# Dynamic Programming | Set 28 (Minimum insertions to form a palindrome)

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

Before we go further, let us understand with few examples:
ab: Number of insertions required is 1. bab
aa: Number of insertions required is 0. aa
abcd: Number of insertions required is 3. dcbabcd
abcda: Number of insertions required is 2. adcbcda which is same as number of insertions in the substring bcd(Why?).
abcde: Number of insertions required is 4. edcbabcde

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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 Solution
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

## C

```// A Naive recursive program to find minimum
// number insertions needed to make a string
// palindrome
#include <stdio.h>
#include <limits.h>
#include <string.h>

// A utility function to find minimum of two numbers
int min(int a, int b)
{  return a < b ? a : b; }

// Recursive function to find minimum number of
// insertions
int findMinInsertions(char str[], int l, int h)
{
// Base Cases
if (l > h) return INT_MAX;
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 subrpoblem(s) to call
return (str[l] == str[h])?
findMinInsertions(str, l + 1, h - 1):
(min(findMinInsertions(str, l, h - 1),
findMinInsertions(str, l + 1, h)) + 1);
}

// Driver program to test above functions
int main()
{
char str[] = "geeks";
printf("%d", findMinInsertions(str, 0, strlen(str)-1));
return 0;
}
```

## 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 subrpoblem(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 program to test above functions
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`

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 to be terminated and the recursion tree cannot originate from there. Substring in the same color indicates overlapping subproblems.

How to reuse solutions of subproblems?
We can create a table to store results of subproblems so that they can be used directly if 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 diagonal fashion. For the string abcde, 0….4, the following should be order 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)```

## C

```// A Dynamic Programming based program to find
// minimum number insertions needed to make a
// string palindrome
#include <stdio.h>
#include <string.h>

// A utility function to find minimum of two integers
int min(int a, int b)
{   return a < b ? a : b;  }

// A DP function to find minimum number of insertions
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[n][n], l, h, gap;

// Initialize all table entries as 0
memset(table, 0, sizeof(table));

// 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] :
(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 program to test above function.
int main()
{
char str[] = "geeks";
printf("%d", findMinInsertionsDP(str, strlen(str)));
return 0;
}
```

## 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 insersions
static int findMinInsertionsDP(char str[], int n)
{
// Create a table of size n*n. table[i][j]
// will store minumum 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 program to test above function.
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 LCS of string and its reverse, we know how many maximum characters can form a palindrome. We need insert remaining characters. Following are the steps.
1) Find the length of LCS of input string and its reverse. Let the length be ‘l’.
2) The minimum number insertions needed is length of input string minus ‘l’.

## C

```// An LCS based program to find minimum number
// insertions needed to make a string palindrome
#include<stdio.h>
#include <string.h>

/* Utility function to get max of 2 integers */
int max(int a, int b)
{   return (a > b)? a : b; }

/* Returns length of LCS for X[0..m-1], Y[0..n-1].
See http://goo.gl/bHQVP for details of this
function */
int lcs( char *X, char *Y, int m, int n )
{
int L[n+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] = 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
int findMinInsertionsLCS(char str[], int n)
{
// Creata another string to store reverse of 'str'
char rev[n+1];
strcpy(rev, str);
strrev(rev);

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

// Driver program to test above functions
int main()
{
char str[] = "geeks";
printf("%d", findMinInsertionsLCS(str, strlen(str)));
return 0;
}
```

## 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[n+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 insersions
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 of this method is also O(n^2) and this method also requires O(n^2) extra space.

Related Article :
Minimum number of Appends needed to make a string palindrome