Java Program To Find Minimum Insertions To Form A Palindrome | DP-28

Given string str, the task is to find the minimum number of characters to be inserted to convert it to a palindrome.

Before we go further, let us understand with a few examples: 

• ab: Number of insertions required is 1 i.e. bab
• aa: Number of insertions required is 0 i.e. aa
• abcd: Number of insertions required is 3 i.e. dcbabcd
• abcda: Number of insertions required is 2 i.e. adcbcda which is the same as the number of insertions in the substring bcd(Why?).
• abcde: Number of insertions required is 4 i.e. edcbabcde

Let the input string be str[l……h]. The problem can be broken down into three parts:  

1. Find the minimum number of insertions in the substring str[l+1,…….h].
2. Find the minimum number of insertions in the substring str[l…….h-1].
3. Find the minimum number of insertions in the substring str[l+1……h-1].

Recursive Approach: The minimum number of insertions in the string str[l…..h] can be given as:  

• minInsertions(str[l+1…..h-1]) if str[l] is equal to str[h]
• min(minInsertions(str[l…..h-1]), minInsertions(str[l+1…..h])) + 1 otherwise

Below is the implementation of the above approach:  

Java

// A Naive recursive Java program to find
// minimum number insertions needed to make
// a string palindrome
class GFG
{
    // Recursive function to find minimum
    // number of insertions
    static int findMinInsertions(char str[],
                                 int l, int h)
    {
        // Base Cases
        if (l > h)
            return Integer.MAX_VALUE;

        if (l == h)
            return 0;

        if (l == h - 1)
            return (str[l] == str[h]) ? 0 : 1;

        // Check if the first and last characters
        // are same. On the basis of the  comparison
        // result, decide which subproblem(s) to call
        return (str[l] == str[h])?
                findMinInsertions(str, l + 1, h - 1):
               (Integer.min(findMinInsertions(str, l, h - 1),
                findMinInsertions(str, l + 1, h)) + 1);
    }

    // Driver code
    public static void main(String args[])
    {
        String str= "geeks";
        System.out.println(
        findMinInsertions(str.toCharArray(),
                          0, str.length()-1));
    }
}
// This code is contributed by Sumit Ghosh

Output:

3

Time Complexity: O(2^n), where n is the length of the input string. This is because for each recursive call, there are two possibilities: either we insert a character at the beginning of the string or at the end of the string. Therefore, the total number of recursive calls made is equal to the number of binary strings of length n, which is 2^n.

Dynamic Programming based Solution
If we observe the above approach carefully, we can find that it exhibits overlapping subproblems. 
Suppose we want to find the minimum number of insertions in string “abcde”:  

                      abcde
            /       |      
           /        |        
           bcde         abcd       bcd  <- case 3 is discarded as str[l] != str[h]
       /   |          /   |   
      /    |         /    |    
     cde   bcd  cd   bcd abc bc
   / |   / |  /| / | 
de cd d cd bc c………………….

The substrings in bold show that the recursion is to be terminated and the recursion tree cannot originate from there. Substring in the same color indicates overlapping subproblems.

How to re-use solutions of subproblems? The memorization technique is used to avoid similar subproblem recalls. We can create a table to store the results of subproblems so that they can be used directly if the same subproblem is encountered again.
The below table represents the stored values for the string abcde. 

a b c d e
----------
0 1 2 3 4
0 0 1 2 3 
0 0 0 1 2 
0 0 0 0 1 
0 0 0 0 0

How to fill the table?
The table should be filled in a diagonal fashion. For the string abcde, 0….4, the following should be ordered in which the table is filled:

Gap = 1: (0, 1) (1, 2) (2, 3) (3, 4)

Gap = 2: (0, 2) (1, 3) (2, 4)

Gap = 3: (0, 3) (1, 4)

Gap = 4: (0, 4)

Below is the implementation of the above approach: 

Java

// A Java solution for Dynamic Programming
// based program to find minimum number
// insertions needed to make a string
// palindrome
import java.util.Arrays;

class GFG
{
    // A DP function to find minimum number
    // of insertions
    static int findMinInsertionsDP(char str[],
                                   int n)
    {
        // Create a table of size n*n. table[i][j]
        // will store minimum number of insertions
        // needed to convert str[i..j] to a palindrome.
        int table[][] = new int[n][n];
        int l, h, gap;

        // Fill the table
        for (gap = 1; gap < n; ++gap)
        for (l = 0, h = gap; h < n; ++l, ++h)
            table[l][h] = (str[l] == str[h])?
                           table[l+1][h-1] :
                          (Integer.min(table[l][h-1],
                                 table[l+1][h]) + 1);

        // Return minimum number of insertions
        // for str[0..n-1]
        return table[0][n-1];
    }

    // Driver code
    public static void main(String args[])
    {
        String str = "geeks";
        System.out.println(
        findMinInsertionsDP(str.toCharArray(),
                            str.length()));
    }
}
// This code is contributed by Sumit Ghosh

Output:

3

Time complexity: O(N^2) 
Auxiliary Space: O(N^2)

Another Dynamic Programming Solution (Variation of Longest Common Subsequence Problem)
The problem of finding minimum insertions can also be solved using Longest Common Subsequence (LCS) Problem. If we find out the LCS of string and its reverse, we know how many maximum characters can form a palindrome. We need to insert the remaining characters. Following are the steps. 

1. Find the length of LCS of the input string and its reverse. Let the length be ‘l’.
2. The minimum number of insertions needed is the length of the input string minus ‘l’.

Below is the implementation of the above approach:  

Java

// An LCS based Java program to find minimum
// number insertions needed to make a string
// palindrome
class GFG
{
    /* Returns length of LCS for X[0..m-1],
       Y[0..n-1]. See http://goo.gl/bHQVP for
       details of this function */
    static int lcs(String X, String Y,
                   int m, int n)
    {
        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.charAt(i-1) ==
                         Y.charAt(j-1))
                    L[i][j] = L[i-1][j-1] + 1;

                else
                    L[i][j] = Integer.max(L[i-1][j],
                                          L[i][j-1]);
            }
        }

        /* L[m][n] contains length of LCS for
           X[0..n-1] and Y[0..m-1] */
        return L[m][n];
    }

    // LCS based function to find minimum number
    // of insertions
    static int findMinInsertionsLCS(String str,
                                    int n)
    {
        // Using StringBuffer to reverse a String
        StringBuffer sb = new StringBuffer(str);
        sb.reverse();
        String revString = sb.toString();

        // The output is length of string minus
        // length of lcs of str and it reverse
        return (n - lcs(str, revString , n, n));
    }

    // Driver program to test above functions
    public static void main(String args[])
    {
        String str = "geeks";
        System.out.println(
        findMinInsertionsLCS(str, str.length()));
    }
}
// This code is contributed by Sumit Ghosh

Output:

3

Time complexity: O(N^2) 
Auxiliary Space: O(N^2) 

Please refer complete article on Minimum insertions to form a palindrome | DP-28 for more details!