# Minimum characters to be added at front to make string palindrome

• Difficulty Level : Hard
• Last Updated : 20 Sep, 2022

Given string str we need to tell minimum characters to be added in front of the string to make string palindrome.

Examples:

```Input  : str = "ABC"
Output : 2
We can make above string palindrome as "CBABC"
by adding 'B' and 'C' at front.

Input  : str = "AACECAAAA";
Output : 2
We can make above string palindrome as AAAACECAAAA
by adding two A's at front of string.
```

Naive approach: Start checking the string each time if it is a palindrome and if not, then delete the last character and check again. When the string gets reduced to either a palindrome or an empty string then the number of characters deleted from the end till now will be the answer as those characters could have been inserted at the beginning of the original string in the order which will make the string a palindrome.

Below is the implementation of the above approach:

## C++

 `// C++ program for getting minimum character to be``// added at front to make string palindrome``#include``using` `namespace` `std;` `// function for checking string is palindrome or not``bool` `ispalindrome(string s)``{``    ``int` `l = s.length();``    ``int` `j;``    ` `    ``for``(``int` `i = 0, j = l - 1; i <= j; i++, j--)``    ``{``        ``if``(s[i] != s[j])``            ``return` `false``;``    ``}``    ``return` `true``;``}` `// Driver code``int` `main()``{``    ``string s = ``"BABABAA"``;``    ``int` `cnt = 0;``    ``int` `flag = 0;``    ` `    ``while``(s.length()>0)``    ``{``        ``// if string becomes palindrome then break``        ``if``(ispalindrome(s))``        ``{``            ``flag = 1;``             ``break``;``        ``}``        ``else``        ``{``        ``cnt++;``        ` `        ``// erase the last element of the string``        ``s.erase(s.begin() + s.length() - 1);``        ``}``    ``}``    ` `    ``// print the number of insertion at front``    ``if``(flag)``        ``cout << cnt;``}`

## Java

 `// Java program for getting minimum character to be``// added at front to make string palindrome` `class` `GFG {` `// function for checking string is palindrome or not``    ``static` `boolean` `ispalindrome(String s) {``        ``int` `l = s.length();` `        ``for` `(``int` `i = ``0``, j = l - ``1``; i <= j; i++, j--) {``            ``if` `(s.charAt(i) != s.charAt(j)) {``                ``return` `false``;``            ``}``        ``}``        ``return` `true``;``    ``}` `// Driver code``    ``public` `static` `void` `main(String[] args) {``        ``String s = ``"BABABAA"``;``        ``int` `cnt = ``0``;``        ``int` `flag = ``0``;` `        ``while` `(s.length() > ``0``) {``            ``// if string becomes palindrome then break``            ``if` `(ispalindrome(s)) {``                ``flag = ``1``;``                ``break``;``            ``} ``else` `{``                ``cnt++;` `                ``// erase the last element of the string``                ``s = s.substring(``0``, s.length() - ``1``);``                ``//s.erase(s.begin() + s.length() - 1);``            ``}``        ``}` `        ``// print the number of insertion at front``        ``if` `(flag == ``1``) {``            ``System.out.println(cnt);``        ``}``    ``}``}` `// This code is contributed by 29AjayKumar`

## Python 3

 `# Python 3 program for getting minimum character``# to be added at front to make string palindrome` `# function for checking string is``# palindrome or not``def` `ispalindrome(s):` `    ``l ``=` `len``(s)``    ` `    ``i ``=` `0``    ``j ``=` `l ``-` `1``    ``while` `i <``=` `j:``    ` `        ``if``(s[i] !``=` `s[j]):``            ``return` `False``        ``i ``+``=` `1``        ``j ``-``=` `1``    ` `    ``return` `True` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``s ``=` `"BABABAA"``    ``cnt ``=` `0``    ``flag ``=` `0``    ` `    ``while``(``len``(s) > ``0``):``    ` `        ``# if string becomes palindrome then break``        ``if``(ispalindrome(s)):``            ``flag ``=` `1``            ``break``        ` `        ``else``:``            ``cnt ``+``=` `1``        ` `            ``# erase the last element of the string``            ``s ``=` `s[:``-``1``]``    ` `    ``# print the number of insertion at front``    ``if``(flag):``        ``print``(cnt)` `# This code is contributed by ita_c`

