CareerCup

I think your solution it's almost complete.. but not quite.
For example if a give you string = "caaca", I should remove the last a, but your algorithm removes the c.. resulting in a non-palindrome

Code for removal of character to make palindrome

#include <iostream>
#include <string>
using namespace std;
int checkForPalindrome(string str,int i,int j)
{
	if(i==j)
		return 1;
	if(str[i]==str[j])
		return 1;
	if(str[i]!=str[j])
		return 0;
	return checkForPalindrome(str,i,j-1)&&checkForPalindrome(str,i+1,j);			
}
string removeChar(string str,int i,int j)
{
	if(i==j)
		return str;
	if(str[i]==str[j]&&i+1==j)
		return str;
	if(str[i]!=str[j]&&i+1==j)
	{
		str.erase(i,1);
		return str;
	}
	if(str[i]==str[j])
		return removeChar(str,i+1,j-1);
	if(str[i]!=str[j])
	{
		if(checkForPalindrome(str,i,j-1))
		{
			str.erase(j,1);
			return str;
		}
		else if(checkForPalindrome(str,i+1,j))
		{
			str.erase(i,1);
			return str;
		}
	}				
}
int main()
{
	string str="fbbfe";
         if(checkForPalindrome(str,0,str.length()-1)!=1)
	         str=removeChar(str,0,str.length()-1);
	cout<<str;
}

Assuming we must delete a character, we can use 2 pointers: from the start and from the end of the string and check if they're equal. If not, we must remove one of them and check if the remaining string is palindrome.
We only do at most 2 checks for palindrome, so this code runs in O(n), n = string.length

bool IsPalindrome(const string&s, int from, int to) {
	for (; from < to; from++, to--)
		if (s[from] != s[to])
			return false;
	return true;
}
int DeleteCharToPalindrome(const string& s, int from, int to) {
	if (from >= to) 	// s is palindrome. if its size is even, we can
		return to;		// just remove one of the middle elements if the size
						// is even or the middle element if size is odd
	if (s[from] == s[to])
		return DeleteCharToPalindrome(s, from+1, to-1);
	else if (IsPalindrome(s, from, to-1)) 
		return to;
	else if (IsPalindrome(s, from+1, to)) 
		return from;
	return -1;
}
// returns -1 if impossible, 0..len-1 otherwise representing the index to delete
int DeleteCharToPalindrome(const string& s) {
	return s == "" ? -1 : DeleteCharToPalindrome(s, 0, s.size()-1);
}

Assuming there exists one character in the string the removal of which turns string into palindrome, we can do as follows:

1. Keep two pointers front=0 and back=n-1.
2. while string[front++] == string[back--] continue;
3. if string[front+1, back] is palindrome then return string without character[front]
else return string without character[back]

Edit: based on gastonbm's reply, changed last step to check for palindrome.

#include<stdio.h>
#include<string.h>
int main()
{
char ch[30],temp[30],i,j=0,ch1[30];
scanf("%s",ch);
do
{
for(i=0;i<strlen(ch);i++)

{
if(i>=j)
{
ch1[j]=ch[j+1];
if(i==strlen(ch)-1)
break;
}
else
ch1[i]=ch[i];
temp[i]=temp[strlen(temp)-i];
temp[i]='\0';
if(!strcmp(temp,ch1))
{
printf("%s",temp);
}j++;
}
while(j<strlen(ch));
return 0;
}

#include<stdio.h>
#include<string.h>
int main()
{
char ch[30],temp[30],i,j=0,ch1[30];
scanf("%s",ch);
do
{
for(i=0;i<strlen(ch);i++)

{
if(i>=j)
{
ch1[j]=ch[j+1];
if(i==strlen(ch)-1)
break;
}
else
ch1[i]=ch[i];
temp[i]=temp[strlen(temp)-i];
temp[i]='\0';
if(!strcmp(temp,ch1))
{
printf("%s",temp);
}j++;
}
while(j<strlen(ch));
return 0;
}

#include<stdio.h>
#include<string.h>
int main()
{
char ch[30],temp[30],i,j=0,ch1[30];
scanf("%s",ch);
do
{
for(i=0;i<strlen(ch);i++)

{
if(i>=j)
{
ch1[j]=ch[j+1];
if(i==strlen(ch)-1)
break;
}
else
ch1[i]=ch[i];
strrev(ch1);

if(!strcmp(temp,ch1))
{
printf("%s",temp);
}j++;
}
while(j<strlen(ch));
return 0;
}