implement Bonus exercises

src/main/java/BonusExercises.java

@@ -5,68 +5,84 @@ import java.util.regex.Pattern;
 
 public class BonusExercises {
 
-    /*
-        complete the method below, so it will validate an email address
-        1. Must have exactly one @ (no more, no less)
-        2. Split into local-part and domain (before @ and after @)
-        3. Local-part rules:
-            - Can't be empty
-            - Can't start or end with dot
-            - Can't have two dots in a row
-        4. Domain rules:
-            - Can't be empty
-            - Can't start or end with hyphen
-            - Can't have underscores
-            - Each segment (between dots) must follow same hyphen rules
-     */
     public boolean validateEmail(String email) {
-        String regex = ""; // todo
+        String regex = "^(?!\\.)(?!.*\\.\\..*)([0-9A-Za-z._]+)(?<!\\.)@(?!-)([0-9A-Za-z.\\-]+)(?<!-)$";
+
         Pattern pattern = Pattern.compile(regex);
         Matcher matcher = pattern.matcher(email);
 
         return matcher.matches();
     }
 
-    /*
-        This method should find and return the first date in a string.
 
-        Supported formats:
-        - American:   MM/DD/YYYY   (e.g., 12/09/2023)
-        - British:    DD/MM/YYYY   (e.g., 12/09/2023 — same pattern, context matters)
-        - ISO:        YYYY-MM-DD   (e.g., 2024-07-15)
-        - Slash variant: YYYY/MM/DD (e.g., 2025/01/01)
-
-        If no match for a date is found in the string, return null.
-    */
     public String findDate(String string) {
-        // todo
+        String regex = "(\\b((1[0-2])|(0[1-9]))\\b/\\b((0[1-9])|([12][0-9])|(3[01]))\\b/\\b(\\d{4})\\b)|(\\b((0[1-9])|([12][0-9])|(3[01]))\\b/\\b((1[0-2])|(0[1-9]))\\b/\\b(\\d{4})\\b)|(\\b(\\d{4})\\b-\\b((1[0-2])|(0[1-9]))\\b-\\b((0[1-9])|([12][0-9])|(3[01]))\\b)|(\\b(\\d{4})\\b/\\b((1[0-2])|(0[1-9]))\\b/\\b((0[1-9])|([12][0-9])|(3[01]))\\b)";
+
+        Pattern pattern = Pattern.compile(regex);
+        Matcher matcher = pattern.matcher(string);
+
+        if (matcher.find()) {
+            return matcher.group();
+        }
+
         return null;
     }
 
-    /*
-        given a string, implement the method to detect all valid passwords
-        then, it should return the count of them
 
-        a valid password has the following properties:
-        - at least 8 characters
-        - has to include at least one uppercase letter, and at least a lowercase
-        - at least one number and at least a special char "!@#$%^&*"
-        - has no white-space in it
-     */
     public int findValidPasswords(String string) {
-        // todo
+        String passwordRegex = "(?=.*[A-Z])(?=.*[a-z])(?=.*\\d)(?=.*\\W).{8,}";
-        return -1;
+        String tokenRegex = "(?<!\\S)[A-Za-z0-9!@#$%^&*_]{8,}(?!\\S)";
+
+        Pattern tokenPattern = Pattern.compile(tokenRegex);
+        Pattern passwordPattern = Pattern.compile(passwordRegex);
+        Matcher matcher = tokenPattern.matcher(string);
+
+        int count = 0;
+        while (matcher.find()) {
+            String token = matcher.group();
+
+            Matcher passwordMatcher = passwordPattern.matcher(token);
+            if (passwordMatcher.matches()){
+                count++;
+            }
         }
 
-    /*
+        return count;
-        you should return a list of *words* which are palindromic
+    }
-        by word we mean at least 3 letters with no whitespace in it
 
-        note: your implementation should be case-insensitive, e.g. Aba -> is palindrome
-     */
     public List<String> findPalindromes(String string) {
         List<String> list = new ArrayList<>();
-        // todo
+        String[] inputList = string.split("\\W+");
+
+        for (String word : inputList)
+        {
+            String regex = "";
+            int wordLength = word.length();
+
+            if (wordLength < 3) continue;
+
+            for (int i = 0; i < wordLength/2; i++) {
+                regex += "(.)";
+            }
+
+            if (wordLength % 2 != 0)
+            {
+                regex += ".";
+            }
+
+            for (int i = wordLength/2; i > 0; i--) {
+                regex += "\\" + i;
+            }
+
+            Pattern pattern = Pattern.compile(regex, Pattern.CASE_INSENSITIVE);
+            Matcher matcher = pattern.matcher(word);
+
+            if (matcher.matches())
+            {
+                list.add(word);
+            }
+        }
+
         return list;
     }
 

src/main/java/MainExercises.java

@@ -1,98 +1,103 @@
+import java.util.ArrayList;
+
 public class MainExercises
 {
-    /*
-    you should create a triangle with "*" and return a two-dimensional array of characters based on that
-    the triangle's area is empty, which means some characters should be " "
-
-    example 1, input = 3:
-    *
-    **
-    ***
-
-    example 2, input = 5:
-    *
-    **
-    * *
-    *  *
-    *****
-
-    the output has to be a two-dimensional array of characters, so don't just print the triangle!
- */
     public char[][] generateTriangle(int n) {
 
