Formatted question description: https://leetcode.ca/all/1030.html

1030. Matrix Cells in Distance Order (Easy)

We are given a matrix with R rows and C columns has cells with integer coordinates (r, c), where 0 <= r < R and 0 <= c < C.

Additionally, we are given a cell in that matrix with coordinates (r0, c0).

Return the coordinates of all cells in the matrix, sorted by their distance from (r0, c0) from smallest distance to largest distance.  Here, the distance between two cells (r1, c1) and (r2, c2) is the Manhattan distance, |r1 - r2| + |c1 - c2|.  (You may return the answer in any order that satisfies this condition.)

 

Example 1:

Input: R = 1, C = 2, r0 = 0, c0 = 0
Output: [[0,0],[0,1]]
Explanation: The distances from (r0, c0) to other cells are: [0,1]

Example 2:

Input: R = 2, C = 2, r0 = 0, c0 = 1
Output: [[0,1],[0,0],[1,1],[1,0]]
Explanation: The distances from (r0, c0) to other cells are: [0,1,1,2]
The answer [[0,1],[1,1],[0,0],[1,0]] would also be accepted as correct.

Example 3:

Input: R = 2, C = 3, r0 = 1, c0 = 2
Output: [[1,2],[0,2],[1,1],[0,1],[1,0],[0,0]]
Explanation: The distances from (r0, c0) to other cells are: [0,1,1,2,2,3]
There are other answers that would also be accepted as correct, such as [[1,2],[1,1],[0,2],[1,0],[0,1],[0,0]].

 

Note:

  1. 1 <= R <= 100
  2. 1 <= C <= 100
  3. 0 <= r0 < R
  4. 0 <= c0 < C

Solution 1.

// OJ: https://leetcode.com/problems/matrix-cells-in-distance-order/

// Time: O(RC)
// Space: O(1)
class Solution {
public:
    vector<vector<int>> allCellsDistOrder(int R, int C, int r0, int c0) {
        vector<vector<int>> ans;
        ans.push_back(vector<int>{ r0, c0 });
        int starts[4][2] = { {-1, 0}, {0, 1}, {1, 0}, {0, -1} };
        int dirs[4][2] = { {1, 1}, {1, -1}, {-1, -1}, {-1, 1} };
        for (int level = 1;; level++) {
            int cnt = 0;
            for (int i = 0; i < 4; ++i) {
                int x = r0 + level * starts[i][0], y = c0 + level * starts[i][1];
                for (int j = 0; j < level; ++j) {
                    if (x >= 0 && x < R && y >= 0 && y < C) {
                        ans.push_back(vector<int>{ x, y });
                        ++cnt;
                    }
                    x += dirs[i][0];
                    y += dirs[i][1];
                }
            }
            if (!cnt) return ans;
        }
    }
};

Java

class Solution {
    public int[][] allCellsDistOrder(int R, int C, int r0, int c0) {
        List<int[]> allCellsList = new ArrayList<int[]>();
        final int WHITE = 0;
        final int GRAY = 1;
        final int BLACK = 2;
        int[][] colors = new int[R][C];
        colors[r0][c0] = GRAY;
        Queue<int[]> queue = new LinkedList<int[]>();
        queue.offer(new int[]{r0, c0});
        int[][] directions = { {-1, 0}, {0, -1}, {1, 0}, {0, 1} };
        while (!queue.isEmpty()) {
            int[] cell = queue.poll();
            allCellsList.add(cell);
            int row = cell[0], column = cell[1];
            for (int[] direction : directions) {
                int newRow = row + direction[0], newColumn = column + direction[1];
                if (newRow >= 0 && newRow < R && newColumn >= 0 && newColumn < C && colors[newRow][newColumn] == WHITE) {
                    colors[newRow][newColumn] = GRAY;
                    queue.offer(new int[]{newRow, newColumn});
                }
            }
            colors[row][column] = BLACK;
        }
        int length = allCellsList.size();
        int[][] allCellsArray = new int[length][2];
        for (int i = 0; i < length; i++) {
            int[] cell = allCellsList.get(i);
            allCellsArray[i][0] = cell[0];
            allCellsArray[i][1] = cell[1];
        }
        return allCellsArray;
    }
}

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