Special String Again HackerRank Solution in Java

Java Solution for Special String Again HackerRank Probelm

A string is said to be a special string if either of two conditions is met:

• All of the characters are the same, e.g. aaa.
• All characters except the middle one are the same, e.g. aadaa.

A special substring is any substring of a string which meets one of those criteria. Given a string, determine how many special substrings can be formed from it.

Example :

str = mnonopoo

str contains the following 12 special substrings {m, n, o, n, o, p, o, o, non, ono, opo, oo}

See full description :

•  HackerRank

Lets see solution and its Explanation

Solution 1 :

import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.regex.*;

public class Solution {

    // Complete the substrCount function below.
    static long substrCount(int n, String str) {
        for (int i = 0; i < str.length(); i++) {
            char iIndexChar = str.charAt(i);
            
            int middleIndexChar = -1;
                 
            for (int j = i+1; j < str.length(); j++) {
                char jIndexChar = str.charAt(j);
                
                // Check for characters except the middle one are the same and
                // All of the characters are the same
                if (iIndexChar == jIndexChar) {
                    if ( (middleIndexChar == -1) || (j - middleIndexChar) == (middleIndexChar - i)) {
                        n++;
                    }
                } else if (middleIndexChar == -1) {
                    middleIndexChar = j;
                } else {
                    break;
                }
            }
        }
        return n;
    }

    private static final Scanner scanner = new Scanner(System.in);

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

        int n = scanner.nextInt();
        scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

        String s = scanner.nextLine();

        long result = substrCount(n, s);

        bufferedWriter.write(String.valueOf(result));
        bufferedWriter.newLine();

        bufferedWriter.close();

        scanner.close();
    }
} 

Solution explanation :

• Traverse through given String from 0 to its length and store current ith index character in iIndexChar variable.
• Take int variable and initialize with -1.
• Now again loop through given string from i+1 to string length and store current jth index string in jIndexChar.
  
• Check if ithIndexChar and jIndexChar are matched or not.
• If middleIndexChar is -1 that means we got consecutive characters (aa, bb). If or condition are matching ((j - middleIndexChar) == (middleIndexChar - i)) that means, we got characters except the middle one are the same. and increment n.
  
• In else if condition, we are storing jth index into middleIndexChar. or else break second for loop (that means string does not satisfy following condition : All characters except the middle one are the same)
  
• return n.

Output explanation :

str = asasd, n = 5

• i = 0
• iIndexChar = a, middleIndexChar = -1
• j = 1 
• jIndexChar = s
• If Condition : iIndexChar == jIndexChar becomes false 
• Else If Condition : middleIndexChar == -1 becomes true | middleIndexChar = 1;
• j = 2, middleIndexChar = 1, n = 5
• jIndexChar = a
• If Condition : a = a becomes true
• (j - middleIndexChar) == (middleIndexChar - i) | 2 - 1 == 1 - 0 becomes true.
• n = 6  
• j = 3, middleIndexChar = 1, n = 6
• jIndexChar = s
• If Condition : a == s becomes false
• Else If : middleIndexChar == -1 becomes false
• Else Condition : break second for loop.
  
  
• i = 1
• iIndexChar = s, middleIndexChar = -1
• j = 2
• jIndexChar = a
• If Condition : s == a becomes false 
• Else If Condition : middleIndexChar == -1 becomes true | middleIndexChar = 2;
• j = 3, middleIndexChar = 2, n = 6
• jIndexChar = s
• If Condition : s = s becomes true
• 3 - 2 == 2 - 1 | 1 == 1 becomes true.
• n = 7
• j = 4, middleIndexChar = 2, n = 7
• jIndexChar =d
• If Condition : s == d becomes false
• Else If : middleIndexChar == -1 becomes false
• Else Condition : break second for loop.
  
  
• i = 2
• iIndexChar = a, middleIndexChar = -1
• j = 3
• jIndexChar = s
• If Condition : a == s becomes false 
• Else If Condition : middleIndexChar == -1 becomes true | middleIndexChar = 3;
• j = 4, middleIndexChar = 3, n = 7
• jIndexChar = d
• If Condition : s = d becomes false
• Else If : middleIndexChar == -1 becomes false
• Else Condition : break second for loop.
  
  
• i = 3
• iIndexChar = a, middleIndexChar = -1
• j = 4
• jIndexChar =d
• If Condition : s == d becomes false 
• Else If Condition : middleIndexChar == -1 becomes true | middleIndexChar = 4;
• Second loop ends
  
  
• i = 4
• First loop also ends
• Return n | 7

 

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