Find maximum distance between any city and station in Python

Sometimes we need to find the maximum distance between any city and its nearest station. Given N cities numbered from 0 to N-1 and a list of cities with stations, we need to calculate the furthest any city is from a station.

So, if the input is like N = 6 and stations = [2, 4], then the output will be 2, because city 0 is distance 2 from the nearest station at city 2.

Algorithm Overview

To solve this problem, we will follow these steps ?

• Create a boolean array to mark which cities have stations

• Initialize distance counter and maximum distance found so far

• Iterate through all cities, calculating distances between stations

• For gaps between stations, the maximum distance is at the middle point

• Handle the distance from the last station to the end

Implementation

Let us see the complete implementation to get better understanding ?

def get_max_dist(N, stations):
    # Mark which cities have stations
    station_present = [False] * N
    for city in stations:
        station_present[city] = True
    
    # Initialize variables
    dist = 0
    maximum_dist = min(stations)  # Distance from city 0 to first station
    
    # Iterate through all cities
    for city in range(N):
        if station_present[city] == True:
            # Found a station, calculate max distance in the gap
            maximum_dist = max((dist + 1) // 2, maximum_dist)
            dist = 0
        else:
            # No station, increment distance
            dist += 1
    
    # Handle distance from last station to end
    return max(maximum_dist, dist)

# Test the function
N = 6
stations = [2, 4]
result = get_max_dist(N, stations)
print(f"Maximum distance: {result}")

Maximum distance: 2

How It Works

The algorithm works by identifying gaps between stations and finding the maximum distance any city can be from its nearest station:

1. Cities 0-1: Distance 2 and 1 from station at city 2

2. City 2: Has a station (distance 0)

3. City 3: Distance 1 from stations at cities 2 or 4

4. City 4: Has a station (distance 0)

5. City 5: Distance 1 from station at city 4

Another Example

Let's test with different station positions ?

def get_max_dist(N, stations):
    station_present = [False] * N
    for city in stations:
        station_present[city] = True
    
    dist = 0
    maximum_dist = min(stations)
    
    for city in range(N):
        if station_present[city] == True:
            maximum_dist = max((dist + 1) // 2, maximum_dist)
            dist = 0
        else:
            dist += 1
    
    return max(maximum_dist, dist)

# Test with stations at positions 1 and 5
N = 8
stations = [1, 5]
result = get_max_dist(N, stations)
print(f"N = {N}, stations = {stations}")
print(f"Maximum distance: {result}")

N = 8, stations = [1, 5]
Maximum distance: 2

Conclusion

This algorithm efficiently finds the maximum distance between any city and its nearest station by tracking gaps between stations. The time complexity is O(N) and space complexity is O(N) for the boolean array.