RegexChecker

import java.util.regex.*;
import java.io.*;

class RegCheck {
    public static void main(String args[]) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String userDefinedPattern = "9.[1,4]...[5-8]..."; //br.readLine();
        System.out.println(userDefinedPattern);
        String phoneNumber = br.readLine();
    //the following line assumes a three character ISD code (e.g. +91), hence the substring trimmer is there.
    //input the 10 digit phone number along with the ISD code. For example: +919876543210
    //note that, userDefinedPattern is just a 10 digit number
        phoneNumber = phoneNumber.substring(3, phoneNumber.length());
        System.out.println("Number at input is: " + phoneNumber);
        int numDigits = 10;
        boolean blockThisNumber = true;
        boolean[] checker = new boolean[numDigits];
        for(int i = 0; i < numDigits; i++)      checker[i] = false;
        for(int i = 0, j = 0; i < phoneNumber.length() && j < userDefinedPattern.length(); i++) {
            char c = phoneNumber.charAt(i);
            int p = Character.getNumericValue(c);
            //System.out.println(p + " " + userDefinedPattern.charAt(j));

            //Case 1: u is a digit, then check if it is equal to p
            if(Character.isDigit(userDefinedPattern.charAt(j))) {
                if(p == Character.getNumericValue(userDefinedPattern.charAt(j))) {
                    checker[i] = true;
                    j++;
                    continue;
                }
                else {
                    j++;
                    continue;
                }
            }

            //Case 2: u is a dot, then no checking just continue
            if(userDefinedPattern.charAt(j) == '.') {
                checker[i] = true;
                j++;
                continue;
            }

            //Case 3: u is a range specified by [a-b] where a, b are digits
            if(userDefinedPattern.charAt(j) == '[' && userDefinedPattern.charAt(j + 2) == '-') {
                //find the next character to [ and name it a,
                //skip the '-', read the next character, name it b
                //check for boundary matchings, manipulate j accordingly

                int a = Character.getNumericValue(userDefinedPattern.charAt(j + 1));
                int b = Character.getNumericValue(userDefinedPattern.charAt(j + 3));
                if(p >= a && p <= b) {
                    checker[i] = true;
                    j = j + 5;
                    continue;
                }
                else {
                    j = j + 5;
                    continue;
                }
            }

            //Case 4: u is a set of digits
            if(userDefinedPattern.charAt(j) == '[' && userDefinedPattern.charAt(j + 2) == ',') {
                while(userDefinedPattern.charAt(j) != ']') {

                    if(Character.isDigit(userDefinedPattern.charAt(j))) {
                        if(p == Character.getNumericValue(userDefinedPattern.charAt(j))) {
                            checker[i] = true;
                            break;
                        }
                    }
                    else    j++;
                }
                //j reached ']', so increment it
                j++;
            }

        }

        for(int i = 0; i < numDigits; i++)     blockThisNumber &= checker[i];

        System.out.println((blockThisNumber ? "Call Blocked!" : "Calling..."));
    }
}