Given a sequence, print a longest palindromic subsequence of it.
Examples :
Input : BBABCBCAB
Output : BABCBAB
The above output is the longest
palindromic subsequence of given
sequence. "BBBBB" and "BBCBB" are
also palindromic subsequences of
the given sequence, but not the
longest ones.
Input : GEEKSFORGEEKS
Output : Output can be either EEKEE
or EESEE or EEGEE, ..
We have discussed a solution in below post to find length of longest palindromic subsequence.
Dynamic Programming | Set 12 (Longest Palindromic Subsequence)
Method 1:
This problem is close to the Longest Common Subsequence (LCS) problem. In fact, we can use LCS as a subroutine to solve this problem. Following is the two step solution that uses LCS.
- Reverse the given sequence and store the reverse in another array say rev[0..n-1]
- LCS of the given sequence and rev[] will be the longest palindromic sequence.
- Once we have found LCS, we can print the LCS.
Below is the implementation of above approach:
C++
/* CPP program to print longest palindromic subsequence */#include<bits/stdc++.h>using namespace std;/* Returns LCS X and Y */string lcs(string &X, string &Y){ int m = X.length(); int n = Y.length(); int L[m+1][n+1]; /* Following steps build L[m+1][n+1] in bottom up fashion. Note that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */ for (int i=0; i<=m; i++) { for (int j=0; j<=n; j++) { if (i == 0 || j == 0) L[i][j] = 0; else if (X[i-1] == Y[j-1]) L[i][j] = L[i-1][j-1] + 1; else L[i][j] = max(L[i-1][j], L[i][j-1]); } } // Following code is used to print LCS int index = L[m][n]; // Create a string length index+1 and // fill it with \0 string lcs(index+1, '\0'); // Start from the right-most-bottom-most // corner and one by one store characters // in lcs[] int i = m, j = n; while (i > 0 && j > 0) { // If current character in X[] and Y // are same, then current character // is part of LCS if (X[i-1] == Y[j-1]) { // Put current character in result lcs[index-1] = X[i-1]; i--; j--; // reduce values of i, j and index index--; } // If not same, then find the larger of // two and go in the direction of larger // value else if (L[i-1][j] > L[i][j-1]) i--; else j--; } return lcs;}// Returns longest palindromic subsequence// of strstring longestPalSubseq(string &str){ // Find reverse of str string rev = str; reverse(rev.begin(), rev.end()); // Return LCS of str and its reverse return lcs(str, rev);}/* Driver program to test above function */int main(){ string str = "GEEKSFORGEEKS"; cout << longestPalSubseq(str); return 0;} |
Java
// Java program to print longest palindromic// subsequenceimport java.io.*;class GFG { /* Returns LCS X and Y */ static String lcs(String a, String b) { int m = a.length(); int n = b.length(); char X[] = a.toCharArray(); char Y[] = b.toCharArray(); int L[][] = new int[m + 1][n + 1]; /* Following steps build L[m+1][n+1] in bottom up fashion. Note that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */ for (int i = 0; i <= m; i++) { for (int j = 0; j <= n; j++) { if (i == 0 || j == 0) { L[i][j] = 0; } else if (X[i - 1] == Y[j - 1]) { L[i][j] = L[i - 1][j - 1] + 1; } else { L[i][j] = Math.max(L[i - 1][j], L[i][j - 1]); } } } // Following code is used to print LCS int index = L[m][n]; // Create a String length index+1 and // fill it with \0 char[] lcs = new char[index + 1]; // Start from the right-most-bottom-most // corner and one by one store characters // in lcs[] int i = m, j = n; while (i > 0 && j > 0) { // If current character in X[] and Y // are same, then current character // is part of LCS if (X[i - 1] == Y[j - 1]) { // Put current character in result lcs[index - 1] = X[i - 1]; i--; j--; // reduce values of i, j and index index--; } // If not same, then find the larger of // two and go in the direction of larger // value else if (L[i - 1][j] > L[i][j - 1]) { i--; } else { j--; } } String ans = ""; for (int x = 0; x < lcs.length; x++) { ans += lcs[x]; } return ans; } // Returns longest palindromic subsequence // of str static String longestPalSubseq(String str) { // Find reverse of str String rev = str; rev = reverse(rev); // Return LCS of str and its reverse return lcs(str, rev); } static String reverse(String str) { String ans = ""; // convert String to character array // by using toCharArray char[] try1 = str.toCharArray(); for (int i = try1.length - 1; i >= 0; i--) { ans += try1[i]; } return ans; } /* Driver program to test above function */ public static void main(String[] args) { String str = "GEEKSFORGEEKS"; System.out.println(longestPalSubseq(str)); }} |
Python3
