Hackerrank - Prepare | Algorithms | Strings | Palindrome Index

 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

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

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 

Input Format

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

Constraints

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

Sample Input

STDIN   Function
-----   --------
3       q = 3
aaab    s = 'aaab' (first query)
baa     s = 'baa'  (second query)
aaa     s = 'aaa'  (third query)

Sample Output

3
0
-1

Explanation

Query 1: "aaab"
Removing 'b' at index  results in a palindrome, so return .

Query 2: "baa"
Removing 'b' at index  results in a palindrome, so return .

Query 3: "aaa"
This string is already a palindrome, so return . Removing any one of the characters would result in a palindrome, but this test comes first.

Note: The custom checker logic for this challenge is available here.

import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;

class Result {


    public static boolean isPalindrome(String s) {
        for (int i = 0; i < s.length() / 2; i++) {
            if (s.charAt(i) != s.charAt(s.length() - i - 1)) {
                return false;
            }
        }
        return true;
    }

    public static int palindromeIndex(String s) {
        StringBuffer sb = new StringBuffer(s);
        for (int i = 0; i < sb.length() / 2; i++) {
            if (sb.charAt(i) != sb.charAt(sb.length() - i - 1)) {
                sb.deleteCharAt(i);

                if (isPalindrome(sb.toString())) {
                    return i;
                } else {
                    sb = new StringBuffer(s);
                    sb.deleteCharAt(sb.length() - i - 1);

                    if (isPalindrome(sb.toString())) {
                        return sb.length() - i - 1 +1;
                    }else{
                        return -1;
                    }

                }
            }
        }
        
        return -1;

    }


}

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        int q = Integer.parseInt(bufferedReader.readLine().trim());

        IntStream.range(0, q).forEach(qItr -> {
            try {
                String s = bufferedReader.readLine();

                int result = Result.palindromeIndex(s);

                bufferedWriter.write(String.valueOf(result));
                bufferedWriter.newLine();
            } catch (IOException ex) {
                throw new RuntimeException(ex);
            }
        });

        bufferedReader.close();
        bufferedWriter.close();
    }
}

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