# Palindrome Index

### Problem Statement :

```Given a string of lowercase letters in the range ascii[a-z], determine the index of a character that can be removed to make the string a palindrome. There may be more than one solution, but any will do. If the word is already a palindrome or there is no solution, return -1. Otherwise, return the index of a character to remove.

Example

s = "bcbc"

Either remove 'b' at index 0 or 'c' at index 3.

Function Description

Complete the palindromeIndex function in the editor below.

palindromeIndex has the following parameter(s):

string s: a string to analyze

Returns

int: the index of the character to remove or -1.

Input Format

The first line contains an integer q, the number of queries.
Each of the next q lines contains a query string s.

Constraints

1  <=   q  <= 20
1  <=   length of s  <=  10^5 + 5

All characters are in the range ascii[a-z].```

### Solution :

```                            ```Solution in C :

In  C++  :

#include<stdio.h>
#include<string.h>
char a;
char b;
int is(char *a, int n){
int i;
for(i=0;i<n/2;i++)if(a[i]!=a[n-1-i])return i;
return -1;
}
int main(){
int t;
scanf("%d",&t);
while(t--){

scanf("%s",&a);
int l = strlen(a);
int x = is(a,l);
if(x==-1){
printf("-1\n");
}
else {
int i;
int j=0;
for(i=0;a[i];i++){
if(i!=x)b[j++] = a[i];

}
if(is(b,l-1)==-1)printf("%d\n",x);
else printf("%d\n",l-1-x);
}
}
}

In  Java  :

import java.util.*;

public class Solution {

static boolean isPalidrom(String s, int i, int j) {
while (i < j) {
if (s.charAt(i++) !=  s.charAt(j--)) {
return false;
}
}
return true;
}

public static void main(String[] args) {

Scanner in = new Scanner(System.in);

int T = in.nextInt();

for (int t = 0; t < T; t++) {
String s = in.next();

int i = 0;
int j = s.length() - 1;
while (i < j) {
if (s.charAt(i) != s.charAt(j)) {
if (isPalidrom(s,i,j-1)) {
System.out.println(j);
} else {
System.out.println(i);
}
break;
}
i ++;
j --;
}

if (i >= j) {
System.out.println(-1);
}
}

}

}

In   C :

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main()
{
int t,i,flag,len,ans,j,stop;
char *str;
str = (char *) malloc(sizeof(char)*100005);
scanf("%d",&t);
while(t>0)
{
scanf("%s",str);
len=strlen(str);
ans=-1;
stop=0;
flag=1;
while(stop==0)
{
i=0;
while(i<len/2 && ans==-1)
{
if(str[i]!=str[len-i-1])
{
ans=i;
j=i+1;
while(j<=len/2 && flag==1)
{
if(str[j]!=str[len-j])
{
flag=0;
}
j++;
}
if(flag==1)
{
stop=1;
}
else
flag=1;
if(stop==0)
{
ans=len-i-1;
j=i;
while(j<=len/2 && flag==1)
{
if(str[j]!=str[len-j-2])
{
flag=0;
}
j++;
}
if(flag==1)
{
stop=1;
}
else
flag=1;
}
}
i++;
}
if(ans==-1)
{
stop=1;
}
}
printf("%d\n",ans);
t--;
}
return 0;
}

In   Python3  :

t = int(input())

def is_palindrome(s):
return s == s[::-1]

def solve(s):
i, j = 0, len(s) - 1
while (i < j) and (s[i] == s[j]):
i += 1
j -= 1
if i == j:
return 0
if is_palindrome(s[i + 1 : j + 1]):
return i
if is_palindrome(s[i:j]):
return j
raise AssertionError

for _ in range(t):
print(solve(input()))```
```

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## Delete a Node

Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo

## Print in Reverse

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