Correct output, but not validated

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Greetings friends, After an ungodly number of hours of figuring out how to break it into Strings without regex (thanks Breakiterator!) and then figuring out how to not lose the period at the end, it's not taking my answer even though it matches the output. It's not obvious to me what circumstances I'm missing that are making my code appear to produce the correct results, but in actuality not. Thanks in advance, Mike

1. In a static block, initialize map with (number, word) pairs from 0 to 12 inclusive.
For example, (0, "zero"), (1, "one"), (2, "two")
2. Read the file name from the console, and then read the contents of the file.
3. Use map to replace all of the numbers with words.
4. Display the result.
5. Close the streams.

Example file data:
This costs 1 dollar, but this is 12. The variable is called file1. 110 is a number.

Example console output:
This costs one dollar, but this is twelve. The variable is called file1. 110 is a number.

Requirements:

The Solution class must have a public static Map<Integer, String> field called words that is initialized immediately.
The program must read the file name from the console (use BufferedReader).
The BufferedReader used for reading input from the console must be closed after use.
The program must read the file's contents (use FileReader).
The file input stream (FileReader) must be closed.
The program should write all of the lines from the file to the console, but any numbers must be replaced with the corresponding words in map.
The Solution class must have a static block that adds thirteen pairs to map.

package com.codegym.task.task19.task1924; import java.io.BufferedReader; import java.io.FileReader; import java.io.IOException; import java.io.InputStreamReader; import java.text.BreakIterator; import java.util.*; /* Replacing numbers 1. In a static block, initialize map with (number, word) pairs from 0 to 12 inclusive. For example, (0, "zero"), (1, "one"), (2, "two") 2. Read the file name from the console, and then read the contents of the file. 3. Use map to replace all of the numbers with words. 4. Display the result. 5. Close the streams. Example file data: This costs 1 dollar, but this is 12. The variable is called file1. 110 is a number. Example console output: This costs one dollar, but this is twelve. The variable is called file1. 110 is a number. Requirements: 1. The Solution class must have a public static Map<Integer, String> field called words that is initialized immediately. 2. The program must read the file name from the console (use BufferedReader). 3. The BufferedReader used for reading input from the console must be closed after use. 4. The program must read the file's contents (use FileReader). 5. The file input stream (FileReader) must be closed. 6. The program should write all of the lines from the file to the console, but any numbers must be replaced with the corresponding words in map. 7. The Solution class must have a static block that adds thirteen pairs to map. */ public class Solution { public static Map<Integer, String> map = new HashMap<>(); static { map.put(0, "zero"); map.put(1, "one"); map.put(2, "two"); map.put(3, "three"); map.put(4, "four"); map.put(5, "five"); map.put(6, "six"); map.put(7, "seven"); map.put(8, "eight"); map.put(9, "nine"); map.put(10, "ten"); map.put(11, "eleven"); map.put(12, "twelve"); } public static void main(String[] args) throws IOException { BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); String fileName = reader.readLine(); reader.close(); BufferedReader fileReader = new BufferedReader(new FileReader(fileName)); while (fileReader.ready()) { String line = fileReader.readLine(); //Now, you should have a list of strings (that look like integers) to replace. ArrayList<String> lineStrings = numberFinder(line); ArrayList<Integer> lineNumbers = new ArrayList<>(); for(String member : lineStrings){ lineNumbers.add(Integer.parseInt(member)); } Collections.sort(lineNumbers); Collections.reverse(lineNumbers); lineStrings = new ArrayList<>(); for(Integer member : lineNumbers){ lineStrings.add(member + ""); } for (String member : lineStrings) { //System.out.println(member); //If line contains a member (which it does since you've already identified it. if (line.contains(member)) { //The issue right now is that it's replacing a single digit rather than the whole number. int integer = Integer.parseInt(member); for (Map.Entry<Integer, String> entry : map.entrySet()) { if (integer == entry.getKey()) { line = line.replace(member, entry.getValue()); } } } } System.out.println(line); } fileReader.close(); } //This method just collects the strings that need to be replaced. public static ArrayList<String> numberFinder(String s) { Locale currentLocale = Locale.US; ArrayList<String> lineArrayList = new ArrayList<>(); ArrayList<String> lineIntegerArrayList = new ArrayList<>(); BreakIterator wordIterator = BreakIterator.getWordInstance(currentLocale); wordIterator.setText(s); int start = wordIterator.first(); int end = wordIterator.next(); while(end != BreakIterator.DONE){ String word = s.substring(start, end); if(Character.isLetterOrDigit(word.charAt(0))){ lineArrayList.add(word); } start = end; end = wordIterator.next(); } //So, now, we should have an arrayList of the numbers that need replacing. Use these numbers to replace their String versions. for(String member : lineArrayList){ try{ int integer = Integer.parseInt(member); lineIntegerArrayList.add(integer + ""); }catch(NumberFormatException e){ e.getMessage(); } } return lineIntegerArrayList; } }

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Thomas

Level 41 , Bayreuth, Germany

15 March 2023, 21:47

I've tried

1-1, 2.2,

separated with a hyphen the output looks good but not for the dot. It still shows 2.2, This task is written to work with regex. If it should validate without, then you need to simulate regex. I tried that in the code below... no clue if it'll validate but as said, it should work similar to the regex word boundaries. The isWordCharacter() method is taking care of that. And it's a quick hack... so it is not really pretty. If you'd try with regex this could be so much easier.