-        // todo
+        char[][] result = new char[n][];
-        return null;
         
+        for (int i = 0; i < n; i++)
+        {
+            char[] line = new char[i+1];
+
+            if (i == n-1 || i == 0)
+            {
+                for (int j = 0; j <= i; j++)
+                {
+                    line[j] = '*';
+                }
+            }
+            else 
+            {
+                line[0] = '*';
+                line[i] = '*';
+
+                for (int j = i - 1; j > 0; j--) {
+                    line[j] = ' ';
+                }
             }
 
+            result[i] = line;
+        }
 
+        return result;
+    }
 
-    /*
-       SPIRAL TRAVERSAL OF A RECTANGULAR MATRIX
-
-       Given a rectangular matrix (2D array) of integers, this method traverses
-       it in a spiral order (clockwise from outside to inside) and returns
-       the elements as a 1D array.
-
-       EXAMPLE:
-       Input matrix:
-           1  2  3
-           4  5  6
-           7  8  9
-
-       Spiral order: start at top-left (1), go right →, then down ↓,
-       then left ←, then up ↑, then repeat inward.
-
-       Result: {1, 2, 3, 6, 9, 8, 7, 4, 5}
-
-       so you should walk in that matrix in a curl and then add the numbers in order you've seen them in a 1D array
-
-       RECTANGULAR MATRIX ASSUMPTION:
-       This method assumes the input matrix is RECTANGULAR (all rows have
-       the same number of columns). In Java, we can verify this because
-       2D arrays might be jagged (rows of different lengths).
-
-       IMPORTANT: In Java, we do NOT need to pass rows and cols!
-       The 2D array 'matrix' knows its own dimensions:
-           - Number of rows:    matrix.length
-           - Number of columns: matrix[0].length (if rectangular)
-   */
     public int[] spiralTraversal(int[][] matrix) {
-        // todo
+        int rows = matrix.length, cols = matrix[0].length;
-        return null;
+        int elementCount = rows * cols, curElement = 0;
+        int[] result = new int[elementCount];
+
+        for (int leyer = 0; curElement < elementCount; leyer++) {
+            int curi = leyer, maxi = rows - 1 - leyer;
+            int curj = leyer, maxj = cols - 1 - leyer;
+
+            for (;curj < maxj && curElement < elementCount; curj++, curElement++)
+            {
+                result[curElement] = matrix[curi][curj];
+            }
+            for (;curi < maxi && curElement < elementCount; curi++, curElement++)
+            {
+                result[curElement] = matrix[curi][curj];
+            }
+            for (;curj > leyer && curElement < elementCount; curj--, curElement++)
+            {
+                result[curElement] = matrix[curi][curj];
+            }
+            for (;curi > leyer && curElement < elementCount; curi--, curElement++)
+            {
+                result[curElement] = matrix[curi][curj];
             }
 
-    /*
+            if (curi == maxi || curj == maxj) {
-        integer partitioning is a combinatorics problem in discreet maths
+                if (curi == maxi && curj == maxj) result[curElement] = matrix[curi][curj];
-        the problem is to generate sum numbers which their summation is the input number
+                break;
+            }
+        }
         
-        e.g. 1 -> all partitions of integer 3 are:
+        return result;
-        3
+    }
-        2, 1
-        1, 1, 1
-
-        e.g. 2 -> for number 4 goes as:
-        4
-        3, 1
-        2, 2
-        2, 1, 1
-        1, 1, 1, 1
-
-        Note: As you can see in the examples, we want to generate distinct partitions,
-        which means 1,2 and 2,1 are not different — they count as the same combination.
-
-        You should generate all partitions of the input number.
-
-        Hint: You can determine the size and order of the arrays by finding the pattern
-        of partitions and their count. Trust me, this one's fun and easy :)
-
-        If you're familiar with Lists and ArrayLists, you can also edit the method's
-        body to use them instead of arrays.
-    */
 
     public int[][] intPartitions(int n) {
-        // todo
+        ArrayList<int[]> result = new ArrayList<>();
-        return null;
+        ArrayList<Integer> curPartition = new ArrayList<>();
+
+        fillPartition(n, n, curPartition, result);
+
+        return result.toArray(new int[result.size()][]);
+    }
+
+    private void fillPartition(int max, int remaining, ArrayList<Integer> curPartition, ArrayList<int[]> result) {
+        if (remaining == 0) {
+            int size = curPartition.size();
+            int[] savingArr = new int[size];
+
+            for (int i = 0; i < size; i++) {
+                savingArr[i] = curPartition.get(i);
+            }
+
+            result.add(savingArr);
+            return;
+        }
+
+        for (int i = Math.min(max, remaining); i >= 1; i--) {
+            curPartition.add(i);
+            int newRemaining = remaining - i;
+
+            fillPartition(i, newRemaining, curPartition, result);
+
+            curPartition.removeLast();
+        }
     }