How do you solve Matrix Cells in Distance Order on LeetCode?

Matrix Cells in Distance Order (LeetCode #1030) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding Manhattan distance is crucial for this problem.

What is the brute force solution for Matrix Cells in Distance Order?

The brute force approach for Matrix Cells in Distance Order is Brute Force, with O(n²) time complexity and O(n) space complexity. In the brute force approach, we generate all possible cell coordinates in the matrix and calculate their distances from the center cell. We then sort these coordinates based on their distances. This is straightforward but inefficient for larger matrices. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Matrix Cells in Distance Order?

Matrix Cells in Distance Order uses the following concepts: Array, Math, Geometry, Sorting, Matrix. The recommended patterns to study are: Layered generation, Sorting.

What are the interview tips for Matrix Cells in Distance Order?

Explain your thought process clearly when coding. Discuss edge cases, such as the smallest and largest matrices. Practice generating coordinates in a systematic way.

What are common mistakes when solving Matrix Cells in Distance Order?

Forgetting to check bounds when generating coordinates. Not understanding how to calculate the distance correctly.

#1030

Matrix Cells in Distance Order

Easy

ArrayMathGeometrySortingMatrixLayered generationSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal solution leverages the fact that we can generate coordinates based on their distance in layers. Instead of sorting after calculating all distances, we can directly generate the cells in order of their distance from the center.

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Algorithm

4 steps

1Step 1: Initialize an empty list to store cell coordinates.
2Step 2: Use a loop to iterate over possible distances from 0 up to the maximum possible distance.
3Step 3: For each distance, calculate the cells that are at that distance from the center and add them to the list.
4Step 4: Return the list of coordinates.

solution.py14 lines

1# Full working Python code
2from typing import List
3
4def all_cells_dist_order(rows: int, cols: int, rCenter: int, cCenter: int) -> List[List[int]]:
5    result = []
6    for d in range(rows + cols):
7        for r in range(max(0, rCenter - d), min(rows, rCenter + d + 1)):
8            c1 = cCenter - (d - abs(r - rCenter))
9            c2 = cCenter + (d - abs(r - rCenter))
10            if 0 <= c1 < cols:
11                result.append([r, c1])
12            if 0 <= c2 < cols and c1 != c2:
13                result.append([r, c2])
14    return result

ℹ

Complexity note: The time complexity is O(n) because we are generating cells based on their distance without needing to sort them afterward. Each cell is processed once, leading to linear complexity.

1Understanding Manhattan distance is crucial for this problem.
2Generating coordinates in layers helps avoid unnecessary sorting.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.