## C#

 `// C# program for getting minimum character to be``// added at front to make string palindrome` `using` `System;``public` `class` `GFG {` `// function for checking string is palindrome or not``    ``static` `bool` `ispalindrome(String s) {``        ``int` `l = s.Length;` `        ``for` `(``int` `i = 0, j = l - 1; i <= j; i++, j--) {``            ``if` `(s[i] != s[j]) {``                ``return` `false``;``            ``}``        ``}``        ``return` `true``;``    ``}` `// Driver code``    ``public` `static` `void` `Main() {``        ``String s = ``"BABABAA"``;``        ``int` `cnt = 0;``        ``int` `flag = 0;` `        ``while` `(s.Length > 0) {``            ``// if string becomes palindrome then break``            ``if` `(ispalindrome(s)) {``                ``flag = 1;``                ``break``;``            ``} ``else` `{``                ``cnt++;` `                ``// erase the last element of the string``                ``s = s.Substring(0, s.Length - 1);``                ``//s.erase(s.begin() + s.length() - 1);``            ``}``        ``}` `        ``// print the number of insertion at front``        ``if` `(flag == 1) {``            ``Console.WriteLine(cnt);``        ``}``    ``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``

Output

`2`

Time complexity: O(n2)
Auxiliary Space: O(1)
Thank you Sanny Kumar for suggesting this approach.

