✅✅✅ SIMPLE AND EASY SOLUTION ✅✅✅||✅✅✅USING JAVA AND PYTHON WITH EXPLANATION ✅✅✅
PLEASE UPVOTE IF YOU FIND IT HELPFULL\n# Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem requires us to traverse a grid in 
# PLEASE UPVOTE IF YOU FIND IT HELPFULL\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem requires us to traverse a grid in a spiral order starting from a given cell. The idea is to simulate the spiral traversal while keeping track of the cells we visit. Since we need to visit every cell in the grid, we can accomplish this by iterating in a controlled manner.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n Direction Array: Define the directions for moving right, down, left, and up.\n### Initialize Variables:\n- num_steps to keep track of the steps taken in the current direction.\n- total_cells to know the total number of cells in the grid.\n- result to store the coordinates of the visited cells.\n- r, c for the starting coordinates, d to keep track of the current direction.\n### Main Loop: \n- Continue until we have visited all cells.\n- For each direction (two passes per loop to ensure the spiral expands outward).\n- Move in the current direction and check if the cell is within bounds.\n- If within bounds, add the cell to the result.\n- Change direction after completing the required steps.\n- Increase the number of steps after completing a full cycle of directions.\n\n# Complexity\n- Time complexity: O(rows\xD7cols)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(rows\xD7cols)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n\n```java []\nimport java.util.*;\n\nclass Solution {\n    public int[][] spiralMatrixIII(int rows, int cols, int rStart, int cStart) {\n        int[][] directions = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};\n        int numSteps = 1;\n        int totalCells = rows * cols;\n        List<int[]> result = new ArrayList<>();\n        int r = rStart, c = cStart;\n        int d = 0;\n\n        while (result.size() < totalCells) {\n            for (int i = 0; i < 2; i++) {\n                for (int j = 0; j < numSteps; j++) {\n                    if (0 <= r && r < rows && 0 <= c && c < cols) {\n                        result.add(new int[]{r, c});\n                    }\n                    if (result.size() == totalCells) {\n                        return convertListToArray(result);\n                    }\n                    r += directions[d][0];\n                    c += directions[d][1];\n                }\n                d = (d + 1) % 4;\n            }\n            numSteps++;\n        }\n\n        return convertListToArray(result);\n    }\n\n    private int[][] convertListToArray(List<int[]> list) {\n        int[][] array = new int[list.size()][2];\n        for (int i = 0; i < list.size(); i++) {\n            array[i] = list.get(i);\n        }\n        return array;\n    }\n\n    public static void main(String[] args) {\n        Solution solution = new Solution();\n        int[][] result = solution.spiralMatrixIII(5, 6, 1, 4);\n        for (int[] coords : result) {\n            System.out.println(Arrays.toString(coords));\n        }\n    }\n}\n\n```\n```Python3 []\nclass Solution:\n    def spiralMatrixIII(self, rows: int, cols: int, rStart: int, cStart: int) -> List[List[int]]:\n        directions=[(0,1),(1,0),(0,-1),(-1,0)]\n        num_steps=1\n        total_cells=rows*cols\n        result=[]\n        r,c=rStart,cStart\n        d=0\n\n        while len(result)<total_cells:\n            for _ in range(2):\n                for _ in range(num_steps):\n                    if 0<=r<rows and 0<=c<cols:\n                        result.append([r,c])\n                    if len(result)==total_cells:\n                        return result\n                    r+=directions[d][0]\n                    c+=directions[d][1]\n                d=(d+1)%4\n            num_steps+=1\n\n        return result\n```\n
['Array', 'Matrix', 'Simulation', 'Java', 'Python3']