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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;
}
}
};

• 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.offer(new int[]{r0, c0});
int[][] directions = { {-1, 0}, {0, -1}, {1, 0}, {0, 1} };
while (!queue.isEmpty()) {
int[] cell = queue.poll();
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;
}
}

############

import java.util.Deque;

class Solution {
public int[][] allCellsDistOrder(int rows, int cols, int rCenter, int cCenter) {
Deque<int[]> q = new ArrayDeque<>();
q.offer(new int[]{rCenter, cCenter});
boolean[][] vis = new boolean[rows][cols];
vis[rCenter][cCenter] = true;
int[][] ans = new int[rows * cols][2];
int[] dirs = {-1, 0, 1, 0, -1};
int idx = 0;
while (!q.isEmpty()) {
for (int n = q.size(); n > 0; --n) {
var p = q.poll();
ans[idx++] = p;
for (int k = 0; k < 4; ++k) {
int x = p[0] + dirs[k], y = p[1] + dirs[k + 1];
if (x >= 0 && x < rows && y >= 0 && y < cols && !vis[x][y]) {
vis[x][y] = true;
q.offer(new int[]{x, y});
}
}
}
}
return ans;
}
}

• // 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;
}
}
};

• class Solution:
def allCellsDistOrder(
self, rows: int, cols: int, rCenter: int, cCenter: int
) -> List[List[int]]:
q = deque([(rCenter, cCenter)])
vis = [[False] * cols for _ in range(rows)]
vis[rCenter][cCenter] = True
ans = []
while q:
for _ in range(len(q)):
i, j = q.popleft()
ans.append([i, j])
for a, b in [[1, 0], [-1, 0], [0, 1], [0, -1]]:
x, y = i + a, j + b
if 0 <= x < rows and 0 <= y < cols and not vis[x][y]:
q.append((x, y))
vis[x][y] = True
return ans

############

class Solution(object):
def allCellsDistOrder(self, R, C, r0, c0):
"""
:type R: int
:type C: int
:type r0: int
:type c0: int
:rtype: List[List[int]]
"""
dis = []
for r in range(R):
for c in range(C):
dis.append((abs(r0 - r) + abs(c0 - c), [r, c]))
dis.sort()
return [x for d, x in dis]

• func allCellsDistOrder(rows int, cols int, rCenter int, cCenter int) (ans [][]int) {
q := [][]int{ {rCenter, cCenter} }
vis := make([][]bool, rows)
for i := range vis {
vis[i] = make([]bool, cols)
}
vis[rCenter][cCenter] = true
dirs := [5]int{-1, 0, 1, 0, -1}
for len(q) > 0 {
for n := len(q); n > 0; n-- {
p := q[0]
q = q[1:]
ans = append(ans, p)
for k := 0; k < 4; k++ {
x, y := p[0]+dirs[k], p[1]+dirs[k+1]
if x >= 0 && x < rows && y >= 0 && y < cols && !vis[x][y] {
vis[x][y] = true
q = append(q, []int{x, y})
}
}
}
}
return
}