CodeGym/Help with Java Tasks/These tasks are hopeless, I don't know what is wrong or h...

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Gijón

21.01.2024
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These tasks are hopeless, I don't know what is wrong or how to fix it.

Question about the task

Phone number verification

Java Multithreading, Level 2, Lesson 13

Under discussion

The checkPhoneNumber method should check whether the phoneNumber argument is a valid phone number.

Verification criteria:
1) if the number starts with '+', then it must contain 12 digits
2) if the number starts with a digit or opening parenthesis, then it must contain 10 digits
3) the number may contain 0-2 non-consecutive '-' characters
4) the number may contain 1 pair of matching parentheses, which must be to the left of the '-' characters
5) the parentheses must contain exactly 3 digits
6) the number cannot contain letters
7) the number must end with a digit

Here are some examples:

+380501234567 - true
+38(050)1234567 - true
+38050123-45-67 - true
050123-4567 - true
+38)050(1234567 - false
+38(050)1-23-45-6-7 - false
050xxx4567 - false
050123456 - false
(0)501234567 - false

Requirements:

The checkPhoneNumber method should return a boolean.
The checkPhoneNumber method must be public.
The checkPhoneNumber method must have one String parameter.
The checkPhoneNumber method must correctly check the validity of the phone number passed as an argument.

package com.codegym.task.task22.task2212; /* Phone number verification */ public class Solution { public static boolean checkPhoneNumber(String phoneNumber) { boolean flag = false; if(phoneNumber.matches("^+[0-9]{12}")) { flag = true; } if(phoneNumber.matches("^[0-9]|^[(][0-9]{10}")) { flag = true; } if(phoneNumber.matches("'-'{1}'-'{1}")) { flag = true; } if(phoneNumber.matches("[()]'-'")) { flag = true; } if(phoneNumber.matches("[(]\\d{3}[)]")) { flag = true; } if(phoneNumber.matches("![a-zB-z]")) { flag = true; } if(phoneNumber.matches("[0-9]$")) { flag = true; } return flag; } public static void main(String[] args) { } }

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Thomas

Level 41 , Bayreuth, Germany

22 January, 10:17

When beginning a task I would not fire up the IDE right away and start typing and then thinking about the problem. Better would be to think about the problem, grasp it and find a solution and that entirely without coding. Take pen and paper and try to solve this task. Remember every step you did. If you can solve it without PC then you can start writing code. Start with the first requirement: 1) if the number starts with '+', then it must contain 12 digits Now take one of the examples or make up your own like: +38050123-45-67 If this does not start with a + you continue, if it starts with a + you need to check if it has 12 digits. How do you check if it has 12 digits? You go through each char and check if it is a digit or not and if it is a digit, you count (of course there are other ways, but that is an easy example). Requirement 2: 2) if the number starts with a digit or opening parenthesis, then it must contain 10 digits It did not start with a +, then the phone number needs to contain 10 digits. How do you do that? You look at each char of the string and check if it is a digit or not. If it is a digit you count it. If you have 10 digits at the end, the requirement is fullfilled. You see what I mean. You need a plan. You need to think about the problems prior to coding. Coding may distrace you from finding a solution. Those regex sequences you have written seem to me wishful thinking. Pick one and test it. Testing is the base of programming. Or try to explain what eg ![a-zB-z] or [0-9]$ shall do? And then TEST it if this really does what you expected (the first will give true if you have a string consisiting of two characters where the first is a ! and the second in the range of ascii values B to z, like ![ or !h, the second will match one character from zero to nine).

Guadalupe Gagnon

Level 37 , Tampa, United States

22 January, 17:41

One thing: I'm sure Felix is confusing the stated goals of the task because he coded this in an attempt to return true if any of the 7 points mentioned in the conditions are true.

FelixUjo

Level 35 , Gijón, Spain

22 January, 21:05

I can take pen and paper and write El Quijote before solving this task.

Thomas

Level 41 , Bayreuth, Germany

22 January, 21:46

And how is that helpful? If you don't really want to think about the task, then the other alternative, which has already been mentioned, would be to test more. You have already delivered some good code. But something like ![a-zB-z] is wishful thinking and nothing that has a foundation. @Lupe That may be. But from the question I do not get what Felix is asking besides that the code doesn't validate. I also don't know what's so surprising about developing an algorithm without programming.

FelixUjo

Level 35 , Gijón, Spain

22 January, 21:53

I only ask for more explanations and examples. With the information provided it is impossible for me to solve the task.

