# Make Palindrome Problem – Dynamic Programming Solutions

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This is a C++ Program that Solves Make Palindrome Problem using Dynamic Programming technique.

Problem Description

You are given a string str. Find the minimum number of characters to be inserted to string str to convert it to a palindrome.

Problem Solution

This problem is a variation of the longest common subsequence problem. First, find the LCS of the given string and its reverse. Then, the required result is simply the length of given string minus the calculated LCS.

Expected Input and Output

Case-1:

```str= ABCDE
result = 4 (ABCDEDCBA)```
Program/Source Code

Here is source code of the C++ Program to Solve Make Palindrome Problem. The C++ program is successfully compiled and run on a Linux system. The program output is also shown below.

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1. `#include<iostream>`
2. `#include<string>`
3. `#include<algorithm>`
4. ` `
5. `using namespace std;`
6. ` `
7. `int longestCommonSubsequece(string str1, string str2)`
8. `{`
9. `    int len1=str1.length(), len2=str2.length();`
10. `    int i, j;`
11. ` `
12. `    //create a matrix of order (len1+1)*(len2+1) to tabulate values`
13. `    int LCS[len1+1][len2+1];`
14. ` `
15. `    //LCS[i][j]=Length of longest common subsequence of str1[0....(i-1)] and str2[0...(j-1)] `
16. ` `
17. `    //initializing`
18. `    for(i=0;i<=len1;i++)`
19. `        LCS[i]=0;    //empty str2`
20. ` `
21. `    for(j=0;j<=len2;j++)`
22. `        LCS[j]=0;   //empty str1`
23. ` `
24. `    //now, start filling the matrix row wise`
25. `    for(i=1;i<=len1;i++)`
26. `    {`
27. `        for(j=1;j<=len2;j++)`
28. `        {`
29. `            //if current character of both strings match`
30. `            if(str1[i-1]==str2[j-1])`
31. `            {`
32. `                LCS[i][j]=1+LCS[i-1][j-1];`
33. `            }`
34. ` `
35. `            //mismatch`
36. `            else`
37. `            {`
38. `                LCS[i][j]=max(LCS[i-1][j],LCS[i][j-1]);`
39. `            }`
40. `        }`
41. `    }`
42. ` `
43. `    //now, return the final value `
44. `    return LCS[len1][len2];`
45. ` `
46. `}`
47. ` `
48. `int main()`
49. `{`
50. `    string str1;`
51. ` `
52. `    cout<<"Enter the string - ";`
53. `    getline(cin,str1);`
54. ` `
55. `    string str2=str1;`
56. `    reverse(str2.begin(),str2.end());`
57. ` `
58. `    cout<<"Minimum number of characters to be inserted in the input string to make it a palindrome is "<<endl;`
59. `    cout<<str1.length()-longestCommonSubsequece(str1,str2);`
60. ` `
61. `    cout<<endl;`
62. `    return 0;`
63. `}`
Program Explanation

In the main function, we will take input for str1 and str2. We will pass these to the function longestCommonSubsequence as parameters. This function will calculate the result using bottom up DP and return the value which is displayed on the standard output.

Runtime Test Cases
```
Case-1:
\$ g++ make_palindrome.cpp
\$ ./a.out
Enter the string - abcde
Minimum number of characters to be inserted in the input string to make it a palindrome is
4```

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