Finding the Shortest Distance to All Buildings: A Comprehensive Guide
Imagine you're designing a new city layout, and you want to ensure that every building has easy access to vital services like fire stations, hospitals, and schools. To achieve this, you need to find the optimal location for these services – a spot that minimizes the total distance residents have to travel to reach them. This is where the "shortest distance from all buildings" problem comes in.
This problem is a classic example of a graph traversal and optimization challenge, often tackled using algorithms like Dijkstra's algorithm or BreadthFirst Search (BFS). Let's break down this concept and explore its applications in detail.
Understanding the Problem
The Problem: Given a grid of cells where '1' represents a building and '0' represents an empty cell, the goal is to find a cell that minimizes the total Manhattan distance to all buildings. The Manhattan distance is the sum of the horizontal and vertical distances between two cells.
Example:
Grid:
[
[1, 0, 2, 0, 1],
[0, 0, 0, 0, 0],
[0, 0, 1, 0, 0]
]
In this example, we need to find the cell with the shortest total distance to the three buildings (marked as '1').
Solving the Problem: A StepbyStep Approach

Initialization:
 Create a distance matrix to store the shortest distance from each cell to all buildings.
 Initialize all distances to infinity.

BuildingWise Iteration:
 Iterate through each building.
 Use BFS to calculate the shortest distance from each building to all other cells.
 Update the distance matrix for each cell.

Finding the Optimal Cell:
 Iterate through the distance matrix.
 Find the cell with the minimum sum of distances to all buildings.
Code Example (Python)  Based on this GitHub repository:
from collections import deque
def shortestDistance(grid):
rows, cols = len(grid), len(grid[0])
total_buildings = 0
distances = [[float('inf') for _ in range(cols)] for _ in range(rows)]
# Count total buildings and mark all buildings as visited
for row in range(rows):
for col in range(cols):
if grid[row][col] == 1:
total_buildings += 1
grid[row][col] = 2 # Mark as visited
# Calculate distances from each building
for row in range(rows):
for col in range(cols):
if grid[row][col] == 2:
bfs(grid, row, col, distances, total_buildings)
# Find cell with the minimum distance
min_distance = float('inf')
min_row, min_col = 1, 1
for row in range(rows):
for col in range(cols):
if distances[row][col] != float('inf') and distances[row][col] < min_distance:
min_distance = distances[row][col]
min_row, min_col = row, col
return [min_row, min_col] if min_distance != float('inf') else [1, 1]
def bfs(grid, row, col, distances, total_buildings):
rows, cols = len(grid), len(grid[0])
queue = deque([(row, col, 0, 1)]) # (row, col, distance, visited buildings)
directions = [(0, 1), (0, 1), (1, 0), (1, 0)]
while queue:
r, c, distance, visited_buildings = queue.popleft()
if visited_buildings == total_buildings:
distances[r][c] = min(distances[r][c], distance)
continue
for dr, dc in directions:
new_r, new_c = r + dr, c + dc
if 0 <= new_r < rows and 0 <= new_c < cols and grid[new_r][new_c] == 0:
queue.append((new_r, new_c, distance + 1, visited_buildings + 1))
grid[new_r][new_c] = 1 # Mark as visited during this iteration
# Test case
grid = [
[1, 0, 2, 0, 1],
[0, 0, 0, 0, 0],
[0, 0, 1, 0, 0]
]
optimal_cell = shortestDistance(grid)
print(f"Optimal cell: {optimal_cell}")
Key Points to Remember
 The shortest distance from all buildings might not always exist. In cases where no cell can reach all buildings, the algorithm will return [1, 1].
 The complexity of the algorithm is O(MNK), where M and N are the dimensions of the grid and K is the total number of buildings.
Applications in the Real World
The concept of finding the shortest distance from all buildings has diverse applications beyond urban planning. Here are some examples:
 Emergency Response: Finding the optimal location for an ambulance or fire station to minimize response time.
 Supply Chain Management: Optimizing the location of distribution centers to minimize transportation costs.
 Network Design: Finding the best location for network hubs to minimize data transfer latency.
 Robotics: Enabling robots to navigate efficiently through a space with multiple objectives.
Conclusion
The "shortest distance from all buildings" problem presents a compelling challenge with practical implications across various domains. By understanding the problem, exploring the solution approach, and applying it to realworld scenarios, we can unlock significant value in efficiency, optimization, and resource allocation. Remember, the key is to find the sweet spot where every building benefits from a central location, minimizing travel time and maximizing accessibility.