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.
Implementation:
// 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 high by both the ends of the string
int low = 0, high = str.length() - 1;
// loop until 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 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 until 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 |
# 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 until 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" )
else if idx = = - 2 :
print ( "Possible without removing any character" )
else :
print ( "Possible by removing character at index" , idx)
# This code is contributed by # sanjeev2552 |
// C# program to check whether // it is possible to make string // palindrome by removing one character using System;
class GFG
{ // Utility method to check if
// substring from low to high is
// palindrome or not.
static bool isPalindrome( string str, int low, int high)
{
while (low < high)
{
if (str[low] != str[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 until 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, 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)
Console.Write( "Not Possible" );
else if (idx == -2)
Console.Write( "Possible without " +
"removing any character" );
else
Console.Write( "Possible by removing" +
" character at index " + idx);
}
} // This code is contributed by shivanisinghss2110 |
<script> // JavaScript 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. function isPalindrome(str, low, 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.
function possiblePalinByRemovingOneChar(str)
{
// Initialize low and right
// by both the ends of the string
var low = 0, high = str.length - 1;
// loop until 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
var str = "abecbea" ;
var idx = possiblePalinByRemovingOneChar(str);
if (idx == -1)
document.write( "Not Possible" );
else if (idx == -2)
document.write( "Possible without " +
"removing any character" );
else
document.write( "Possible by removing" +
" character at index " + idx);
// this code is contributed by shivanisinghss2110 </script> |
Not Possible
Time complexity : O(N)
Space Complexity: O(1)