Imprimir subsecuencia palindrómica más larga

Dada una secuencia, imprima una subsecuencia palindrómica más larga de ella. 

Ejemplos: 

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, ..

Hemos discutido una solución en la publicación a continuación para encontrar la longitud de la subsecuencia palindrómica más larga. 
Programación Dinámica | Conjunto 12 (Subsecuencia palindrómica más larga)
En esta publicación se analiza una solución para imprimir la subsecuencia palindrómica más larga. 

Este problema está cerca del problema de la subsecuencia común más larga (LCS) . De hecho, podemos usar LCS como una subrutina para resolver este problema. La siguiente es la solución de dos pasos que utiliza LCS. 

1. Invierta la secuencia dada y almacene el reverso en otra array, digamos rev[0..n-1] 
2. LCS de la secuencia dada y rev[] será la secuencia palindrómica más larga. 
3. Una vez que hayamos encontrado LCS, podemos imprimir el LCS .

A continuación se muestra la implementación del enfoque anterior:

/* 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 str
string 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 program to print longest palindromic
//subsequence
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 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 Code
if __name__ == "__main__" :
 
    string = "GEEKSFORGEEKS";
    print(longestPalSubseq(string))
 
# This code is contributed by Ryuga

// 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

<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>

Producción: 
 

EEGEE