Find maximum distance between any city and station

Given the number of cities n numbered from 0 to n-1 and the cities in which stations are located, the task is to find the maximum distance between any city and its nearest station. Note that the cities with stations can be given in any order.

Examples:

Input: numOfCities = 6, stations = [1, 4]
Output: 1

Input: numOfCities = 6, stations = [3]
Output: 3

Input: numOfCities = 6, stations = [3, 1]
Output: 2


  1. The below figure represents the first example containing 6 cities and the cities with stations highlighted with green color. In this case, the farthest cities from their nearest stations are 0, 2, 3 and 5 each at a distance of 1. Hence maximum distance is 1.

  2. In the second example, the farthest city from its nearest station is 0 which is at a distance of 3. Hence maximum distance is 3.

  3. In the third example, the farthest city from its nearest station is 5 which is at a distance of 2. Hence maximum distance is 2.

Approach: There are three possible cases in this problem:

  1. When the farthest city is between two stations.
  2. When the farthest city is on the left side of the first station.
  3. When the farthest city is on the right side of the last station.

Below is the algorithm to solve the above problem:

  • Initialize a boolean array of size n (number of cities) with False. Then mark the values of cities with stations as True
  • Initialize a variable dist with 0. Initialize another variable maxDist with value which is equal to the first city with station (used for case 2).
  • Start looping through all the cities ony by one.
  • If the current city has a station, then assign the maximum of (dist+1)//2 and maxDist to maxDist (used for case 1). Also, assign 0 to dist.
  • Else, increment dist.
  • At the end, return the maximum of dist and maxDist (used for case 3).

Below is the implementation of the above approach:

C++

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// C++ program to calculate the maximum 
// distance between any city
// and its nearest station
#include<bits/stdc++.h>
  
using namespace std;
  
// Function to calculate the maximum 
// distance between any city and its nearest station
int findMaxDistance(int numOfCities,int station[],int n)
    // Initialize boolean list
    bool hasStation[numOfCities + 1] = {false};
      
    // Assign True to cities containing station
    for (int city = 0; city < n; city++)
    {
        hasStation[station[city]] = true;
    }
          
    int dist = 0;
    int maxDist = INT_MAX;
  
    for(int i = 0; i < n; i++)
    {
        maxDist = min(station[i],maxDist);
    }
  
    for (int city = 0; city < numOfCities; city++)
    {
        if (hasStation[city] == true)
        {
            maxDist = max((dist + 1) / 2, maxDist);
            dist = 0;
        }
        else
            dist += 1;
    }
    return max(maxDist, dist);
  
//Driver code
int main()
    int numOfCities = 6;
    int station[] = {3, 1};
    int n = sizeof(station)/sizeof(station[0]);
  
    cout << "Max Distance:" << findMaxDistance(numOfCities,
                                                station, n);
}
  
//This code is contributed by Mohit Kumar 29

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Python

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# Python3 code to calculate the maximum 
# distance between any city and its nearest station
  
# Function to calculate the maximum 
# distance between any city and its nearest station
def findMaxDistance(numOfCities, station):
      
    # Initialize boolean list
    hasStation = [False] * numOfCities
    # Assign True to cities containing station
    for city in station:
        hasStation[city] = True
          
    dist, maxDist = 0, min(station)
  
    for city in range(numOfCities):
        if hasStation[city] == True:
            maxDist = max((dist + 1) // 2, maxDist)
            dist = 0
              
        else:
            dist += 1
              
    return max(maxDist, dist)
      
numOfCities = 6
station = [3, 1]
print("Max Distance:", findMaxDistance(numOfCities, station))

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Output:

Max Distance: 2

Time Complexity: O(n)
Space Complexity: O(n)



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Improved By : mohit kumar 29