1030. Matrix Cells in Distance Order

Description

You are given four integers row, cols, rCenter, and cCenter. There is a rows x cols matrix and you are on the cell with the coordinates (rCenter, cCenter).

Return the coordinates of all cells in the matrix, sorted by their distance from (rCenter, cCenter) from the smallest distance to the largest distance. You may return the answer in any order that satisfies this condition.

The distance between two cells (r1, c1) and (r2, c2) is |r1 - r2| + |c1 - c2|.

Example 1:

Input: rows = 1, cols = 2, rCenter = 0, cCenter = 0
Output: [[0,0],[0,1]]
Explanation: The distances from (0, 0) to other cells are: [0,1]


Example 2:

Input: rows = 2, cols = 2, rCenter = 0, cCenter = 1
Output: [[0,1],[0,0],[1,1],[1,0]]
Explanation: The distances from (0, 1) 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: rows = 2, cols = 3, rCenter = 1, cCenter = 2
Output: [[1,2],[0,2],[1,1],[0,1],[1,0],[0,0]]
Explanation: The distances from (1, 2) 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]].


Constraints:

• 1 <= rows, cols <= 100
• 0 <= rCenter < rows
• 0 <= cCenter < cols

Solutions

• 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;
}
}

• class Solution {
public:
vector<vector<int>> allCellsDistOrder(int rows, int cols, int rCenter, int cCenter) {
queue<pair<int, int>> q;
q.emplace(rCenter, cCenter);
vector<vector<int>> ans;
bool vis[rows][cols];
memset(vis, false, sizeof(vis));
vis[rCenter][cCenter] = true;
int dirs[5] = {-1, 0, 1, 0, -1};
while (!q.empty()) {
for (int n = q.size(); n; --n) {
auto [i, j] = q.front();
q.pop();
ans.push_back({i, j});
for (int k = 0; k < 4; ++k) {
int x = i + dirs[k];
int y = j + dirs[k + 1];
if (x >= 0 && x < rows && y >= 0 && y < cols && !vis[x][y]) {
vis[x][y] = true;
q.emplace(x, y);
}
}
}
}
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)):
p = q.popleft()
ans.append(p)
for a, b in pairwise((-1, 0, 1, 0, -1)):
x, y = p[0] + a, p[1] + b
if 0 <= x < rows and 0 <= y < cols and not vis[x][y]:
vis[x][y] = True
q.append([x, y])
return ans


• 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
}