[08 Feb 2020] Minimum number of platforms required for a railway station

Problem:

You are given arrival and departure time of trains reaching to a particular station. You need to find minimum number of platforms required to accommodate the trains at any point of time.
For example:

arrival[] = {1:00, 1:40, 1:50, 2:00, 2:15, 4:00} 

departure[] = {1:10, 3:00, 2:20, 2:30, 3:15, 6:00}

No. of platforms required in above scenario = 4

Please note that arrival time is in chronological order.

Solution :

If you notice we need to find maximum number of trains that can be on the station with the help of arrival and departure time.

Solution 1:

You can iterate over all intervals and check how many other intervals are overlapping with it but that will require o(N^2) time complexity.

 Solution 2:

We will use logic very much similar to merge sort.

• Sort both arrival(arr) and departure(dep) arrays.
• Compare current element in arrival and departure array and pick smaller one among both.
• If element is pick up from arrival array then increment platform_needed.
• If element is pick up from departure array then decrement platform_needed.
• While performing above steps, we need track count of maximum value reached for platform_needed.
• In the end, we will return maximum value reached for platform_needed.

Time complexity : O(NLogN)

Below diagram will make you understand above code better:

Java program for Minimum number of platform required for a railway station

import java.util.Arrays;

 

public class TrainPlatformMain {

public static void main(String args[])

{

// arr[] = {1:00, 1:40, 1:50, 2:00, 2:15, 4:00}

// dep[] = {1:10, 3:00, 2:20, 2:30, 3:15, 6:00}

 

int arr[] = {100, 140, 150, 200, 215, 400};

int dep[] = {110, 300, 210, 230,315, 600};

System.out.println("Minimum platforms needed:"+findPlatformsRequiredForStation(arr,dep,6));

}

 

static int findPlatformsRequiredForStation(int arr[], int dep[], int n)

{

int platform_needed = 0, maxPlatforms = 0;

Arrays.sort(arr);

Arrays.sort(dep);

int i = 0, j = 0;

 

// Similar to merge in merge sort

while (i < n && j < n)

{

if (arr[i] < dep[j])

{

platform_needed++;

i++;

if (platform_needed > maxPlatforms) 

maxPlatforms = platform_needed;

}

else 

{

platform_needed--;

j++;

}

}

return maxPlatforms;

}

}

 

When you run above program, you will get below output:

Minimum platforms needed:4