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. **b**ab

aa: Number of insertions required is 0. aa

abcd: Number of insertions required is 3. **dcb**abcd

abcda: Number of insertions required is 2. a**dc**bcda

abcde: Number of insertions required is 4. **edcb**abcde

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 :

- if l > h (we crossed pointers) return INT_MAX
- if l == h return 0;

(only one character, which is already palindrome, 0 insertion is required to make it palindrome) - 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 “**b**ab” or “ab**a**“

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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