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Remove a character from a string to make it a palindrome
• Difficulty Level : Medium
• Last Updated : 20 Apr, 2021

Given a string, we need to check whether it is possible to make this string a palindrome after removing exactly one character from this.
Examples:

```Input  : str = “abcba”
Output : Yes
we can remove character ‘c’ to make string palindrome

Input  : str = “abcbea”
Output : Yes
we can remove character ‘e’ to make string palindrome

Input : str = “abecbea”
It is not possible to make this string palindrome
just by removing one character ```

We can solve this problem by finding the position of mismatch. We start looping in the string by keeping two pointers at both the ends which traverse towards mid position after each iteration, this iteration will stop when we find a mismatch, as it is allowed to remove just one character we have two choices here,
At mismatch, either remove character pointed by left pointer or remove character pointed by right pointer.
We will check both the cases, remember as we have traversed equal number of steps from both sides, this mid string should also be a palindrome after removing one character, so we check two substrings, one by removing left character and one by removing right character and if one of them is palindrome then we can make complete string palindrome by removing corresponding character, and if both substrings are not palindrome then it is not possible to make complete string a palindrome under given constraint.

C/C++

``````
// C/C++ program to check whether it is possible to make
// string palindrome by removing one character
#include <bits/stdc++.h>
using namespace std;

// Utility method to check if substring from low to high is
// palindrome or not.
bool isPalindrome(string::iterator low, string::iterator high)
{
while (low < high)
{
if (*low != *high)
return false;
low++;
high--;
}
return true;
}

// This method returns -1 if it is not possible to make string
// a palindrome. It returns -2 if string is already a palindrome.
// Otherwise it returns index of character whose removal can
// make the whole string palindrome.
int possiblePalinByRemovingOneChar(string str)
{
//  Initialize low and right by both the ends of the string
int low = 0, high = str.length() - 1;

//  loop untill low and high cross each other
while (low < high)
{
// If both characters are equal then move both pointer
// towards end
if (str[low] == str[high])
{
low++;
high--;
}
else
{
/*  If removing str[low] makes the whole string palindrome.
We basically check if substring str[low+1..high] is
palindrome or not. */
if (isPalindrome(str.begin() + low + 1, str.begin() + high))
return low;

/*  If removing str[high] makes the whole string palindrome
We basically check if substring str[low+1..high] is
palindrome or not. */
if (isPalindrome(str.begin() + low, str.begin() + high - 1))
return high;

return -1;
}
}

//  We reach here when complete string will be palindrome
//  if complete string is palindrome then return mid character
return -2;
}

