CodeGym/Help with Java Tasks/Don't try this at home...

Level 50

Kraków

07.02.2023
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13comments

Don't try this at home...

Question about the task

Under discussion

Maybe someone find inspiration in my code. After almost seven hours non-stop working (not counting WC visits, and coffe/pizza break) I did this Frankenstain. I start this task yesterday and I tried did it with Pattern and Matcher help. It's not work, and I ending on help section. I find beauty code from Evgen but I don't understand fully regex with (?=) questionmark, I could copy this and do the rest, but somewhere else I have read comment's that it's possible do this without it, use knowledge whose we have. It's unbelievable, I was in trance reading it, I couldn't do this one more time from scrach. I know that posting complete solutions is not allowed but I need more experienced eyes looking for my algorithm and tell me what I can do to simplify it! Because I write it for success, and I know my solutions is too long, and maybe not so neat and optimal.

while (reader.ready()) {
String line = reader.readLine();
                StringBuilder sb = new StringBuilder();
                char[] chars = line.toCharArray();
                for (int i = 0; i < chars.length; i++) {
                    // is Number, different than 1, first element   (2 edf) 2edf 2.r
                    if (Character.isDigit(chars[i]) && !String.valueOf(chars[i]).equals("1") && i == 0) {
                        if (i < chars.length - 1) {
                            // if next isn't number and alphabetic
                            if (!Character.isDigit(chars[i + 1]) && !Character.isAlphabetic(chars[i + 1])) {
                                String string = String.valueOf(chars[i]);
                                int data = Integer.parseInt(string);
                                sb.append(map.get(data));
                            } else {
                                sb.append(chars[i]);
                            }
                        } else {
                            String string = String.valueOf(chars[i]);
                            int data = Integer.parseInt(string);
                            sb.append(map.get(data));
                        }
                        // is Number 1 , first element  (1 ede)  1we 123 12,  12
                    } else if (Character.isDigit(chars[i]) && String.valueOf(chars[i]).equals("1") && i == 0) {
                        if (i < chars.length - 1) {
                            if (!Character.isDigit(chars[i + 1]) && !Character.isAlphabetic(chars[i + 1])) {
                                String string = String.valueOf(chars[i]);
                                int data = Integer.parseInt(string);
                                sb.append(map.get(data));
                            } else if (String.valueOf(chars[i + 1]).equals("0") || String.valueOf(chars[i + 1]).equals("1") || String.valueOf(chars[i + 1]).equals("2")) {
                                if (i + 1 < chars.length) {
                                    if (!Character.isDigit(chars[i + 2]) && !Character.isAlphabetic(chars[i + 2])) {
                                        String string = String.valueOf(chars[i]) + String.valueOf(chars[i + 1]);
                                        int data = Integer.parseInt(string);
                                        sb.append(map.get(data));
                                        i++;
                                    } else {
                                        sb.append(chars[i]).append(chars[i + 1]).append(chars[i + 2]);
                                        i += 2;
                                    }
                                } else {
                                    String string = String.valueOf(chars[i]) + String.valueOf(chars[i + 1]);
                                    int data = Integer.parseInt(string);
                                    sb.append(map.get(data));
                                    i++;
                                }
                            }
                        } else {
                            String string = String.valueOf(chars[i]);
                            int data = Integer.parseInt(string);
                            sb.append(map.get(data));
                        }
                        // if Number different than 1, last element   2  2ed2  2.2 fuck 2  12
                    } else if (Character.isDigit(chars[i]) && !String.valueOf(chars[i]).equals("1") && i == chars.length - 1) {
                        if (!Character.isDigit(chars[i - 1]) && !Character.isAlphabetic(chars[i - 1])) {
                            String string = String.valueOf(chars[i]);
                            int data = Integer.parseInt(string);
                            sb.append(map.get(data));
                        } else {
                            sb.append(chars[i]);
                        }
                        // if Number == 1, last element   1  2ed1  2.1 fuck 1  11
                    } else if (Character.isDigit(chars[i]) && String.valueOf(chars[i]).equals("1") && i == chars.length - 1) {
                        if (!Character.isDigit(chars[i - 1]) && !Character.isAlphabetic(chars[i - 1])) {
                            String string = String.valueOf(chars[i]);
                            int data = Integer.parseInt(string);
                            sb.append(map.get(data));
                        } else {
                            sb.append(chars[i]);
                        }
                        // is Number, different than 1, middle elements   (2 edf) 2edf 2.r 12.
                    } else if (Character.isDigit(chars[i]) && !String.valueOf(chars[i]).equals("1")) {
                        if (!Character.isDigit(chars[i + 1]) && !Character.isAlphabetic(chars[i + 1]) &&
                                !Character.isDigit(chars[i - 1]) && !Character.isAlphabetic(chars[i - 1])) {
                            String string = String.valueOf(chars[i]);
                            int data = Integer.parseInt(string);
                            sb.append(map.get(data));
                        } else {
                            sb.append(chars[i]);
                        }
                        // is Number 1 , middle elements       (1 ede)  1we 123 12,  12.
                    } else if (Character.isDigit(chars[i]) && String.valueOf(chars[i]).equals("1")) {
                        if (!Character.isDigit(chars[i + 1]) && !Character.isAlphabetic(chars[i + 1]) &&
                                !Character.isDigit(chars[i - 1]) && !Character.isAlphabetic(chars[i - 1])) {
                            String string = String.valueOf(chars[i]);
                            int data = Integer.parseInt(string);
                            sb.append(map.get(data));
                        } else if (String.valueOf(chars[i + 1]).equals("0") || String.valueOf(chars[i + 1]).equals("1") || String.valueOf(chars[i + 1]).equals("2")) {
                            if (i + 1 < chars.length) {
                                if (!Character.isDigit(chars[i + 2]) && !Character.isAlphabetic(chars[i + 2])) {
                                    String string = String.valueOf(chars[i]) + String.valueOf(chars[i + 1]);
                                    int data = Integer.parseInt(string);
                                    sb.append(map.get(data));
                                    i++;
                                } else {
                                    sb.append(chars[i]).append(chars[i + 1]).append(chars[i + 2]);
                                    i += 2;
                                }
                            } else {
                                String string = String.valueOf(chars[i]) + String.valueOf(chars[i + 1]);
                                int data = Integer.parseInt(string);
                                sb.append(map.get(data));
                                i++;
                            }
                        } else {
                            sb.append(chars[i]);
                        }
                    } else {
                        sb.append(chars[i]);
                    }
                    if (debugMode()) {
                        System.out.println(sb);
                    }
                }
                System.out.println(sb.toString());
}

