CodeGym/Help with Java Tasks/Tell me am I thinking right?

Level 50

Kraków

25.02.2023
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Tell me am I thinking right?

Question about the task

Word search

Java Core, Level 10, Lesson 10

Resolved

Now I have method that return one word when I search for it , e.g

System.out.println(detectAllWords(wordSearch, "same"));

I get back only one directories, but I have three words "same" in table now. It's the problem? Have I good understanding, that I need return every word from crossword? If you see other problems in my code give me a hint :)

{'f', 'd', 'e', 'r', 'l', 'k'},
                {'s', 's', 'a', 'm', 'e', 'o'},
                {'a', 'n', 'a', 'r', 'o', 'v'},
                {'m', 's', 'a', 'm', 'e', 'h'},
                {'e', 'o', 'e', 'e', 'e', 'j'}

1. We have a two-dimensional array that contains lowercase letters.
2. The detectAllWords method must find all the words in the wordSearch array.
3. The element(startX, startY) must correspond to the first letter of the word, while the element(endX, endY) corresponds to the last letter.
text is the word itself, located between the start and end elements
4. All the words are in the array.
5. The words can be arranged horizontally, vertically, and diagonally, both forward and backward.
6. The main method is not tested.

Requirements:

The Solution class must have a detectAllWords method.
The Solution class must contain a static Word class.
The Solution class must not have static fields.
The detectAllWords method must be static.
The detectAllWords method must be public.
The detectAllWords method should return the list of all words in the word search (according to the task conditions).