private static String parseLine(String s) {
    StringBuilder word = new StringBuilder();
    StringBuilder line = new StringBuilder();
    char[] c = s.toCharArray();

    for (int i = -1; i < c.length - 1; ) {
        while (++i != c.length && isWordCharacter(c[i])) {
            word.append(c[i]);
        }

        if (word.length() > 0) {
            String currWord = word.toString();
            word.setLength(0);
            i--;

            int num;
            if (isNumUpTwelve(currWord)) {
                line.append(map.getOrDefault(Integer.parseInt(currWord), currWord));
            } else {
                line.append(word);
            }
        } else {
            line.append(c[i]);
        }
    }
    return line.toString();
}

private static boolean isWordCharacter(char c) {
    return c >= 'a' && c <= 'z' || c >= 'A' && c <= 'Z' || c >= '0' && c <= '9' || c == '_';
}

private static boolean isNumUpTwelve(String s) {
    if (s == null || s.isEmpty() || s.length() > 2)
        return false;

    for (char c : s.toCharArray())
        if (c < '0' || c > '9')
            return false;
    return true;
}

Mike S

Level 28 , Saint Louis, United States

16 March 2023, 14:47

Thank you, Thomas. Going back to the regex point. You say that regex makes it "so easy" and I've tried a couple of things with regex and it's not obvious to me what the advantage is there. Yes, you can parse out the strings that you want to work with "1, 2, 3, etc.", but you're still left with my original question of figuring out how to "put punctuation back in". It is to say, what I achieved a couple of different ways, one with regex was to split the String like this string.split("\\W+"). So, that gave me the items to work with, but no obvious way of "putting them back". I'll look at your code closely, but again, it's really not obvious to me what regex achieves here.

Thomas

Level 41 , Bayreuth, Germany

16 March 2023, 15:12

The code above is working without any regex. It is just trying to do behave similar to regex \b (word boundaries). Working with regex and pattern-matcher you can do it like this:

public class Solution {
    public static Map<Integer, String> map;

    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));
             Stream<String> lines = Files.lines(Paths.get(console.readLine()))) {

            System.out.println(lines.map(Solution::parseLine)
                    .collect(Collectors.joining("\n")));
        }
    }

    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 currInt = Integer.parseInt(m.group());
            if (map.containsKey(currInt)) m.appendReplacement(sb, map.get(currInt));
        }
        m.appendTail(sb);
        return sb.toString();
    }
}

The regex "\\b\\d{1,2}\\b" matches all "words" that consists of one or two digits. \b describes a word start and end. Inside this word the allowed characters are A-Ba-b0-9_ All other chars start and end a word (compare to my first code -> isWordCharacter()) Just try to exchange the parseLine methods... thy should (hopefully) behave the same.

Thomas

Level 41 , Bayreuth, Germany

16 March 2023, 15:33

Another approach using regex would be to loop over the map and look for the keys inside the line using Strings replaceAll() method. If a key is found, replace it with the value.

private static String parseLine(String line) {
    for (Map.Entry<Integer, String> entry : map.entrySet()) {
        line = line.replaceAll("\\b" + entry.getKey() + "\\b", entry.getValue());
    }
    return line;
}

I like the pattern-matcher more as it comes with less overhead. But I guess you now see the power of regex. It can be very helpful in times.

Guadalupe Gagnon

Level 37 , Tampa, United States

16 March 2023, 15:49

solution

The word boundary that Thomas is talking about is specifically referencing the beginning or end of a run of alphanumeric characters. Typically each unique character in a regex String equals a character in the String it is being matched against, word boundary does not equal a single character. It is hard to explain, but easier to show in an example. Take this bit of code:

public static void main(String[] args) {
   String test = "15 asflkj15aslflk 15, 15.15 something yada yada15 15";
   String result = test.replaceAll("\\b15\\b", "*REPLACED*");

   System.out.println("Original String:");
   System.out.println("\t" + test);

   System.out.println("Resulting String after replacing '15' with '*REPLACED*':");
   System.out.println("\t" + result);
}

You can see that each "15" by itself has been replaced if it fits the pattern of: word boundary15word boundary You can see (when you run the code and check the output) that each 15 has been replaced, but not the character before it or the character after it. Once again word boundary is not a character. You can also see that the 15 in the middle of "asflkj15aslflk" is not changed as well as the 15 in "yada15". This is because those two examples are not surrounded by word boundaries. I hope this clears things a little bit.

Mike S

Level 28 , Saint Louis, United States

16 March 2023, 16:27

Thanks very much gentlemen. Yes, Guadlupe's example helped clear up the nature of "\\b" and WAY oversimplified things. My head was stuck on the concept of "regex" in terms of the split method. In fact, I did try something similar to the replace() at one point but couldn't figure out how to search for the keyword without understanding the nature of the \\b better. Onward!

Guadalupe Gagnon

Level 37 , Tampa, United States

16 March 2023, 17:38

I had trouble with word boundary too. It didn't make sense to me how it wasn't counting that as a character like everything else in the regex. If you used a pattern like this ""\\W15\\W"" and applied it to replace all such as like this: "a 15 a".replaceAll(pattern, "*REPLACED*"); you would end up with "a*REPLACED*a", where the "\\W" was part of what was replaced.