You are given the arrival and departure time of trains reaching a particular station.

You need to find a minimum number of platforms required to accommodate the trains at any point in 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 the arrival time is in chronological order.

### Solution :

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

#### 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