package com.codegym.task.task20.task2027; import java.util.ArrayList; import java.util.List; /* Word search */ public class Solution { public static void main(String[] args) { int[][] wordSearch2 = new int[][]{ //0 1 2 3 4 5 6 {'f', 'd', 'e', 'r', 'l', 'k', 'x'},// 0 {'u', 's', 'a', 'm', 'e', 'o', 'd'},// 1 {'l', 'n', 'g', 'r', 'o', 'v', 'z'},// 2 {'m', 'l', 'p', 'r', 'r', 'h', 'w'},// 3 {'p', 'o', 'e', 'e', 'a', 'a', 'c'},// 4 {'u', 's', 'a', 'm', 'a', 'a', 'd'},// 5 {'l', 'n', 'g', 'r', 'a', 'a', 'z'},// 6 {'m', 'l', 'p', 'r', 'a', 'a', 'w'},// 7 {'p', 'o', 'e', 'q', 's', 'a', 'c'},// 8 {'f', 'd', 'e', 'r', 'a', 'a', 'x'},// 9 }; /* Expected result qsa - (3, 8) - (5, 8) {'f', 'd', 'e', 'r', 'l', 'k'}, {'u', 's', 'a', 'm', 'e', 'o'}, // (j = 1, i = 1) (y = 1, x = 2) --> x only one direction {'l', 'n', 'a', 'r', 'o', 'v'}, {'m', 'l', 'p', 'm', 'r', 'h'}, {'p', 'o', 'e', 'e', 'e', 'j'} */ int[][] wordSearch = new int[][]{ {'f', 'd', 'e', 'r', 'l', 'k'}, {'u', 's', 'a', 'm', 'e', 'o'}, // (j = 1, i = 1) (y = 1, x = 2) --> x only one direction {'l', 'n', 'a', 'r', 'o', 'v'}, {'m', 'l', 'p', 'm', 'r', 'h'}, {'p', 'o', 'e', 'e', 'e', 'j'} }; System.out.println(detectAllWords(wordSearch, "poe", "ok", "uff")); // System.out.println(detectAllWords(wordSearch2, "qsa")); /* Expected result home - (5, 3) - (2, 0) same - (1, 1) - (4, 1) */ System.out.println("test"); } public static List<Word> detectAllWords(int[][] wordSearch, String... words) { // search for each word separately List<Word> listOfWords = new ArrayList<>(); for (String word : words) { Word word1 = new Word(word); word1.text = word; char[] ch = word.toCharArray(); int lengthThisWord = word.length(); // firstly going through matrix for (int j = 0; j < wordSearch.length; j++) { for (int i = 0; i < wordSearch[j].length; i++) { int f = 0; // when found first letter from this word if (ch[f] == wordSearch[j][i]) { // check if maybe word have only one letter (btw it's word at all in that situation?) if (f == lengthThisWord - 1) { // break; word1.setStartPoint(i, j); word1.setEndPoint(i, j); listOfWords.add(word1); } else { // until I got whole word! // now going through neighbors looking for another good characters Coordinates coordinates; int y = j; int x = i; // going to next character until method searchNeighbor() return null f++; if (searchNeighbor(wordSearch, new Coordinates(y, x), ch[f]) != null) { coordinates = searchNeighbor(wordSearch, new Coordinates(y, x), ch[f]); if (f == lengthThisWord - 1) { word1.setStartPoint(i, j); word1.setEndPoint(coordinates.x, coordinates.y); listOfWords.add(word1); } else { f++; // now we have 2 found characters, and we'll must see for another, // but we know direction if (y == coordinates.y && x == coordinates.x - 1) { // searchRight y = coordinates.y; x = coordinates.x + 1; while (f < lengthThisWord && x < wordSearch[0].length && wordSearch[y][x] == ch[f]) { if (f == lengthThisWord - 1) { word1.setStartPoint(i, j); word1.setEndPoint(x, y); listOfWords.add(word1); } f++; x++; } } if (y == coordinates.y && x == coordinates.x + 1) { // searchLeft y = coordinates.y; x = coordinates.x - 1; while (f < lengthThisWord && x >= 0 && wordSearch[y][x] == ch[f]) { if (f == lengthThisWord - 1) { word1.setStartPoint(i, j); word1.setEndPoint(x, y); listOfWords.add(word1); } f++; x--; } } if (y == coordinates.y - 1 && x == coordinates.x) { // searchDown y = coordinates.y + 1; x = coordinates.x; while (f < lengthThisWord && y < wordSearch.length && wordSearch[y][x] == ch[f]) { if (f == lengthThisWord - 1) { word1.setStartPoint(i, j); word1.setEndPoint(x, y); listOfWords.add(word1); } f++; y++; } } if (y == coordinates.y + 1 && x == coordinates.x) { // searchUp y = coordinates.y - 1; x = coordinates.x; while (f < lengthThisWord && y >= 0 && wordSearch[y][x] == ch[f]) { if (f == lengthThisWord - 1) { word1.setStartPoint(i, j); word1.setEndPoint(x, y); listOfWords.add(word1); } f++; y--; } } if (y == coordinates.y + 1 && x == coordinates.x + 1) { // searchUpLeft y = coordinates.y - 1; x = coordinates.x - 1; while (f < lengthThisWord && x >= 0 && y >= 0 && wordSearch[y][x] == ch[f]) { if (f == lengthThisWord - 1) { word1.setStartPoint(i, j); word1.setEndPoint(x, y); listOfWords.add(word1); } f++; y--; x--; } } if (y == coordinates.y + 1 && x == coordinates.x - 1) { // searchUpRight y = coordinates.y - 1; x = coordinates.x + 1; while (f < lengthThisWord && y >= 0 && x < wordSearch[0].length && wordSearch[y][x] == ch[f]) { if (f == lengthThisWord - 1) { word1.setStartPoint(i, j); word1.setEndPoint(x, y); listOfWords.add(word1); } f++; y--; x++; } } if (y == coordinates.y - 1 && x == coordinates.x + 1) { // searchDownLeft y = coordinates.y + 1; x = coordinates.x - 1; while (f < lengthThisWord && y < wordSearch.length && x >= 0 && wordSearch[y][x] == ch[f]) { if (f == lengthThisWord - 1) { word1.setStartPoint(i, j); word1.setEndPoint(x, y); listOfWords.add(word1); } f++; y++; x--; } } if (y == coordinates.y - 1 && x == coordinates.x - 1) { // searchDownRight y = coordinates.y + 1; x = coordinates.x + 1; while (f < lengthThisWord && x < wordSearch[0].length && y < wordSearch.length && wordSearch[y][x] == ch[f]) { if (f == lengthThisWord - 1) { word1.setStartPoint(i, j); word1.setEndPoint(x, y); listOfWords.add(word1); } f++; y++; x++; } } } } } } } } } return listOfWords; } // method searching for accurate neighbor public static Coordinates searchNeighbor(int[][] wordSearch, Coordinates co, char ch) { for (int y = co.y - 1; y <= co.y + 1; y++) { for (int x = co.x - 1; x <= co.x + 1; x++) { if (y < 0 || y >= wordSearch.length) continue; if (x < 0 || x >= wordSearch[0].length) continue; if (wordSearch[y][x] == ch) { return new Coordinates(y, x); } } } return null; } public static class Word { private String text; private int startX; private int startY; private int endX; private int endY; public Word(String text) { this.text = text; } public void setStartPoint(int i, int j) { startX = i; startY = j; } public void setEndPoint(int i, int j) { endX = i; endY = j; } @Override public String toString() { return String.format("%s - (%d, %d) - (%d, %d)", text, startX, startY, endX, endY); } } }

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Thomas

Level 41 , Bayreuth, Germany

25 February 2023, 12:40

useful

I think that my be the problem. You need to return all occurrences. You have a lot of redundancies inside your code. Maybe you want to remove them. That is a nice little task and I like it. Have now tried to write a solution for this and here is what I came up with. Probably not the best code but you can get an idea on how you might possibly improve your program. I added a good bunch of comments so it shouldn't be a problem for you to get what I mean.

public static List<Word> detectAllWords(int[][] wordSearch, String... words) {
    // list that holds all the results
    List<Word> results = new ArrayList<>();

    // loop over the words array and test each word
    for (String word : words) {
        char firstLetter = word.charAt(0);

