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# Minimum characters required to add to given string to make it palindrome

Last Updated on March 25, 2022 by Ria Pathak ### Concepts Used

Strings, LPS array KPS Algorithm

Hard

### Problem Statement (Simplified):

Find the minimum number of characters required to add to the given string to make it a palindrome.

See original problem statement here

#### Test Case

``````Input:
2
aab
abc

Output:
1
2

Explanation:
Case-1:
We can make string 'aab' a palindrome by adding 'b' in front of the string. Adding 'b' converts string to 'baab'. So the answer is 1.

Case-2:
We can make string 'abc' a palindrome by adding 'cb' in front of the string. Adding 'cb' converts string to 'cbabc'. So the answer is 2.``````

### Solving Approach :

LPS Array: A `LPS` array (Longest Prefix that is suffix) is an array which provide length and index of prefix of string which is also suffix of array. For example, if string is "`abcbdab`", `LPS` array for this string would be `[0,0,0,0,0,1,2]` suggesting the values at index `5` and `6` are `1ˢᵗ` and `2ⁿᵈ` character of string respectively.

1) We use `LPS` array from `KPS Algorithm` to solve this problem.
2) We concatenate the reverse of the string with any random non-alphabetic character in last to the current string.
3) `LPS` array defines the largest `prefix` which is also `suffix` in the string.
4) Here we are only interested in the last value of this `LPS` array because it shows us the largest suffix of the reversed string that matches the prefix of the original string i.e these many characters already satisfy the palindrome property.
5) Finally the minimum number of characters needed to make the string a palindrome is the length of the input string minus last entry of our `LPS` array.

### Example

• Lets assume the string be "`codec`", its reverse would be "`cedoc`", so we concatenate both strings using a random character in between creating a new string i.e. "codec@cedoc".
• LPS array for the above string will be `[0,0,0,0,1,0,1,0,0,0,1]` which means last character of above string is `1ˢᵗ`
character of array. By `LPS` we can see, first and last element of given string is matching. Hence we need to insert `lengthOfString``lastIndexOfLPSArray` elements in string to make it palindrome. Hence `5-1` i.e. `4` elements are required.

### Solutions

```#include<stdio.h>
#include<string.h>

int main(){

int test;
scanf("%d", &test);

while(test--){

char a;
scanf("%s",a);

int sizeOfString = strlen(a);

//Concatenating reverse of string to string
int n=sizeOfString;

a[n] = '`';
a[n+1] = '\0';

n = strlen(a);

for(int i=0; i<n-1; i++){
a[i+n] = a[n-i-2];
}

a[(2*n) + 1] = '\0';
int lps[strlen(a)];

//Creating lps array for the concatenated string
lps = 0;
int i = 1, len=0;
while (i < strlen(a))
{
if (a[i] == a[len])
{
len++;
lps[i] = len;
i++;
}
else
{
if (len != 0)
{
len = lps[len-1];
}
else
{
lps[i] = 0;
i++;
}
}
}

printf("%d\n",sizeOfString-lps[strlen(a)-1]);
}

}
```
```#include<bits/stdc++.h>
using namespace std;

int main(){

int test;
cin>>test;

while(test--){

string a;
cin>>a;

int sizeOfString = a.length();

//Concatenating reverse of string to string
string b = a;
reverse(b.begin(), b.end());
a += a +'@' + b;

int lps[a.length()];

//Creating lps array for the concatenated string
lps = 0;
int i = 1, len=0;
while (i < a.length())
{
if (a[i] == a[len])
{
len++;
lps[i] = len;
i++;
}
else
{
if (len != 0)
{
len = lps[len-1];
}
else
{
lps[i] = 0;
i++;
}
}
}

cout<<sizeOfString-lps[a.length()-1]<<endl;
}

}
```
```import java.util.*;
import java.io.*;

public class Main {
public static void main(String args[]) throws IOException {

Scanner sc = new Scanner(System.in);
int test = sc.nextInt();

while(test!=0){

String a = sc.next();

int sizeOfString = a.length();

//Concatenating reverse of string to string

StringBuilder sb = new StringBuilder(a);
sb.reverse();

a += '@';

a += sb.toString();

int lps[] = new int[a.length()];

//Creating lps array for the concatenated string
lps = 0;
int i = 1, len=0;
while (i < a.length())
{
if (a.charAt(i) == a.charAt(len))
{
len++;
lps[i] = len;
i++;
}
else
{
if (len != 0)
{
len = lps[len-1];
}
else
{
lps[i] = 0;
i++;
}
}
}

System.out.println(sizeOfString-lps[a.length()-1]);

test--;
}
}
}
```

[forminator_quiz id="873"]

This article tried to discuss Strings, LPS array KPS Algorithm. Hope this blog helps you understand and solve the problem. To practice more problems on Strings, LPS array KPS Algorithm you can check out MYCODE | Competitive Programming.