// Driver code to test above methods
int main()
{
string str = "abecbea";
int idx = possiblePalinByRemovingOneChar(str);
if (idx == -1)
cout << "Not Possible \n";
else if (idx == -2)
cout << "Possible without removing any character";
else
cout << "Possible by removing character"
<< " at index " << idx << "\n";
return 0;
}
``````

## Java

 `// Java program to check whether``// it is possible to make string``// palindrome by removing one character``import` `java.util.*;` `class` `GFG``{` `    ``// Utility method to check if``    ``// substring from low to high is``    ``// palindrome or not.``    ``static` `boolean` `isPalindrome(String str,``                       ``int` `low, ``int` `high)``    ``{``        ``while` `(low < high)``        ``{``            ``if` `(str.charAt(low) != str.charAt(high))``                ``return` `false``;``            ``low++;``            ``high--;``        ``}``        ``return` `true``;``    ``}` `    ``// This method returns -1 if it is``    ``// not possible to make string a palindrome.``    ``// It returns -2 if string is already``    ``// a palindrome. Otherwise it returns``    ``// index of character whose removal can``    ``// make the whole string palindrome.``    ``static` `int` `possiblePalinByRemovingOneChar(String str)``    ``{` `        ``// Initialize low and right``        ``// by both the ends of the string``        ``int` `low = ``0``, high = str.length() - ``1``;` `        ``// loop untill low and``        ``// high cross each other``        ``while` `(low < high)``        ``{` `            ``// If both characters are equal then``            ``// move both pointer towards end``            ``if` `(str.charAt(low) == str.charAt(high))``            ``{``                ``low++;``                ``high--;``            ``}``            ``else``            ``{` `                ``/*``                ``* If removing str[low] makes the``                ``* whole string palindrome. We basically``                ``* check if substring str[low+1..high]``                ``* is palindrome or not.``                ``*/``                ``if` `(isPalindrome(str, low + ``1``, high))``                    ``return` `low;` `                ``/*``                ``* If removing str[high] makes the whole string``                ``* palindrome. We basically check if substring``                ``* str[low+1..high] is palindrome or not.``                ``*/``                ``if` `(isPalindrome(str, low, high - ``1``))``                    ``return` `high;``                ``return` `-``1``;``            ``}``        ``}` `        ``// We reach here when complete string``        ``// will be palindrome if complete string``        ``// is palindrome then return mid character``        ``return` `-``2``;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String str = ``"abecbea"``;``        ``int` `idx = possiblePalinByRemovingOneChar(str);` `        ``if` `(idx == -``1``)``            ``System.out.println(``"Not Possible"``);``        ``else` `if` `(idx == -``2``)``            ``System.out.println(``"Possible without "` `+``                          ``"removing any character"``);``        ``else``            ``System.out.println(``"Possible by removing"` `+``                               ``" character at index "` `+ idx);``    ``}``}` `// This code is contributed by``// sanjeev2552`

## Python3

 `# Python program to check whether it is possible to make``# string palindrome by removing one character` `# Utility method to check if substring from``# low to high is palindrome or not.``def` `isPalindrome(string: ``str``, low: ``int``, high: ``int``) ``-``> ``bool``:``    ``while` `low < high:``        ``if` `string[low] !``=` `string[high]:``            ``return` `False``        ``low ``+``=` `1``        ``high ``-``=` `1``    ``return` `True` `# This method returns -1 if it``# is not possible to make string``# a palindrome. It returns -2 if``# string is already a palindrome.``# Otherwise it returns index of``# character whose removal can``# make the whole string palindrome.``def` `possiblepalinByRemovingOneChar(string: ``str``) ``-``> ``int``:` `    ``# Initialize low and right by``    ``# both the ends of the string``    ``low ``=` `0``    ``high ``=` `len``(string) ``-` `1` `    ``# loop untill low and high cross each other``    ``while` `low < high:` `        ``# If both characters are equal then``        ``# move both pointer towards end``        ``if` `string[low] ``=``=` `string[high]:``            ``low ``+``=` `1``            ``high ``-``=` `1``        ``else``:` `            ``# If removing str[low] makes the whole string palindrome.``            ``# We basically check if substring str[low+1..high] is``            ``# palindrome or not.``            ``if` `isPalindrome(string, low ``+` `1``, high):``                ``return` `low` `            ``# If removing str[high] makes the whole string palindrome``            ``# We basically check if substring str[low+1..high] is``            ``# palindrome or not``            ``if` `isPalindrome(string, low, high ``-` `1``):``                ``return` `high``            ``return` `-``1` `    ``# We reach here when complete string will be palindrome``    ``# if complete string is palindrome then return mid character``    ``return` `-``2` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``string ``=` `"abecbea"``    ``idx ``=` `possiblepalinByRemovingOneChar(string)` `    ``if` `idx ``=``=` `-``1``:``        ``print``(``"Not possible"``)``    ``elif` `idx ``=``=` `-``2``:``        ``print``(``"Possible without removig any character"``)``    ``else``:``        ``print``(``"Possible by removing character at index"``, idx)` `# This code is contributed by``# sanjeev2552`
Output:

`Not Possible `

Time complexity : O(N)

Space Complexity : O(1)

This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.