Minimum Number of Platforms Required for a Railway/Bus Station | Set 2 (Set based approach)

• Difficulty Level : Medium
• Last Updated : 06 Sep, 2021

Given arrival and departure times of all trains that reach a railway station, find the minimum number of platforms required for the railway station so that no train waits. We are given two arrays which represent 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
There are at-most three trains at a time
(time between 11:00 to 11:20)

We have already discussed its simple and sorting based solutions in below post.
Minimum Number of Platforms Required for a Railway/Bus Station.
In this post, we are inserting all the arrival and departure times in a multiset. The first value of element in multiset tells the arrival/departure time and second value tells whether it’s arrival or departure represented by ‘a’ or ‘d’ respectively.
If its arrival then do increment by 1 otherwise decrease value by 1. If we are taking the input from STDIN then we can directly insert the times in the multiset and no need to store the times in the array.

C++

 // C++ program to find minimum number of platforms // required on a railway station #include using namespace std;   int findPlatform(int arr[], int dep[], int n) {     // Insert all the times (arr. and dep.)     // in the multiset.     multiset > order;     for (int i = 0; i < n; i++) {           // If its arrival then second value         // of pair is 'a' else 'd'         order.insert(make_pair(arr[i], 'a'));         order.insert(make_pair(dep[i], 'd'));     }       int result = 0;     int plat_needed = 0;       // Start iterating the multiset.     for (auto it : order) {           // If its 'a' then add 1 to plat_needed         // else minus 1 from plat_needed.         if (it.second == 'a')             plat_needed++;         else             plat_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);     cout << "Minimum Number of Platforms Required = "          << findPlatform(arr, dep, n);     return 0; }

Java

 // Java program to find minimum number // of platforms required on a railway station import java.io.*; import java.util.*;   class pair {     int first;     char second;           pair(int key1, char key2)     {         this.first = key1;         this.second = key2;     } }   class GFG{       public static int findPlatform(int arr[], int dep[],                                int n) {           // Insert all the times (arr. and dep.)     // in the ArrayList of pairs.     ArrayList order = new ArrayList<>();     for(int i = 0; i < n; i++)     {         order.add(new pair(arr[i], 'a'));         order.add(new pair(dep[i], 'd'));     }       // Custom sort order ArrayList, first     // by time than by arrival or departure     Collections.sort(order, new Comparator()     {         public int compare(pair p1, pair p2)         {             if (p1.first == p2.first)                 return new Character(p1.second)                     .compareTo(                         new Character(p2.second));                                       return p1.first - p2.first;         }     });           int result = 1;     int plat_needed = 0;           for(int i = 0; i < order.size(); i++)     {         pair p = order.get(i);                   // If its 'a' then add 1 to plat_needed         // else minus 1 from plat_needed.         if (p.second == 'a')             plat_needed++;         else             plat_needed--;                       if (plat_needed > result)             result = plat_needed;     }     return result; }   // Driver Code public static void main(String[] args) {     int arr[] = { 900, 940, 950, 1100, 1500, 1800 };     int dep[] = { 910, 1200, 1120, 1130, 1900, 2000 };     int n = 6;           System.out.println("Minimum Number of " +                        "Platforms Required = " +                        findPlatform(arr, dep, n)); } }   // This code is contributed by RohitOberoi

Python3

 # Python3 program to find minimum number # of platforms required on a railway station def findPlatform(arr, dep, n):           # Inserting all the arr. and dep. times     # in the array times     times = []     for i in range(n):         times.append([dep[i], 'd'])         times.append([arr[i], 'a'])               # Sort the array     times = sorted(times, key = lambda x: x)     times = sorted(times, key = lambda x: x)           result, plat_needed = 0, 0       for i in range(2 * n):                   # If its 'a' then add 1 to plat_needed         # else minus 1 from plat_needed.         if times[i] == 'a':             plat_needed += 1             result = max(plat_needed, result)         else:             plat_needed -= 1           return result   # Driver code arr = [ 900, 940, 950, 1100, 1500, 1800 ] dep = [ 910, 1200, 1120, 1130, 1900, 2000 ] n = len(arr)   print("Minimum Number of Platforms Required =",       findPlatform(arr, dep, n))   # This code is contributed by Tharun Reddy

Output:

Minimum Number of Platforms Required = 3

Complexity Analysis:

• Time Complexity: O( N* LogN).

