Minimum insertions to form a palindrome

Given a string, find the minimum number of characters to be inserted to form Palindrome string out of given string

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
abcde: Number of insertions required is 4. edcbabcde

Lets first think How to create recursive solution

Case 1 : String abcda : first and last characters are matching so they can create palindrom, so we need to think for rest of the string only ie. bcd.
Case 2 : String abce : first and last characters are not matching so we can think either we can insert character into “abc” to form Palindrom, or We can insert inbetween “bce” to form abce as palindrom.

So recursive formula created is

                    abcea
	       /        |      \
	      /         |        \
          bcea          abce      bce  <- case 3 is only working as str[l] == str[h]
       /   |   \      /   |   \
      /    |    \    /    |    \
    cea  bce    ce  abc   bce      bc  <- case 3 is discrded as str[l] != str[h]

So we can think of recursive solution as follows

return (str[l] == str[h])? findMinInsertions(str, l + 1, h - 1) :
                               (min(findMinInsertions(str, l, h - 1),  findMinInsertions(str, l + 1, h)) + 1);

Base Case For Recursive Solution :

1. if  l > h (we crossed pointers) return INT_MAX
2. if l == h  return 0;
   (only one character, which is already palindrome, 0 insertion is required to make it palindrome)
3. if l == h-1 and if str[l] == str[h]  return 0;
   (if two length string example “aa or ab” and both characters are same ie : “aa” , its already palindrome,
   so return 0)
   if l == h-1 and if str[l] != str[h] return 1;
   (if two length string is there example “ab” and both characters are different , we need 1 insertion to make this string a palindrome one ie “bab” or “aba“

How to derive Recursive solution and from there How to create Dynamic programming solution , we can learn from Video Below

Code :

#include <limits.h>
#include <iostream>
#include <string>
#include <string.h>

using namespace std;

int min(int a, int b){  return a < b ? a : b; }

int findMinInsertionsRec(char str[], int l, int h)
{
    if (l > h) return INT_MAX;
    if (l == h) return 0;
    if (l == h - 1) return (str[l] == str[h])? 0 : 1;
 
    return (str[l] == str[h])? findMinInsertionsRec(str, l + 1, h - 1):
                               (min(findMinInsertionsRec(str, l, h - 1),
                                   findMinInsertionsRec(str, l + 1, h)) + 1);
}
//Find Minimum Insertion to Form Palindrome Dynamic Programming solution top down Approach
int findMinInsertionsDP(char str[], int n)
{
    int table[n][n], l, h, gap;
    memset(table, 0, sizeof(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 table[0][n-1];
}
////Find Minimum Insertion to Form Palindrome Dynamic Programming solution bottom up Approach
int findMinInsertionsDP2(string& str)
{
	int n = str.size();
	if( n == 0 )
		return 0;
	int table[n][n];
	int i,j;
	for( i = n-1; i >= 0; i-- )
	{
		for( j = i+1; j < n; j++ )
		{
			if( str[i] == str[j] ) 
			{
				if( j-i > 1 )
					table[i][j] = table[i+1][j-1];
			}
			else
			{
				table[i][j] = 1;
				if( j-i > 1 ) 
					table[i][j] = min(table[i][j-1], table[i+1][j])+1;
			}
		}
	}
	return table[0][n-1];
}

int main()
{
    char str[] = "abcd";
    string str1="abcd";
    printf("%d\n", findMinInsertionsRec(str, 0, strlen(str)-1));
    cout << findMinInsertionsDP(str,strlen(str))<<endl;
    cout << findMinInsertionsDP2(str1)<<endl;
    
    return 0;
}

Solution 2 :

You can see in above examples

• AB   –   1 insertion required   Inverse of String : BA
• ABA  –  0 insertion required   Inverse of String : ABA
• ABC  –  2 insertion required    Inverse of String : CBA
• ABCAD  – 3 insertion required    Inverse of String : DACBA

You can see

• LCS of AB and BA is 1 : so strlen( AB ) – LCS( AB, BA ) = 1 is number of insertion required
• LCS of ABA and ABA is 3 : so strlen(ABA) – LCS(ABA,ABA) = 0  is number of insertion required
• LCS of ABC and CBA is 1 : so strlen(ABC) – LCS(ABC,CBA) = 2 is number of insertion required
• LCS of ABCAD and DACBA is 2 : so strlen(ABCAD) – LCS(ABCAD,DACBA) = 3 is number of insertion required

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