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

3

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 C++ implementation of above idea –

// 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;
}

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.

This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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