implement spiralTraversal method in MainExercises

src/main/java/MainExercises.java

@@ -23,41 +23,39 @@ public class MainExercises
         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) {
+                if (curi == maxi && curj == maxj) result[curElement] = matrix[curi][curj];
+                break;
+            }
+        }
+        
+        return result;
     }
 
     /*