# Leetcode – Matrix Cells in Distance Order Solution

The problem of “Matrix Cells in Distance Order” from Leetcode is a fascinating study into matrix manipulations and algorithmic efficiencies. This challenge pushes one to think spatially and to consider the distances between cells in a structured manner. This article will provide a deep dive into the problem’s statement, its implications, potential solution avenues, and Pythonic implementations.

## Contents:

1. Problem Statement
2. The Essence of the Problem
3. Solution Approaches
4. Python Implementations
5. Testing the Solution
6. Final Thoughts

## 1. Problem Statement:

You are given four integers: R, C, r0, and c0. We are required to return the coordinates of all cells in a R x C matrix, sorted by their distance from the cell (r0, c0) where the distance is calculated using the Manhattan distance (not the traditional Euclidean distance).

### Example:

Input: R = 1, C = 2, r0 = 0, c0 = 0
Output: [[0,0],[0,1]]

## 2. The Essence of the Problem:

The challenge is essentially a sorting problem but not in the traditional sense of sorting numbers. Instead, we’re sorting matrix cells based on their distance to a given cell.

## 3. Solution Approaches:

1. Brute-force Approach: Calculate the Manhattan distance for each cell in the matrix to (r0, c0) and then sort based on this distance.
2. Bucket Sort Approach: Since we know the maximum possible distance in advance, we can make use of bucket sort for a more efficient solution.

## 4. Python Implementations:

### 1. Brute-force Approach:

def allCellsDistOrder(R: int, C: int, r0: int, c0: int):
# Calculate the Manhattan distance
def distance(r1, c1, r2, c2):
return abs(r1 - r2) + abs(c1 - c2)

# List all cells
cells = [(r, c) for r in range(R) for c in range(C)]

# Sort cells by their distance to (r0, c0)
cells.sort(key=lambda x: distance(r0, c0, x, x))

return cells

### 2. Bucket Sort Approach:

def allCellsDistOrderBucket(R: int, C: int, r0: int, c0: int):
# Calculate the Manhattan distance
def distance(r1, c1, r2, c2):
return abs(r1 - r2) + abs(c1 - c2)

# The maximum possible distance is R+C-2, so create buckets for each distance
buckets = [[] for _ in range(R + C - 1)]

# Place cells in appropriate buckets based on their distance to (r0, c0)
for r in range(R):
for c in range(C):
d = distance(r0, c0, r, c)
buckets[d].append([r, c])

# Flatten the list of buckets
return [cell for bucket in buckets for cell in bucket]

This bucket sort approach optimizes the sorting step, resulting in a more efficient solution compared to the brute-force method.

## 5. Testing the Solution:

After implementation, it’s crucial to test the solution for correctness:

print(allCellsDistOrder(1, 2, 0, 0))  # Expected output: [[0,0],[0,1]]
print(allCellsDistOrderBucket(1, 2, 0, 0))  # Expected output: [[0,0],[0,1]]

print(allCellsDistOrder(2, 3, 1, 2))  # Possible output: [[1,2],[0,2],[1,1],[0,1],[1,0],[0,0]]
print(allCellsDistOrderBucket(2, 3, 1, 2))  # Possible output: [[1,2],[0,2],[1,1],[0,1],[1,0],[0,0]]

## 6. Final Thoughts:

The “Matrix Cells in Distance Order” problem serves as an excellent example of the power of optimized sorting techniques in solving computational problems. While a brute-force approach provides an intuitive understanding, the optimized bucket sort method showcases the efficiency gains possible with more advanced algorithms.