885. Spiral Matrix III LeetCode Solution

In this guide, you will get 885. Spiral Matrix III LeetCode Solution with the best time and space complexity. The solution to Spiral Matrix III problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Spiral Matrix III solution in C++
Spiral Matrix III solution in Java
Spiral Matrix III solution in Python
Additional Resources

885. Spiral Matrix III LeetCode Solution image

Problem Statement of Spiral Matrix III

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

Complexity Analysis

Time Complexity: O(\max^2(R, C))
Space Complexity: O(R \cdot C)

885. Spiral Matrix III LeetCode Solution in C++

class Solution {
 public:
  vector<vector<int>> spiralMatrixIII(int rows, int cols, int rStart,
                                      int cStart) {
    const vector<int> dx{1, 0, -1, 0};
    const vector<int> dy{0, 1, 0, -1};
    vector<vector<int>> ans{{rStart, cStart}};

    for (int i = 0; ans.size() < rows * cols; ++i)
      for (int step = 0; step < i / 2 + 1; ++step) {
        rStart += dy[i % 4];
        cStart += dx[i % 4];
        if (0 <= rStart && rStart < rows && 0 <= cStart && cStart < cols)
          ans.push_back({rStart, cStart});
      }

    return ans;
  }
};
/* code provided by PROGIEZ */

885. Spiral Matrix III LeetCode Solution in Java

class Solution {
  public int[][] spiralMatrixIII(int rows, int cols, int rStart, int cStart) {
    final int[] dx = {1, 0, -1, 0};
    final int[] dy = {0, 1, 0, -1};
    List<int[]> ans = new ArrayList<>(List.of(new int[] {rStart, cStart}));

    for (int i = 0; ans.size() < rows * cols; ++i)
      for (int step = 0; step < i / 2 + 1; ++step) {
        rStart += dy[i % 4];
        cStart += dx[i % 4];
        if (0 <= rStart && rStart < rows && 0 <= cStart && cStart < cols)
          ans.add(new int[] {rStart, cStart});
      }

    return ans.stream().toArray(int[][] ::new);
  }
}
// code provided by PROGIEZ

885. Spiral Matrix III LeetCode Solution in Python

class Solution:
  def spiralMatrixIII(self, rows: int, cols: int, rStart: int, cStart: int) -> list[list[int]]:
    dx = [1, 0, -1, 0]
    dy = [0, 1, 0, -1]
    ans = [[rStart, cStart]]
    i = 0

    while len(ans) < rows * cols:
      for _ in range(i // 2 + 1):
        rStart += dy[i % 4]
        cStart += dx[i % 4]
        if 0 <= rStart < rows and 0 <= cStart < cols:
          ans.append([rStart, cStart])
      i += 1

    return ans
 # code by PROGIEZ