][] t = new int[n+1][m+1];
// step-2 initialise the matrix with base condition
for(int i=0; i<n+1; i++)
{
for(int j=0; j<m+1; j++)
{
// string length is 0, NO LCS found
if(i == 0 || j == 0)
t[i][j] = 0;
}
}
// step-3 iterative code to find length of LCS
for(int i=1; i<n+1; i++)
{
for(int j=1; j<m+1; j++)
{
// characters match
if(s.charAt(i-1) == s2.charAt(j-1))
t[i][j] = 1 + t[i-1][j-1];
// characters do not match
else
t[i][j] = Math.max(t[i-1][j], t[i][j-1]);
}
}
// The bottom-right cell contains the length of LPS
int lcsLength = t[n][m];
return lcsLength;
}
}

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class Solution {
    // idea: longest palindromic subsequence will be length of LCS
    
    public int longestPalindromeSubseq(String s) {
        int n = s.length();

// reverse the string and stroe it in new string
        String s2 = new StringBuilder(s).reverse().toString();
        int m = s2.length();

// step-1 create a matrix
        // t[i][j] = LCS of first i chars of s and first j chars of reversed string
        int[][] t = new int[n+1][m+1];

// step-2 initialise the matrix with base condition
        for(int i=0; i<n+1; i++)
        {
            for(int j=0; j<m+1; j++)
            {
                // string length is 0, NO LCS found
                if(i == 0 || j == 0)
                    t[i][j] = 0;
            }
        }

// step-3 iterative code to find length of LCS
        for(int i=1; i<n+1; i++)
        {
            for(int j=1; j<m+1; j++)
            {
                // characters match
                if(s.charAt(i-1) == s2.charAt(j-1))
                    t[i][j] = 1 + t[i-1][j-1];

// characters do not match
                else
                    t[i][j] = Math.max(t[i-1][j], t[i][j-1]);
            }
        }
        // The bottom-right cell contains the length of LPS
        int lcsLength = t[n][m];
        return lcsLength;
    }
}