Find if string is K-Palindrome or not | Set 2

Given a string, find out if the string is K-Palindrome or not. A K-palindrome string transforms into a palindrome on removing at most k characters from it.
Examples: 

Input : String - abcdecba, k = 1
Output : Yes
String can become palindrome by removing
1 character i.e. either d or e

Input  : String - abcdeca, K = 2
Output : Yes
Can become palindrome by removing
2 characters b and e (or b and d).

Input : String - acdcb, K = 1
Output : No
String can not become palindrome by
removing only one character.

We have discussed a DP solution in previous post where we saw that the problem is basically a variation of Edit Distance problem. In this post, another interesting DP solution is discussed.
The idea is to find the longest palindromic subsequence of the given string. If the difference between longest palindromic subsequence and the original string is less than equal to k, then the string is k-palindrome else it is not k-palindrome.
For example, longest palindromic subsequence of string abcdeca is acdca(or aceca). The characters which do not contribute to longest palindromic subsequence of the string should be removed in order to make the string palindrome. So on removing b and d (or e) from abcdeca, string will transform into a palindrome.
Longest palindromic subsequence of a string can easily be found using LCS. Following is the two step solution for finding longest palindromic subsequence that uses LCS. 

1. Reverse the given sequence and store the reverse in another array say rev[0..n-1]
2. LCS of the given sequence and rev[] will be the longest palindromic sequence.

Below is the implementation of above idea –

CPP

// C++ program to find if given string is K-Palindrome
// or not
#include <bits/stdc++.h>
using namespace std;

/* Returns length of LCS for X[0..m-1], Y[0..n-1] */
int lcs( string X, string Y, int m, int n )
{
    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]);
        }
    }
    // L[m][n] contains length of LCS for X and Y
    return L[m][n];
}

// find if given string is K-Palindrome or not
bool isKPal(string str, int k)
{
    int n = str.length();

    // Find reverse of string
    string revStr = str;
    reverse(revStr.begin(), revStr.end());

    // find longest palindromic subsequence of
    // given string
    int lps = lcs(str, revStr, n, n);

    // If the difference between longest palindromic
    // subsequence and the original string is less
    // than equal to k, then the string is k-palindrome
    return (n - lps <= k);
}

// Driver program
int main()
{
    string str = "abcdeca";
    int k = 2;
    isKPal(str, k) ? cout << "Yes" : cout << "No";

    return 0;
}

Java

// Java program to find if given
// String is K-Palindrome or not
import java.util.*;
import java.io.*;

class GFG
{

    /* Returns length of LCS for
    X[0..m-1], Y[0..n-1] */
    static int lcs(String X, String Y,
                        int m, int n)
    {
        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.charAt(i - 1) == Y.charAt(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]);
                }
            }
        }
        // L[m][n] contains length
        // of LCS for X and Y
        return L[m][n];
    }

    // find if given String is
    // K-Palindrome or not
    static boolean isKPal(String str, int k)
    {
        int n = str.length();

        // Find reverse of String
        StringBuilder revStr = new StringBuilder(str);
        revStr = revStr.reverse();

        // find longest palindromic
        // subsequence of given String
        int lps = lcs(str, revStr.toString(), n, n);

        // If the difference between longest
        // palindromic subsequence and the
        // original String is less than equal
        // to k, then the String is k-palindrome
        return (n - lps <= k);
    }

    // Driver code
    public static void main(String[] args)
    {
        String str = "abcdeca";
        int k = 2;
        if (isKPal(str, k))
        {
            System.out.println("Yes");
        }
        else
            System.out.println("No");
    }
}

// This code is contributed by Rajput-JI

Python3

# Python program to find
# if given string is K-Palindrome
# or not

# Returns length of LCS
# for X[0..m-1], Y[0..n-1]
def lcs(X, Y, m, n ):

    L = [[0]*(n+1) for _ in range(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(m+1):
        for j in range(n+1):
            if not i or not j:
                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])

    # L[m][n] contains length
        # of LCS for X and Y
    return L[m][n]

# find if given string is
# K-Palindrome or not
def isKPal(string, k):

    n = len(string)

    # Find reverse of string
    revStr = string[::-1]

    # find longest palindromic
        # subsequence of
    # given string
    lps = lcs(string, revStr, n, n)

    # If the difference between
        # longest palindromic
    # subsequence and the original
        # string is less
    # than equal to k, then
        # the string is k-palindrome
    return (n - lps <= k)

# Driver program
string = "abcdeca"
k = 2

print("Yes" if isKPal(string, k) else "No")

# This code is contributed
# by Ansu Kumari.

C#

// C# program to find if given
// String is K-Palindrome or not
using System;

class GFG
{

    /* Returns length of LCS for
    X[0..m-1], Y[0..n-1] */
    static int lcs(String X, String Y,
                        int m, int n)
    {
        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]);
                }
            }
        }

        // L[m,n] contains length
        // of LCS for X and Y
        return L[m, n];
    }

    // find if given String is
    // K-Palindrome or not
    static bool isKPal(String str, int k)
    {
        int n = str.Length;

        // Find reverse of String
        str = reverse(str);

        // find longest palindromic
        // subsequence of given String
        int lps = lcs(str, str, n, n);

        // If the difference between longest
        // palindromic subsequence and the
        // original String is less than equal
        // to k, then the String is k-palindrome
        return (n - lps <= k);
    }
    static String reverse(String input)
    {
        char[] temparray = input.ToCharArray();
        int left, right = 0;
        right = temparray.Length - 1;

        for (left = 0; left < right; left++, right--)
        {

            // Swap values of left and right
            char temp = temparray[left];
            temparray[left] = temparray[right];
            temparray[right] = temp;
        }
        return String.Join("",temparray);
    }

    // Driver code
    public static void Main(String[] args)
    {
        String str = "abcdeca";
        int k = 2;
        if (isKPal(str, k))
        {
            Console.WriteLine("Yes");
        }
        else
            Console.WriteLine("No");
    }
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// JavaScript program to find
// if given string is K-Palindrome
// or not

// Returns length of LCS
// for X[0..m-1], Y[0..n-1]
function lcs(X, Y, m, n ){

    let L = new Array(m+1);
    for(let i=0;i<m+1;i++){
        L[i] = new Array(n+1).fill(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 + 1; i++)
    {
        for(let j = 0; j < n + 1; j++)
        {
            if(!i || !j)
                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])
        }
    }

    // L[m][n] contains length
        // of LCS for X and Y
    return L[m][n]
}

// find if given string is
// K-Palindrome or not
function isKPal(string, k){

    let n = string.length

    // Find reverse of string
    let revStr = string.split("").reverse().join("")

    // find longest palindromic
        // subsequence of
    // given string
    let lps = lcs(string, revStr, n, n)

    // If the difference between
        // longest palindromic
    // subsequence and the original
        // string is less
    // than equal to k, then
        // the string is k-palindrome
    return (n - lps <= k)
}

// Driver program
let string = "abcdeca"
let k = 2

document.write(isKPal(string, k)?"Yes" : "No")

// This code is contributed by shinjanpatra
</script>

Output

Yes

Time complexity of above solution is O(n²). 
Auxiliary space used by the program is O(n²). It can further be reduced to O(n) by using Space Optimized Solution of LCS.
Thanks to Ravi Teja Kaveti for suggesting above solution.