implement spiralTraversal function in MainExercises

src/main/java/MainExercises.java

@@ -1,23 +1,6 @@
 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][];
         for (int i = 0; i < n; i++) {
@@ -34,70 +17,31 @@ public class MainExercises
         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
-        return null;
+        int[] result = new int[matrix.length * matrix[0].length];
+        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;
     }
 
-    /*
-        integer partitioning is a combinatorics problem in discreet maths
-        the problem is to generate sum numbers which their summation is the input number
+    private int[][] rotateMatrix(int[][] matrix) {
+        int m  = matrix.length;
+        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