Minimum number of deletions to make a string palindrome | Set 2

Given a string A, compute the minimum number of characters you need to delete to make resulting string a palindrome.

Examples:

Input : baca
Output : 1

Input : geek
Output : 2



We have discussed one approach in below post.
Minimum number of deletions to make a string palindrome

Below approach will use modified Levenshtein distance. We consider modified Levensthein (considering only deletions) both original string and its reverse.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP program to find minimum deletions to make
// palindrome.
#include <bits/stdc++.h>
using namespace std;
  
  
int getLevenstein(string const& input)
{
    // Find reverse of input string
    string revInput(input.rbegin(), input.rend());
  
    // Create a DP table for storing edit distance
    // of string and reverse.
    int n = input.size();
    vector<vector<int> > dp(n + 1, vector<int>(n + 1, -1));
    for (int i = 0; i <= n; ++i) {
        dp[0][i] = i;
        dp[i][0] = i;
    }
  
    // Find edit distance between input and revInput 
    // considering only delete operation.
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            if (input[i - 1] == revInput[j - 1]) 
                dp[i][j] = dp[i - 1][j - 1];            
            else 
                dp[i][j] = 1 + min({ dp[i - 1][j], dp[i][j - 1] });            
        }
    }
  
    /*Go from bottom left to top right and find the minimum*/
    int res = numeric_limits<int>::max();
    for (int i = n, j = 0; i >= 0; --i, ++j) {
        res = min(res, dp[i][j]);
        if (i < n) 
            res = min(res, dp[i + 1][j]);        
        if (i > 0) 
            res = min(res, dp[i - 1][j]);        
    }
    return res;
}
  
// Driver code
int main()
{
    string input("myfirstgeekarticle");
    cout << getLevenstein(input);
    return 0;
}

chevron_right






                        filter_none









Python3














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















# Python3 program to find minimum deletions  


# to make palindrome. 


INT_MAX = 99999999999


  


def getLevenstein(inpt): 


  


    # Find reverse of input string 


    revInput = inpt[::-1] 


  


    # Create a DP table for storing   


    # edit distance of string and reverse. 


    n = len(inpt) 


    dp = [[-1 for _ in range(n + 1)]  


              for __ in range(n + 1)] 


    for i in range(n + 1): 


        dp[0][i] = i 


        dp[i][0] = i 


  


    # Find edit distance between  


    # input and revInput considering 


    # only delete operation. 


    for i in range(1, n + 1): 


        for j in range(1, n + 1): 


            if inpt[i - 1] == revInput[j - 1]: 


                dp[i][j] = dp[i - 1][j - 1] 


            else: 


                dp[i][j] = 1 + min(dp[i - 1][j],  


                                   dp[i][j - 1]) 


  


    # Go from bottom left to top right 


    # and find the minimum 


    res = INT_MAX 


    i, j = n, 0


    while i >= 0: 


        res = min(res, dp[i][j]) 


        if i < n: 


            res = min(res, dp[i + 1][j]) 


        if i > 0: 


            res = min(res, dp[i - 1][j]) 


        i -= 1


        j += 1


    return res 


  


# Driver Code 


if __name__ == "__main__": 


    inpt = "myfirstgeekarticle"


    print(getLevenstein(inpt)) 


  


# This code is contributed 


# by vibhu4agarwal 



















                        chevron_right







                        filter_none










Output:


12



Time complexity: O( $n^{2}$ )

Space complexity: O( $n^{2}$ )

where $n$ is length of string


Why is it working?

To understand it we need to start from the very beginning of how we create dp[][], for example for word “geek”, it initially looks like this:


   \begin{array}{c c c c c c} & null & g & e & e & k\\ null & 0 & 1 & 2 & 3 & 4\\ k & 1 & -1 & -1 & -1 & -1\\ e & 2 & -1 & -1 & -1 & -1\\ e & 3 & -1 & -1 & -1 & -1\\ g & 4 & -1 & -1 & -1 & -1\\ \end{array} 


Both 1st row and 1st column are filled with number 1..4 as this is the number of modifications needed to create empty string, i.e:

[0][1] == 1, to create empty string from letter ‘g’ remove this one letter

[0][2] == 2, to create empty string from letters “ge”, remove both those letters, etc.

The same story for first column:

[1][0] == 1, to create empty string from letter ‘k’ remove this one letter

[2][0] == 2, to create empty string from letters “ke”, remove both those letters, etc.


Now, we are using dynamic programming approach to get the minimum number of modifications to get every other substring to become second substring, and at the end out dp[][] looks like this:


   \begin{array}{c c c c c c} & null & g & e & e & k\\ null & 0 & 1 & 2 & 3 & 4\\ k & 1 & 2 & 3 & 4 & 3\\ e & 2 & 3 & 2 & 3 & 4\\ e & 3 & 4 & 3 & 2 & 3\\ g & 4 & 3 & 4 & 3 & 4\\ \end{array} 


So for example minimum number of modifications to get substring ‘gee’ from ‘kee’ is 2. So far so good but this algorithm is doing two things, it is both inserting and deleting characters, and we are only interested in number of removals. So let’s one more time take a look at our resulting array, for example at entry [4][1], this entry states:

[4][1] – to make string ‘g’ from string “keeg” we need to perform 3 modifications(which is just delete chars “kee”)

[3][2] – to make string “ge” from “kee” we need to perform 3 modifications also by removing from first string ‘g’ and from second ‘ke’

So basically every time we will be moving diagonally up, from left corner we will get number of removals to get the same substring backwards. Thing to notice here is that it is like having on string two pointers, one moving from beginning and other from end. Very important spot is that strings do not necessary has to have even number of characters, so this is the reason we also has to check upper and lower values in dp[][].



          
          
          
          

            

    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts:
Mateusz_Wojtczak
Check out this Author's contributed articles.
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.
Please Improve this article if you find anything incorrect by clicking on  the "Improve Article" button below.


Improved By :  vibhu4agarwal