        // loop through the char field
        for (int y = 0; y < wordSearch.length; y++) {
            for (int x = 0; x < wordSearch[y].length; x++) {
                if (wordSearch[y][x] == firstLetter) {
                    // if the current char of the field matches the first char of the search word then inspect
                    // further inside the checkWord method
                    checkWord(x, y, wordSearch, word, results);
                }
            }
        }
    }
    return results;
}

public static boolean inField(int x, int y, int[][] arr) {
    return x >= 0 && y >= 0 && x < arr[0].length && y < arr.length;
}

Thomas

Level 41 , Bayreuth, Germany

25 February 2023, 12:41

solution

public static void checkWord(int startX, int startY, int[][] arr, String s, List<Word> results) {
    char[] word = s.toCharArray();

    // dirX, dirY specify the direction to test
    for (int dirY = -1; dirY <= 1; dirY++) {
        for (int dirX = -1; dirX <= 1; dirX++) {

            // skip if x == 0 and y == 0 as this is no valid direction
            if (dirX == 0 && dirY == 0)
                continue;

            // x,y: the position of the current char to test against word, also the end position
            int
                    x = startX,
                    y = startY;

            // index [0] of word already matched inside the detectAllWords lool.
            // Now we check each other char of the word
            int pos = 1; // start index of the char to test
            while (pos < word.length) {

                // the direction to test is x/y, eg, in the first iteration -1/-1
                // that direction we add to the start positions x and y and have a position to test
                x += dirX;
                y += dirY;

                // check if this position is still inside the array
                if (!inField(x, y, arr))
                    break;

                // check if the char matches
                if (arr[y][x] != word[pos])
                    break;

                // check if we reached the end of the word, if yes, add it to the results list
                if (++pos == word.length) {
                    results.add(new Word(s)
                            .setStartPoint(startX, startY)
                            .setEndPoint(x, y));
                }
            }
        }
    }
}

I didn't like the Word class and setting positions so I made that fluent. Don't know if it'll validate this

Thomas

Level 41 , Bayreuth, Germany

25 February 2023, 12:47

useful

Another possibility for making that direction checking is to use position arrays like

int dirX[] = new int[] {-1, 0, 1, -1, 1, -1, 0, 1};
int dirY[] = new int[] {-1, -1, -1, 0, 0, 1, 1, 1};

for (int i = 0; i < dirX.length; i++) {
    System.out.format("direction: x=%d, y=%d%n", dirX[i], dirY[i]);
}

matemate123

Level 50 , Kraków, Poland

25 February 2023, 15:06

Thanks a lot Thomas, like always! Yeah I saw something similar like your code on GeeksForGeeks, firstly I'll try accomplish multiple occurence, and then I' ll thoroughly analize your code and try improve my code. PS: I have multtiple occurence but in this example I found why second same don't show... I have a lot to work.

matemate123

Level 50 , Kraków, Poland

26 February 2023, 10:46

I'm really impressed your Solution. So powerfull, yet so easy to read code. And static method inField really helps narrow my code!

Thomas

Level 41 , Bayreuth, Germany

27 February 2023, 09:31

useful

That's great. This was a really hard task for me when I coded it the first time. I knew that there has to be a possibility to avoid coding the same stuff for each direction but I didn't have any idea how to do it. After fiddling around for a few hours I had that super cool array idea and solved it that way. Still it was a mess with nested loops over and over. Hard to read, hard to understand. When you have more experience with coding you'll do some things without thinking about them anymore. Like that above approach. Instead of a lot of nesting you can make use of methods. And, again instead of nesting you can check some conditions at the start of a code block and drop out of the block (here break and continue). That makes it easier readable. When using methods I find it also useful to do some basic checking at the beginning and use the return statement instead of using if-else clauses. Most important is, that you solve the task. It's not about writing pretty code but to understand arrays and how to work with them. And now you also have seen how to check array borders. This is really very useful sometimes ;)

matemate123

Level 50 , Kraków, Poland

27 February 2023, 10:00

This method for check array borders I will use henceforth when I step on tasks similar to that. I never do this without trying before minesweaper game. Sincerely, it's very hard learning programming when I'm 31 old, and my math education (and other technical) was only good untill I was 12-13 years old. Sincerely 2x I'm not proud of myself in this task becouse, I start doing it 5:43 AM and I little looking for inspiration for your so ingenious simple solution, I needed little peek for hint :( I need many many months to work with my brain, many practice but I'm not giving up I hope. I conclude from what you write, that brain someway adapt for programming thinking, maybe it's something like subconciousness ? :D

Thomas

Level 41 , Bayreuth, Germany

27 February 2023, 12:30

useful

I'm sure that's the case. With constant practice, your skills develop further and it also becomes easier for you to recognize connections and use what you have already learned in a meaningful way. That's what it means to think like a programmer ;) It just takes so much time. The age is also not a reason. I am quite a bit older than you, not to say a real old fart. Whereas you are a teenager, hahaha. Motivation counts and you have it. Now just keep at it and have patience. As I said, it takes time.