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Guadalupe Gagnon

Level 37 , Tampa, United States

7 February 2023, 21:59

useful

Some thoughts: #1 a valid number will have this specific 3 part pattern: start of String OR a space; 1 of any digit OR two digits where the first is a 1 and the next is either '0', '1', or '2'; anything that is not a letter or a number including end of line Valid Examples: "12 " <-- this is a start of line; two digits where the first is a 1; and a space is not a number or a letter " 11." <-- there is a space before the digit; two digits where the first is a '1'; and a period is not a number or a letter " 5" <-- there is a space before the digit; its only one digit; and its the end of a line #2 It maybe simpler to split the line up. You can split the lines up into the individual "words" that are separated by spaces. Because the substrings will have no spaces to consider you can then just focus on the simpler pattern: line starts with one of any digit OR line starts with a '1' and then '0', '1', or '2'; anything that is not a letter or a number including end of line AND the string can't be more than 3 chars long Valid Examples: "12" <-- this String starts with a 1 and then has another digit; and the String ends thereafter "16." <-- this String starts with a 1 and then has another digit; and then has one char not alphanumeric and is length 3 or less "5" <-- this only has one digit #3 If you take the second tip then you just need to put the String back together by adding in the spaces between the words. I think I gave you my tip on doing that, though I can go over it again: start by putting first part then append each additional part with a leading space. Example:

