.idea/misc.xml

@@ -8,7 +8,7 @@
       </list>
     </option>
   </component>
-  <component name="ProjectRootManager" version="2" project-jdk-name="25" project-jdk-type="JavaSDK">
+  <component name="ProjectRootManager" version="2" languageLevel="JDK_21" project-jdk-name="21" project-jdk-type="JavaSDK">
     <output url="file://$PROJECT_DIR$/out" />
   </component>
 </project>

pom.xml

@@ -9,8 +9,8 @@
     <version>1.0-SNAPSHOT</version>
 
     <properties>
-        <maven.compiler.source>25</maven.compiler.source>
+        <maven.compiler.source>21</maven.compiler.source>
-        <maven.compiler.target>25</maven.compiler.target>
+        <maven.compiler.target>21</maven.compiler.target>
         <project.build.sourceEncoding>UTF-8</project.build.sourceEncoding>
     </properties>
 
@@ -21,6 +21,13 @@
             <version>5.11.4</version>
             <scope>test</scope>
         </dependency>
+
+        <dependency>
+            <groupId>org.junit.platform</groupId>
+            <artifactId>junit-platform-launcher</artifactId>
+            <version>1.11.4</version>
+            <scope>test</scope>
+        </dependency>
     </dependencies>
 
     <build>

src/main/java/BonusExercises.java

@@ -5,71 +5,68 @@ 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 = "^[a-zA-Z0-9]+[a-zA-Z0-9._]*@[a-zA-Z0-9]+[a-zA-Z0-9-]*(\\.[a-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
+        if (string == null) {return null;}
+        String regex = "\\b(\\d{4}-(0[1-9]|1[0-2])-(0[1-9]|[12][0-9]|3[01])|\\d{4}/(0[1-9]|1[0-2])/(0[1-9]|[12][0-9]|3[01])|(0[1-9]|[12][0-9]|3[01])/(0[1-9]|1[0-2])/\\d{4})\\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 "!@#$%^&*"
-        - has no white-space in it
-     */
     public int findValidPasswords(String string) {
-        // todo
+        String regex = "^(?=.*[a-z])(?=.*[A-Z])(?=.*\\d)(?=.*[^A-Za-z0-9])\\S{8,}$";
-        return -1;
+        Pattern pattern = Pattern.compile(regex);
+        Matcher matcher = pattern.matcher(string);
+        String[] passwords = string.split(" ");
+        int count = 0;
+        for (String pass: passwords){
+            if (pattern.matcher(pass).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 regex = "\\b[a-zA-Z]{3,}\\b";
+        Pattern pattern = Pattern.compile(regex);
+        Matcher matcher = pattern.matcher(string);
+        while (matcher.find()) {
+            if (CheckPalindrome(matcher.group())) {
+                list.add(matcher.group());
+            }
+        }
         return list;
     }
 
+    private boolean CheckPalindrome(String str) {
+        str = str.toLowerCase();
+        int i = 0;
+        int j = str.length() - 1;
+        while (i < str.length() / 2 ) {
+            if (str.charAt(i) != str.charAt(j)) {
+                return false;
+            }
+            i++;
+            j--;
+
+        }
+        return true;
+    }
+
     public static void main(String[] args) {
         // you can test your code here
     }

src/main/java/MainExercises.java

@@ -1,98 +1,75 @@
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.List;
 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) {
-
+        char[][] triangle = new char[n][];
-        // todo
+        for (int i = 0; i < n; i++) {
-        return null;
+            triangle[i] = new char[ i+1 ];
-
+            for (int j = 0; j <= i; j++) {
+                if(j == 0 || j == i || i == n - 1) {
+                    triangle[i][j] = '*';
+                }
+                else {
+                    triangle[i][j] = ' ';
+                }
+            }
+        }
+        return triangle;
     }
 
-
-
-    /*
-       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[] result = new int[matrix.length * matrix[0].length];
-        return null;
+        int index = 0;
+        while (matrix.length > 0) {
+            for (int i = 0; i < matrix[0].length; i++) {
+                result[index++] = matrix[0][i];
+            }
+            matrix = rotateMatrix(matrix);
+
+        }
+        return result;
     }
 
-    /*
+    private int[][] rotateMatrix(int[][] matrix) {
-        integer partitioning is a combinatorics problem in discreet maths
+        int m  = matrix.length;
-        the problem is to generate sum numbers which their summation is the input number
+        int n = matrix[0].length;
+        int[][] newMatrix = new int[n][m -1];
+        for (int i = 0; i < n; i++) {
+            for (int j = 1; j < m; j++) {
+                newMatrix[i][j -1] = matrix[j][n -1 -i];
+            }
+        }
+        return newMatrix;
+    }
 
-        e.g. 1 -> all partitions of integer 3 are:
-        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
+        List<int[]> result = new ArrayList<>();
-        return null;
+        int[] partition = new int[n];
+        int length = 0;
+        partition[length] = n;
+        while (true) {
+            result.add(Arrays.copyOf(partition, length+1));
+            int rem = 0;
+            while (length >= 0 && partition[length] == 1){
+                rem += partition[length];
+                length--;
+            }
+            if ( length < 0 ) break;
+            partition[length]--;
+            rem++;
+            while (rem > partition[length]) {
+                partition[length+1] = partition[length];
+                rem -= partition[length];
+                length++;
+            }
+            partition[length+1] = rem;
+            length++;
+        }
+        return result.toArray(new int[0][]);
     }