Thomas

Level 41 , Bayreuth, Germany

23 January, 08:56

When you use pen and paper you can end up with something like this (not using regex and leaving the main act to you): A) the phone numbers you're supposed to check have 10 digits eg. 1234567890 they may start with a parenthesis. If so there must be a matching closing parenthesis eg. (123)4567890 there may be up to two (non-consecutive) dashes (not at the end and after the first opening parenthesis) B) if the phone number has 12 digits the above rules apply and in addition it has a leading + and to following digits first you do the standard checks null, then forbidden characters (if the passed argument contains anything else than 0-9 () +- ) then you return false So how would you approach the rest. If the number has 10 digits you check A and if it has 12 digits you check B and A. 1. you need the number of digits numberOfDigits = getDigitCount(); if: numberOfDigits == 10 checkNumber; else if: numberOfDigits == 12 checkCoutryCode; checkNumber; getDigitCount(String): loop over String and count digits checkCoutryCode(String): is index0 + AND is index1 and 2 a digit; return true; return false; If you need explanations about regex (what's not necessary here but helpful) then check out some reources in the internet. That's to much to write down here and has been done hundreds of times before. As said, test. TEST a lot. The minimum is to use the test cases codegym tells you. But it is up to you to find edge cases and these to the list, too. Means before you write your code you might want to write the test (tdd, test driven development)

String[] phoneNumbers = {"+380501234567", "+38(050)1234567", "+38050123-45-67", "050123-4567", "+38)050(1234567", "+38(050)1-23-45-6-7", "050xxx4567", "050123456", "(0)501234567"};
for (String phoneNumber : phoneNumbers)
    System.out.format("%s - %b%n", phoneNumber, checkPhoneNumber(phoneNumber));

Thomas

Level 41 , Bayreuth, Germany

23 January, 09:51

Version 1 without regex

private static boolean checkPhoneNumber(String phoneNumber) {
    if (phoneNumber == null || !isDigits(phoneNumber))
        return false;

    int length = getPhoneNumberLength(phoneNumber);

    if (length == 10) {
        return check10DigitPhoneNumber(phoneNumber);
    } else if (length == 12) {
        return check12DigitPhoneNumber(phoneNumber);
    }
    return false;
}

private static boolean check10DigitPhoneNumber(String phoneNumber) {
    return checkParenthesis(phoneNumber) && checkDashes(phoneNumber);
}

private static boolean check12DigitPhoneNumber(String phoneNumber) {
    return checkCountryCode(phoneNumber) && check10DigitPhoneNumber(phoneNumber.substring(3));
}

private static boolean checkCountryCode(String phoneNumber) {
    return phoneNumber.charAt(0) == '+' && Character.isDigit(phoneNumber.charAt(1)) && Character.isDigit(phoneNumber.charAt(2));
}

private static boolean checkParenthesis(String phoneNumber) {
    if (phoneNumber.startsWith("(")) {
        return phoneNumber.charAt(4) == ')';
    } else {
        return !phoneNumber.contains(")");
    }
}

Thomas

Level 41 , Bayreuth, Germany

23 January, 09:52

part 2 of version 1:

private static boolean checkDashes(String phoneNumber) {
    int count = 0;

    for (int i = 0; i < phoneNumber.length(); i++) {
        if (phoneNumber.charAt(i) == ')' && count > 0) {
            return false;
        }

        if (phoneNumber.charAt(i) == '-') {
            count++;

            if (i == 0 || i == phoneNumber.length() - 1)
                return false;

            if (phoneNumber.charAt(i - 1)  < '0' || phoneNumber.charAt(i - 1) > '9' ||
                    phoneNumber.charAt(i + 1) < '0' || phoneNumber.charAt(i + 1) > '9')
                return false;
        }
    }

    return count <= 2;
}

private static int getPhoneNumberLength(String phoneNumber) {
    int length = 0;
    for (char c : phoneNumber.toCharArray()) {
        if (c >= '0' && c <= '9')
            length++;
    }
    return length;
}

private static boolean isDigits(String phoneNumber) {
    for (char c : phoneNumber.toCharArray()) {
        if(!(c >= '0' && c <= '9' || c == '+' || c == '-' || c == '(' || c == ')'))
            return false;
    }
    return true;
}

Thomas

Level 41 , Bayreuth, Germany

23 January, 09:54

A regex version:

public static boolean checkPhoneNumber(String phoneNumber) {
    if (phoneNumber == null)
        return false;

    int length = phoneNumber.replaceAll("\\D" , "").length();
    String regexExclCountryCode = "\\(?\\d{3}\\)?\\d+-?\\d+-?\\d+";
    return length == 12 ? phoneNumber.matches("\\+\\d{2}" + regexExclCountryCode) : length == 10 && phoneNumber.matches(regexExclCountryCode);
}

I needed longer for that two versions than I thought (1h+) Imho you learn more if you go the non regex route. Also the non regex version is more robust as it checks for a closing parenthesis if there is an opening one. So this may not validate and you need to add a check if there is only one parenthesis and return false in that case. There is a lot of optimization potential in my code. As said it is quickly written and not the best demonstration material. Especially the checkDashes method is butt-ugly.