String[] test = "this is only a test".split(" ");

StringBuilder sb = new StringBuilder();
int i = 0;

// put in first part
sb.append(test[i++]);

// put each additional parts with leading space
while (i < test.length) sb.append(" " + test[i++]);

System.out.println("--" + sb + "--");

Guadalupe Gagnon

Level 37 , Tampa, United States

7 February 2023, 22:00

if you run that example code you will see the output has no leading or trailing spaces

matemate123

Level 50 , Kraków, Poland

8 February 2023, 10:42

Yeah, I'm sure that your way is neater and more smooth, my was something like challenge, do it character after character, I was curious whether is possible for me. It's really absorb task, I don't even know how much work guys from CG must have to do it and prepare validator for every possibilities.

Thomas

Level 41 , Bayreuth, Germany

8 February 2023, 11:38

useful

Just few comments to your first thought. There's no need for regex lookaheads if you followed your initial intention and used pattern-matcher. Here (and for String.replaceAll()) you could have used regex word boundaries. That's pretty simple as it describes a word start and a word end

\\bHello World\\b

That regex would find the 4 Hello Worlds in the following sentence. Hello World, something else Hello World and Hello World!!! as well as Hello World If you place something inside for matching numbers, then it'll return all numbers in word boundaries, eg. a regex for every integer in word boundaries

\\b\\d+\\b

It'll match 1, 2!! 3333, 123456789999000 and so on. You can improve it a little bit to only match 1 and 2 digit numbers in word boundaries like

\\b\\d{1,2}\\b

where the meaning of the values in the parenthesis is {min,max} repetitions So you just build your pattern-matcher, maybe in a method (for parsing huge files it'll be not the best approach, instead I'd reset the matcher).

private static String parseLine(String line) {
    Matcher m = Pattern.compile("\\b\\d{1,2}\\b").matcher(line);
    StringBuffer sb = new StringBuffer();
    while (m.find()) {
        int currNum = Integer.parseInt(m.group());
        if (map.containsKey(currNum))
            m.appendReplacement(sb, map.get(currNum));
    }
    m.appendTail(sb);
    return sb.toString();
}

If there is a match, all the non matching parts before are added to the sb. The match you have to take care of yourself with appendReplacement. Finally you just append the rest (appendTail) When the map is initialized you just need to read each line, send it to the method and then output the result.

Thomas

Level 41 , Bayreuth, Germany

8 February 2023, 11:41

useful

public static Map<Integer, String> map = new HashMap<Integer, String>();
    // ignore the init of the map, I was to lazy to do this one by one
    static {
        String[] numStrings = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven", "twelve"};
        map = IntStream.range(0, 13)
                .boxed()
                .collect(Collectors.toMap(k -> k, v -> numStrings[v]));
    }

    public static void main(String[] args) throws IOException {
        try (BufferedReader console = new BufferedReader(new InputStreamReader(System.in));
             BufferedReader in = new BufferedReader(new FileReader(console.readLine()))) {

/*          // edited to something easier understandable below
            in.lines()
                    .map(Solution::parseLine)
                    .forEach(System.out::println);
*/
            while (in.ready()) {
                System.out.println(parseLine(in.readLine()));
            }
        }
    }

matemate123

Level 50 , Kraków, Poland

8 February 2023, 15:46

I don't know whether I understand. If you use this boundaries and like your example: \\bHello World\\b - that mean if part of string in file be like "HHello World", "Hello World2" , or "HeHello World" , this matcher don't grasp that wrongly strings? It find "Hello World" but discard "1Hello World" ? And Could I do something like \\b\\ d{0,1,2,3,4,5,6,7,8,9,10,11,12 }\\b ?? I must read about that Matcher and maybe some youtube tutorial on this topic.

