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Problem 885 Spiral Matrix III

Problem Statement

Link - Problem 885

Question

You start at the cell (rStart, cStart) of an rows x cols grid facing east. The northwest corner is at the first row and column in the grid, and the southeast corner is at the last row and column.

You will walk in a clockwise spiral shape to visit every position in this grid. Whenever you move outside the grid's boundary, we continue our walk outside the grid (but may return to the grid boundary later.). Eventually, we reach all rows * cols spaces of the grid.

Return an array of coordinates representing the positions of the grid in the order you visited them.

Example 1:

Input: rows = 1, cols = 4, rStart = 0, cStart = 0
Output: [[0,0],[0,1],[0,2],[0,3]]

Example 2:

Input: rows = 5, cols = 6, rStart = 1, cStart = 4
Output: [[1,4],[1,5],[2,5],[2,4],
[2,3],[1,3],[0,3],[0,4],
[0,5],[3,5],[3,4],[3,3],
[3,2],[2,2],[1,2],[0,2],
[4,5],[4,4],[4,3],[4,2],
[4,1],[3,1],[2,1],[1,1],
[0,1],[4,0],[3,0],[2,0],
[1,0],[0,0]]

Constraints:

• 1 <= rows, cols <= 100
• 0 <= rStart < rows
• 0 <= cStart < cols

Solution

class Solution {
public:
    vector<vector<int>> spiralMatrixIII(int rows, int cols, int rStart, int cStart) {
        vector<vector<int>> directions = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}}; // East, South, West, North
        vector<vector<int>> result = {{rStart, cStart}};
        int steps = 0, d = 0;

        while (result.size() < rows * cols) {
            if (d == 0 || d == 2) steps++;

            for (int i = 0; i < steps; i++) {
                rStart += directions[d][0];
                cStart += directions[d][1];

                if (rStart >= 0 && rStart < rows && cStart >= 0 && cStart < cols) {
                    result.push_back({rStart, cStart});
                }

                if (result.size() == rows * cols) return result;
            }

            d = (d + 1) % 4;
        }

        return result;
    }
};

Complexity Analysis

| Algorithm               | Time Complexity | Space Complexity |
| ----------------------- | --------------- | ---------------- |
| Spiral Matrix Traversal | O(rows cols)    | O(rows cols)     |

Explanation

Intial Thoughts

To approach this problem, consider the spiral path as a series of connected line segments that turn right at each corner. The key is to track the number of steps taken in each direction and the current direction. Initially, think about the grid as a cartesian plane and visualize how the spiral would unfold. The given code defines four directions: east, south, west, and north, which corresponds to moving right, down, left, and up in the grid. Start by identifying the current position and the total number of steps required to visit every cell in the grid.

Intuitive Analysis

Intuitively, to solve this problem, think about the spiral path as layers of concentric rectangles, where each layer represents a part of the spiral that can fit within the grid. Each layer consists of four line segments, and the number of steps in each segment increases by one after every four turns. The solution iterates through the grid, adjusting the direction and steps as necessary, ensuring that it visits every cell exactly once. Consider visualizing the grid with the given start and end points and imagine tracing the spiral path to better understand the problem’s solution.

1. Intuition

• The problem requires traversing a grid in a clockwise spiral order, starting from a given cell.
• To achieve this, we need to keep track of the current direction and the number of steps to take in that direction.
• The spiral pattern can be divided into four directions: east, south, west, and north, which can be represented by the directions array.
• We need to increase the number of steps every two directions to maintain the spiral pattern.
• The key insight is to check if the current cell is within the grid boundaries before adding it to the result.
• The algorithm should continue until all cells in the grid have been visited.
• The use of the modulo operator (%) ensures that the direction index wraps around to the start of the directions array.

2. Implementation

• The directions array is defined as {{0, 1}, {1, 0}, {0, -1}, {-1, 0}} to represent the four directions.
• The steps variable is used to keep track of the number of steps to take in the current direction, and is incremented every two directions.
• The d variable is used as an index into the directions array to determine the current direction.
• The line rStart += directions[d][0]; cStart += directions[d][1]; updates the current cell coordinates based on the current direction.
• The conditional statement if (rStart >= 0 && rStart < rows && cStart >= 0 && cStart < cols) checks if the current cell is within the grid boundaries.
• The line result.push_back({rStart, cStart}); adds the current cell to the result if it is within the grid boundaries.
• The loop condition while (result.size() < rows * cols) ensures that the algorithm continues until all cells in the grid have been visited.

Complexity Analysis

Time Complexity:

• The time complexity of the given algorithm is O(rows _ cols) because in the while loop, we traverse the entire matrix once. The loop runs rows _ cols times in total, where each iteration performs a constant amount of work.
• The dominant operations are the push_back operation on the result vector and the conditional checks in the if statement. These operations are performed within the loop, making the overall time complexity linear with respect to the size of the input matrix.
• The Big O classification of O(rows _ cols) is justified because we drop lower-order terms and ignore constants. Therefore, the time complexity can be simplified to O(rows _ cols), where the input size is rows * cols.

Space Complexity:

• The space complexity of the given algorithm is O(rows _ cols) due to the storage requirements of the result vector, which stores the coordinates of the visited cells. In the worst-case scenario, the result vector will store rows _ cols coordinates.
• The directions vector and other variables only occupy a constant amount of space and do not impact the overall space complexity. The result vector is the primary data structure that affects the space complexity of the algorithm.
• Justification of the space complexity is based on the fact that we need to store all the coordinates of the cells in the matrix. This requires O(rows * cols) space to store the result vector, making the space complexity linear with respect to the size of the input matrix.

Footnote

This question is rated as Medium difficulty.

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