LeetCode #1030 — EASY

Matrix Cells in Distance Order

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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|.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math

Example 1

Example 2

Example 3

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1030: Matrix Cells in Distance Order
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;
    }
}

// Accepted solution for LeetCode #1030: Matrix Cells in Distance Order
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
}

# Accepted solution for LeetCode #1030: Matrix Cells in Distance Order
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

// Accepted solution for LeetCode #1030: Matrix Cells in Distance Order
struct Solution;

impl Solution {
    fn all_cells_dist_order(r: i32, c: i32, r0: i32, c0: i32) -> Vec<Vec<i32>> {
        let mut cells: Vec<Vec<i32>> = vec![];
        for i in 0..r {
            for j in 0..c {
                cells.push(vec![i, j]);
            }
        }
        cells.sort_unstable_by_key(|v| (v[0] - r0).abs() + (v[1] - c0).abs());
        cells
    }
}

#[test]
fn test() {
    let r = 1;
    let c = 2;
    let r0 = 0;
    let c0 = 0;
    let res: Vec<Vec<i32>> = vec_vec_i32![[0, 0], [0, 1]];
    assert_eq!(Solution::all_cells_dist_order(r, c, r0, c0), res);
    let r = 2;
    let c = 2;
    let r0 = 0;
    let c0 = 1;
    let res: Vec<Vec<i32>> = vec_vec_i32![[0, 1], [0, 0], [1, 1], [1, 0]];
    assert_eq!(Solution::all_cells_dist_order(r, c, r0, c0), res);
    let r = 2;
    let c = 3;
    let r0 = 1;
    let c0 = 2;
    let res: Vec<Vec<i32>> = vec_vec_i32![[1, 2], [0, 2], [1, 1], [0, 1], [1, 0], [0, 0]];
    assert_eq!(Solution::all_cells_dist_order(r, c, r0, c0), res);
}

// Accepted solution for LeetCode #1030: Matrix Cells in Distance Order
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1030: Matrix Cells in Distance Order
// 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;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.