Thomas

Level 41 , Bayreuth, Germany

8 February 2023, 16:30

useful

String row = "hello,  hello?? hhello helloo helo helLo hello";

Matcher m = Pattern.compile("\\bhello\\b").matcher(row);
while (m.find()) {
    System.out.println(m.group() + " : " + m.start() + "-" + m.end());
}

that finds three hellos, the first, second and last. The other do not match the word boundaries (hhello, helloo) or have not enough or wrong letters in it. you create a pattern like Pattern.compile("e") to match every e in the search string. if you want to find an e, an ee or even an eee or any number of e in one match, then you write Pattern.compile("e+") (find minimum 1 e or any number of e). You also can write this as e{1,}. If you want an upper boundary for the quantifier to eg. match 1 to max 10 e, then you set the upper boundary like e{1, 10}. If you want to match instead of an 'e' a zero, then it is 0{1, 10}. If you want not only a zero but you want to match one zero or one one or one two, then you write [012] or [0-2]. If the search string is 01234567, then you'd have three matches. If you add word boundaries //b[0-2]//b then you'd have no match at all. If you change the search string to "0 1 2 3 4 5 6 7 8 9" then you'd had again three matches. [0-9] finds all numbers -> 654 would give three matches 6, 5 and 4. If you add the mentioned + quantifier above [0-9]+, then you'd have one match 654. So no, it does not work as you wished ;) you can eg. do smth like this [01][0-9], that'll match 00 , 01, 02 - 19. But that's not what you want. But there's an or symbol... you could eg. try grouping with the or "\\b(1|2|3|4|5|6|7|8|9|10|11|12)\\b" or without grouping you'd have to repeat the word boundaries like \\b1\\b|\\b2\\b|\\b3\\b....

matemate123

Level 50 , Kraków, Poland

8 February 2023, 16:37

Thanks for amazing reply. It's convenient method, I like it.

Thomas

Level 41 , Bayreuth, Germany

8 February 2023, 16:49

useful

Yep, I like it, too... for easier things. If it gets more difficult, then regex can be a pain. Later there are tasks with eg. phone number matching. That can easily be done using regex. You also can use String methods but that is a lot more code. String also has some convenience regex methods so that you do not need to go the pattern-matcher path. A very easy approach for the above problem can be done with String.replaceAll and word boundaries like:

String[] replacement = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven", "twelve"};

String row = "This costs 1 dollar, but this is 12.";
for (int i = 0; i <= 12; i++) {
    row = row.replaceAll("\\b" + i + "\\b", replacement[i]);
}
System.out.println(row);

ahh... man, you of course can write the regex in my above post even easier ;) \\b([0-9]|10|11|12)\\b that'll also match 0 - 12

matemate123

Level 50 , Kraków, Poland

9 February 2023, 05:03

I sit and try now this two options.

matemate123

Level 50 , Kraków, Poland

9 February 2023, 05:36

It was so easy with your last String replacement example! Knowledge is power in programing. One \\b and do so powerfull thinks, unbeliveable. 6 hours work vs 3 minutes work...

Thomas

Level 41 , Bayreuth, Germany

9 February 2023, 07:34

Yes, it is amazing how easy it can be sometimes. However exercise is the key to success and that is what you're doing. And what you coded above not loosing the track is amazing. Just one final note. That Leela task you tried few days back you can solve with very little regex and a nested Pattern-Matcher. Lisa posted that solution some month back and I like it a lot. You start a matcher for the start tag. When found you start a innner matcher from that point on for the end tag. If it is found, then you cut out the string between start tag and found end tag. To check if the nesting matches she cut out the result and split it once using the start tag and once using the end tag and compared the resulting array length. If these are equal, then of course the number of starting/ closing tags is equal and you have a match. https://codegym.cc/help/16644 (use the lazy regex she mentioned in the text and replace the start pattern with that)

matemate123

Level 50 , Kraków, Poland

9 February 2023, 09:14

Thanks a lot man, I had hard times with that Leela task. I would didn't never solved it without Guadalupe help, and this solution from Lisa are very neat and ingeniuous (both of her Solutions). I don't know how many time and programing experience need to write pretty solutions like that, it'll be long jurney. I'm happy to read and understand at least.