# Python3 program to print longest # palindromic subsequence# Returns LCS X and Y def lcs_(X, Y) : m = len(X) n = len(Y) L = [[0] * (n + 1)] * (m + 1) # Following steps build L[m+1][n+1] # in bottom up fashion. Note that # L[i][j] contains length of LCS of # X[0..i-1] and Y[0..j-1] for i in range(n + 1) : for j in range(n + 1) : if (i == 0 or j == 0) : L[i][j] = 0; elif (X[i - 1] == Y[j - 1]) : L[i][j] = L[i - 1][j - 1] + 1; else : L[i][j] = max(L[i - 1][j], L[i][j - 1]); # Following code is used to print LCS index = L[m][n]; # Create a string length index+1 and # fill it with \0 lcs = ["\n "] * (index + 1) # Start from the right-most-bottom-most # corner and one by one store characters # in lcs[] i, j= m, n while (i > 0 and j > 0) : # If current character in X[] and Y # are same, then current character # is part of LCS if (X[i - 1] == Y[j - 1]) : # Put current character in result lcs[index - 1] = X[i - 1] i -= 1 j -= 1 # reduce values of i, j and index index -= 1 # If not same, then find the larger of # two and go in the direction of larger # value elif(L[i - 1][j] > L[i][j - 1]) : i -= 1 else : j -= 1 ans = "" for x in range(len(lcs)) : ans += lcs[x] return ans# Returns longest palindromic # subsequence of str def longestPalSubseq(string) : # Find reverse of str rev = string[: : -1] # Return LCS of str and its reverse return lcs_(string, rev)# Driver Codeif __name__ == "__main__" : string = "GEEKSFORGEEKS"; print(longestPalSubseq(string))# This code is contributed by Ryuga |
C#
// C# program to print longest palindromic //subsequence using System;public class GFG { /* Returns LCS X and Y */ static String lcs(String a, String b) { int m = a.Length; int n = b.Length; char []X = a.ToCharArray(); char []Y = b.ToCharArray(); int [,]L = new int[m + 1,n + 1]; int i, j; /* Following steps build L[m+1,n+1] in bottom up fashion. Note that L[i,j] contains length of LCS of X[0..i-1] and Y[0..j-1] */ for (i = 0; i <= m; i++) { for (j = 0; j <= n; j++) { if (i == 0 || j == 0) { L[i,j] = 0; } else if (X[i - 1] == Y[j - 1]) { L[i,j] = L[i - 1,j - 1] + 1; } else { L[i,j] = Math.Max(L[i - 1,j], L[i,j - 1]); } } } // Following code is used to print LCS int index = L[m,n]; // Create a String length index+1 and // fill it with \0 char[] lcs = new char[index + 1]; // Start from the right-most-bottom-most // corner and one by one store characters // in lcs[] i = m; j = n; while (i > 0 && j > 0) { // If current character in X[] and Y // are same, then current character // is part of LCS if (X[i - 1] == Y[j - 1]) { // Put current character in result lcs[index - 1] = X[i - 1]; i--; j--; // reduce values of i, j and index index--; } // If not same, then find the larger of // two and go in the direction of larger // value else if (L[i - 1,j] > L[i,j - 1]) { i--; } else { j--; } } String ans = ""; for (int x = 0; x < lcs.Length; x++) { ans += lcs[x]; } return ans; } // Returns longest palindromic subsequence // of str static String longestPalSubseq(String str) { // Find reverse of str String rev = str; rev = reverse(rev); // Return LCS of str and its reverse return lcs(str, rev); } static String reverse(String str) { String ans = ""; // convert String to character array // by using toCharArray char[] try1 = str.ToCharArray(); for (int i = try1.Length - 1; i >= 0; i--) { ans += try1[i]; } return ans; } /* Driver program to test above function */ public static void Main() { String str = "GEEKSFORGEEKS"; Console.Write(longestPalSubseq(str)); }}// This code is contributed by 29AjayKumar |
Javascript
<script> // Javascript program to print longest palindromic subsequence /* Returns LCS X and Y */ function lcs(a, b) { let m = a.length; let n = b.length; let X = a.split(''); let Y = b.split(''); let L = new Array(m + 1); for (let i = 0; i <= m; i++) { L[i] = new Array(n + 1); for (let j = 0; j <= n; j++) { L[i][j] = 0; } } /* Following steps build L[m+1][n+1] in bottom up fashion. Note that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */ for (let i = 0; i <= m; i++) { for (let j = 0; j <= n; j++) { if (i == 0 || j == 0) { L[i][j] = 0; } else if (X[i - 1] == Y[j - 1]) { L[i][j] = L[i - 1][j - 1] + 1; } else { L[i][j] = Math.max(L[i - 1][j], L[i][j - 1]); } } } // Following code is used to print LCS let index = L[m][n]; // Create a String length index+1 and // fill it with \0 let lcs = new Array(index + 1); lcs.fill(''); // Start from the right-most-bottom-most // corner and one by one store characters // in lcs[] let i = m, j = n; while (i > 0 && j > 0) { // If current character in X[] and Y // are same, then current character // is part of LCS if (X[i - 1] == Y[j - 1]) { // Put current character in result lcs[index - 1] = X[i - 1]; i--; j--; // reduce values of i, j and index index--; } // If not same, then find the larger of // two and go in the direction of larger // value else if (L[i - 1][j] > L[i][j - 1]) { i--; } else { j--; } } let ans = ""; for (let x = 0; x < lcs.length; x++) { ans += lcs[x]; } return ans; } // Returns longest palindromic subsequence // of str function longestPalSubseq(str) { // Find reverse of str let rev = str; rev = reverse(rev); // Return LCS of str and its reverse return lcs(str, rev); } function reverse(str) { let ans = ""; // convert String to character array // by using toCharArray let try1 = str.split(''); for (let i = try1.length - 1; i >= 0; i--) { ans += try1[i]; } return ans; } let str = "GEEKSFORGEEKS"; document.write(longestPalSubseq(str)); // This code is contributed by suresh07.</script> |
Output:
EEGEE
Time Complexity: O(n*m)
Auxiliary Space: O(n*m)
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