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1030. Matrix Cells in Distance Order

  • Time: $O(n)$
  • Space: $O(n)$
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class Solution {
 public:
  vector<vector<int>> allCellsDistOrder(int rows, int cols, int rCenter,
                                        int cCenter) {
    const vector<int> dirs{0, 1, 0, -1, 0};
    vector<vector<int>> ans;
    vector<vector<int>> seen(rows, vector<int>(cols));
    queue<pair<int, int>> q{{{rCenter, cCenter}}};
    seen[rCenter][cCenter] = true;

    while (!q.empty()) {
      const auto [i, j] = q.front();
      q.pop();
      ans.push_back({i, j});
      for (int k = 0; k < 4; ++k) {
        const int x = i + dirs[k];
        const int y = j + dirs[k + 1];
        if (x < 0 || x == rows || y < 0 || y == cols)
          continue;
        if (seen[x][y])
          continue;
        seen[x][y] = true;
        q.emplace(x, y);
      }
    }

    return ans;
  }
};
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class Solution {
  public int[][] allCellsDistOrder(int rows, int cols, int rCenter, int cCenter) {
    final int[] dirs = {0, 1, 0, -1, 0};
    List<int[]> ans = new ArrayList<>();
    boolean[][] seen = new boolean[rows][cols];
    Queue<Pair<Integer, Integer>> q = new LinkedList<>(Arrays.asList(new Pair<>(rCenter, cCenter)));
    seen[rCenter][cCenter] = true;

    while (!q.isEmpty()) {
      final int i = q.peek().getKey();
      final int j = q.poll().getValue();
      ans.add(new int[] {i, j});
      for (int k = 0; k < 4; ++k) {
        final int x = i + dirs[k];
        final int y = j + dirs[k + 1];
        if (x < 0 || x == rows || y < 0 || y == cols)
          continue;
        if (seen[x][y])
          continue;
        seen[x][y] = true;
        q.offer(new Pair<>(x, y));
      }
    }

    return ans.toArray(int[][] ::new);
  }
}
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class Solution:
  def allCellsDistOrder(self, rows: int, cols: int, rCenter: int, cCenter: int) -> List[List[int]]:
    dirs = [0, 1, 0, -1, 0]
    ans = []
    q = collections.deque([(rCenter, cCenter)])
    seen = {(rCenter, cCenter)}

    while q:
      i, j = q.popleft()
      ans.append([i, j])
      for k in range(4):
        x = i + dirs[k]
        y = j + dirs[k + 1]
        if x < 0 or x == rows or y < 0 or y == cols:
          continue
        if (x, y) in seen:
          continue
        seen.add((x, y))
        q.append((x, y))

    return ans