Efficient approach: We can solve this problem efficiently in O(n) time using lps array of KMP algorithm
First, we concat string by concatenating the given string, a special character and reverse of given string then we will get lps array for this concatenated string, recall that each index of lps array represents the longest proper prefix which is also a suffix. We can use this lps array to solve the problem.

```For string = AACECAAAA
Concatenated String = AACECAAAA\$AAAACECAA
LPS array will be {0, 1, 0, 0, 0, 1, 2, 2, 2,
0, 1, 2, 2, 2, 3, 4, 5, 6, 7}```

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. Finally minimum number of characters needed to make the string a palindrome is the length of the input string minus the last entry of our lps array.

Below is the implementation of the above approach:

## C++

 `// C++ program for getting minimum character to be``// added at front to make string palindrome``#include ``using` `namespace` `std;` `// returns vector lps for given string str``vector<``int``> computeLPSArray(string str)``{``    ``int` `M = str.length();``    ``vector<``int``> lps(M);` `    ``int` `len = 0;``    ``lps[0] = 0; ``// lps[0] is always 0` `    ``// the loop calculates lps[i] for i = 1 to M-1``    ``int` `i = 1;``    ``while` `(i < M)``    ``{``        ``if` `(str[i] == str[len])``        ``{``            ``len++;``            ``lps[i] = len;``            ``i++;``        ``}``        ``else` `// (str[i] != str[len])``        ``{``            ``// This is tricky. Consider the example.``            ``// AAACAAAA and i = 7. The idea is similar``            ``// to search step.``            ``if` `(len != 0)``            ``{``                ``len = lps[len-1];` `                ``// Also, note that we do not increment``                ``// i here``            ``}``            ``else` `// if (len == 0)``            ``{``                ``lps[i] = 0;``                ``i++;``            ``}``        ``}``    ``}``    ``return` `lps;``}` `// Method returns minimum character to be added at``// front to make string palindrome``int` `getMinCharToAddedToMakeStringPalin(string str)``{``    ``string revStr = str;``    ``reverse(revStr.begin(), revStr.end());` `    ``// Get concatenation of string, special character``    ``// and reverse string``    ``string concat = str + ``"\$"` `+ revStr;` `    ``//  Get LPS array of this concatenated string``    ``vector<``int``> lps = computeLPSArray(concat);` `    ``// By subtracting last entry of lps vector from``    ``// string length, we will get our result``    ``return` `(str.length() - lps.back());``}` `// Driver program to test above functions``int` `main()``{``    ``string str = ``"AACECAAAA"``;``    ``cout << getMinCharToAddedToMakeStringPalin(str);``    ``return` `0;``}`

## Java

 `// Java program for getting minimum character to be``// added at front to make string palindrome``import` `java.util.*;``class` `GFG``{` `// returns vector lps for given string str``public` `static` `int``[] computeLPSArray(String str)``{``    ``int` `n = str.length();``    ``int` `lps[] = ``new` `int``[n];``    ``int` `i = ``1``, len = ``0``;``    ``lps[``0``] = ``0``; ``// lps[0] is always 0``    ` `    ``while` `(i < n)``    ``{``        ``if` `(str.charAt(i) == str.charAt(len))``        ``{``            ``len++;``            ``lps[i] = len;``            ``i++;``        ``}``        ``else``        ``{``            ``// This is tricky. Consider the example.``            ``// AAACAAAA and i = 7. The idea is similar``            ``// to search step.``            ``if` `(len != ``0``)``            ``{``                ``len = lps[len - ``1``];``                ` `                ``// Also, note that we do not increment``                ``// i here``            ``}``            ``else``            ``{``                ``lps[i] = ``0``;``                ``i++;``            ``}``        ``}``    ``}``    ``return` `lps;``}` `// Method returns minimum character to be added at``// front to make string palindrome``static` `int` `getMinCharToAddedToMakeStringPalin(String str)``{``    ``StringBuilder s = ``new` `StringBuilder();``    ``s.append(str);``    ` `    ``// Get concatenation of string, special character``    ``// and reverse string``    ``String rev = s.reverse().toString();``    ``s.reverse().append(``"\$"``).append(rev);``    ` `    ``// Get LPS array of this concatenated string``    ``int` `lps[] = computeLPSArray(s.toString());``    ``return` `str.length() - lps[s.length() - ``1``];``}` `// Driver Code``public` `static` `void` `main(String args[])``{``    ``String str = ``"AACECAAAA"``;``    ``System.out.println(getMinCharToAddedToMakeStringPalin(str));``}``}` `// This code is contributed by Sparsh Singhal`

## Python3

 `# Python3 program for getting minimum``# character to be added at the front``# to make string palindrome` `# Returns vector lps for given string str``def` `computeLPSArray(string):` `    ``M ``=` `len``(string)``    ``lps ``=` `[``None``] ``*` `M` `    ``length ``=` `0``    ``lps[``0``] ``=` `0` `# lps[0] is always 0` `    ``# the loop calculates lps[i]``    ``# for i = 1 to M-1``    ``i ``=` `1``    ``while` `i < M:``    ` `        ``if` `string[i] ``=``=` `string[length]:``        ` `            ``length ``+``=` `1``            ``lps[i] ``=` `length``            ``i ``+``=` `1``        ` `        ``else``: ``# (str[i] != str[len])``        ` `            ``# This is tricky. Consider the example.``            ``# AAACAAAA and i = 7. The idea is``            ``# similar to search step.``            ``if` `length !``=` `0``:``            ` `                ``length ``=` `lps[length ``-` `1``]` `                ``# Also, note that we do not``                ``# increment i here``            ` `            ``else``: ``# if (len == 0)``            ` `                ``lps[i] ``=` `0``                ``i ``+``=` `1` `    ``return` `lps` `# Method returns minimum character``# to be added at front to make``# string palindrome``def` `getMinCharToAddedToMakeStringPalin(string):` `    ``revStr ``=` `string[::``-``1``]` `    ``# Get concatenation of string,``    ``# special character and reverse string``    ``concat ``=` `string ``+` `"\$"` `+` `revStr` `    ``# Get LPS array of this``    ``# concatenated string``    ``lps ``=` `computeLPSArray(concat)` `    ``# By subtracting last entry of lps``    ``# vector from string length, we``    ``# will get our result``    ``return` `len``(string) ``-` `lps[``-``1``]` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``string ``=` `"AACECAAAA"``    ``print``(getMinCharToAddedToMakeStringPalin(string))``    ` `# This code is contributed by Rituraj Jain`

## C#

 `// C# program for getting minimum character to be ``// added at front to make string palindrome ``using` `System;``using` `System.Text;` `public` `class` `GFG{` `  ``// returns vector lps for given string str ``  ``public` `static` `int``[] computeLPSArray(``string` `str)``  ``{``    ``int` `n = str.Length;``    ``int``[] lps = ``new` `int``[n];``    ``int` `i = 1, len = 0;``    ``lps[0] = 0; ``// lps[0] is always 0 ` `    ``while` `(i < n) ``    ``{``      ``if` `(str[i] == str[len])``      ``{``        ``len++;``        ``lps[i] = len;``        ``i++;``      ``}``      ``else``      ``{``        ``// This is tricky. Consider the example. ``        ``// AAACAAAA and i = 7. The idea is similar ``        ``// to search step.``        ``if` `(len != 0)``        ``{``          ``len = lps[len - 1];` `          ``// Also, note that we do not increment ``          ``// i here ``        ``}``        ``else``        ``{``          ``lps[i] = 0;``          ``i++;``        ``}``      ``}``    ``}``    ``return` `lps;``  ``}` `  ``// Method returns minimum character to be added at ``  ``// front to make string palindrome ``  ``static` `int` `getMinCharToAddedToMakeStringPalin(``string` `str) ``  ``{ ``    ``char``[] s = str.ToCharArray();` `    ``// Get concatenation of string, special character ``    ``// and reverse string ``    ``Array.Reverse( s );``    ``string` `rev = ``new` `string``(s);` `    ``string` `concat=  str + ``"\$"` `+ rev;` `    ``// Get LPS array of this concatenated string``    ``int``[] lps = computeLPSArray(concat);``    ``return` `str.Length - lps[concat.Length - 1];``  ``}``  ` `  ``// Driver Code``  ``static` `public` `void` `Main (){``    ``string` `str = ``"AACECAAAA"``; ``    ``Console.WriteLine(getMinCharToAddedToMakeStringPalin(str));``  ``}``}` `// This code is contributed by avanitrachhadiya2155`

## Javascript

 ``

Output

`2`

Time Complexity: O(n)
Auxiliary Space: O(n)

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