Number of Platforms Required

Software Development
#1

Given the arrival and departure times of all trains that reach a railway station, the task is to find the minimum number of platforms required for the railway station so that no train waits.
We are given two arrays that represent the arrival and departure times of trains that stop.

Examples:

Input : arr[] = {9:00, 9:40, 9:50, 11:00, 15:00, 18:00}
dep[] = {9:10, 12:00, 11:20, 11:30, 19:00, 20:00}
Output : 3
Explanation: There are at-most three trains at a time (time between 11:00 to 11:20)

Input : arr[] = {9:00, 9:40}
dep[] = {9:10, 12:00}
Output : 1
Explanation: Only one platform is needed.

Naive Solution:

  • Approach: The idea is to take every interval one by one and find the number of intervals that overlap with it. Keep track of the maximum number of intervals that overlap with an interval. Finally, return the maximum value.
  • Algorithm:
    1. Run two nested loops the outer loop from start to end and the inner loop from i+1 to end.
    2. For every iteration of the outer loop find the count of intervals that intersect with the current interval.
    3. Update the answer with the maximum count of overlap in each iteration of the outer loop.
    4. Print the answer.

Implementation:

// Program to find minimum number of platforms

// required on a railway station

#include <algorithm>

#include <iostream>

using namespace std;

// Returns minimum number of platforms required

int findPlatform( int arr[], int dep[], int n)

{

`` // plat_needed indicates number of platforms

`` // needed at a time

`` int plat_needed = 1, result = 1;

`` int i = 1, j = 0;

`` // run a nested loop to find overlap

`` for ( int i = 0; i < n; i++) {

`` // minimum platform

`` plat_needed = 1;

`` for ( int j = i + 1; j < n; j++) {

`` // check for overlap

`` if ((arr[i] >= arr[j] && arr[i] <= dep[j]) ||

`` (arr[j] >= arr[i] && arr[j] <= dep[i]))

`` plat_needed++;

`` }

`` // update result

`` result = max(result, plat_needed);

`` }

`` return result;

}

// Driver Code

int main()

{

`` int arr[] = { 900, 940, 950, 1100, 1500, 1800 };

`` int dep[] = { 910, 1200, 1120, 1130, 1900, 2000 };

`` int n = sizeof (arr) / sizeof (arr[0]);

`` cout << "Minimum Number of Platforms Required = "

`` << findPlatform(arr, dep, n);

`` return 0;

}

Output

Minimum Number of Platforms Required = 3

Complexity Analysis:

  • Time Complexity: O(n^2).
    Two nested loops traverse the array, so the time complexity is O(n^2).
  • Space Complexity: O(1).
    As no extra space is required.

Efficient Solution:

  • Approach: The idea is to consider all events in sorted order. Once the events are in sorted order, trace the number of trains at any time keeping track of trains that have arrived, but not departed.

For example, consider the above example.

arr[] = {9:00, 9:40, 9:50, 11:00, 15:00, 18:00} dep[] = {9:10, 12:00, 11:20, 11:30, 19:00, 20:00} All events are sorted by time. Total platforms at any time can be obtained by subtracting total departures from total arrivals by that time. Time Event Type Total Platforms Needed at this Time 9:00 Arrival 1 9:10 Departure 0 9:40 Arrival 1 9:50 Arrival 2 11:00 Arrival 3 11:20 Departure 2 11:30 Departure 1 12:00 Departure 0 15:00 Arrival 1 18:00 Arrival 2 19:00 Departure 1 20:00 Departure 0 Minimum Platforms needed on railway station = Maximum platforms needed at any time = 3

Note: This approach assumes that trains are arriving and departing on the same date.

Algorithm:

  1. Sort the arrival and departure times of trains.
  2. Create two pointers i=0, and j=0 and a variable to store ans and current count plat
  3. Run a loop while i<n and j<n and compare the ith element of arrival array and jth element of departure array.
  4. If the arrival time is less than or equal to departure then one more platform is needed so increase the count, i.e. plat++ and increment i
  5. Else if the arrival time greater than departure then one less platform is needed so decrease the count, i.e. plat– and increment j
  6. Update the ans, i.e ans = max(ans, plat).

Implementation: This doesn’t create a single sorted list of all events, rather it individually sorts arr[] and dep[] arrays, and then uses the merge process of merge sort to process them together as a single sorted array.

// Program to find minimum number of platforms

// required on a railway station

#include <algorithm>

#include <iostream>

using namespace std;

// Returns minimum number of platforms reqquired

int findPlatform( int arr[], int dep[], int n)

{

`` // Sort arrival and departure arrays

`` sort(arr, arr + n);

`` sort(dep, dep + n);

`` // plat_needed indicates number of platforms

`` // needed at a time

`` int plat_needed = 1, result = 1;

`` int i = 1, j = 0;

`` // Similar to merge in merge sort to process

`` // all events in sorted order

`` while (i < n && j < n) {

`` // If next event in sorted order is arrival,

`` // increment count of platforms needed

`` if (arr[i] <= dep[j]) {

`` plat_needed++;

`` i++;

`` }

`` // Else decrement count of platforms needed

`` else if (arr[i] > dep[j]) {

`` plat_needed--;

`` j++;

`` }

`` // Update result if needed

`` if (plat_needed > result)

`` result = plat_needed;

`` }

`` return result;

}

// Driver code

int main()

{

`` int arr[] = { 900, 940, 950, 1100, 1500, 1800 };

`` int dep[] = { 910, 1200, 1120, 1130, 1900, 2000 };

`` int n = sizeof (arr) / sizeof (arr[0]);

`` cout << "Minimum Number of Platforms Required = "

`` << findPlatform(arr, dep, n);

`` return 0;

}

Output

Minimum Number of Platforms Required = 3

Complexity Analysis:

  • Time Complexity: O(N * log N).
    One traversal O(n) of both the array is needed after sorting O(N * log N), so the time complexity is O(N * log N).
  • Space Complexity: O(1).
    As no extra space is required.