Since we are inserting into multiset and it maintain elements in sorted order. So N insert operations in multiset requires N*LogN time complexity.

• Space Complexity: O(N).

We are using multiset which will have 2*N elements .

C++

 // C++ program to find minimum number of platforms // required on a railway station #include using namespace std;   int minPlatform(int arrival[], int departure[], int n) {       // as time range from 0 to 2359 in 24 hour clock,     // we declare an array for values from 0 to 2360     int platform = {0};     int requiredPlatform = 1;     for (int i = 0; i < n; i++) {           // increment the platforms for arrival         ++platform[arrival[i]];            // once train departs we decrease the platform count         --platform[departure[i] + 1];     }       // We are running loop till 2361 because maximum time value     // in a day can be 23:60     for (int i = 1; i < 2361; i++) {           // taking cumulative sum of platform give us required         // number of platform fro every minute         platform[i] = platform[i] + platform[i - 1];         requiredPlatform = max(requiredPlatform, platform[i]);     }     return requiredPlatform; }   // 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);     cout << "Minimum Number of Platforms Required = "          << minPlatform(arr, dep, n);     return 0; }

Java

 // Java program to find minimum number // of platforms required on a railway // station import java.util.*; import java.lang.*;   class GFG{   static int minPlatform(int arrival[],                        int departure[],                        int n) {           // As time range from 0 to 2359 in     // 24 hour clock, we declare an array     // for values from 0 to 2360     int[] platform = new int;     int requiredPlatform = 1;           for(int i = 0; i < n; i++)     {                   // Increment the platforms for arrival         ++platform[arrival[i]];            // Once train departs we decrease          // the platform count         --platform[departure[i] + 1];     }           // We are running loop till 2361 because     // maximum time value in a day can be 23:60     for(int i = 1; i < 2361; i++)     {                   // Taking cumulative sum of platform         // give us required number of         // platform fro every minute         platform[i] = platform[i] +                       platform[i - 1];         requiredPlatform = Math.max(requiredPlatform,                                     platform[i]);     }     return requiredPlatform; }   // Driver code public static void main(String[] args) {     int arr[] = { 900, 940, 950, 1100, 1500, 1800 };     int dep[] = { 910, 1200, 1120, 1130, 1900, 2000 };     int n = arr.length;           System.out.println("Minimum Number of " +                        "Platforms Required = " +                        minPlatform(arr, dep, n)); } }   // This code is contributed by offbeat

Python3

 # Python3 program to find minimum number # of platforms required on a railway station def minPlatform(arrival, departure, n):       # As time range from 0 to 2359 in     # 24 hour clock, we declare an array     # for values from 0 to 2360     platform =  * 2631     requiredPlatform = 1           for i in range(n):           # Increment the platforms for arrival         platform[arrival[i]] += 1           # Once train departs we decrease the         # platform count         platform[departure[i] + 1] -= 1       # We are running loop till 2361 because     # maximum time value in a day can be 23:60     for i in range(1, 2631):           # Taking cumulative sum of         # platform give us required         # number of platform fro every minute         platform[i] = platform[i] + platform[i - 1]         requiredPlatform = max(requiredPlatform,                                platform[i])               return requiredPlatform   # Driver code arr = [ 900, 940, 950, 1100, 1500, 1800 ] dep = [ 910, 1200, 1120, 1130, 1900, 2000 ] n = len(arr)   print("Minimum Number of Platforms Required = ",        minPlatform(arr, dep, n))   # This code is contributed by PawanJain1

C#

 // C# program to find minimum number // of platforms required on a railway // station using System; class GFG {           static int minPlatform(int[] arrival, int[] departure, int n)     {         // As time range from 0 to 2359 in         // 24 hour clock, we declare an array         // for values from 0 to 2360         int[] platform = new int;         int requiredPlatform = 1;                    for(int i = 0; i < n; i++)         {                            // Increment the platforms for arrival             ++platform[arrival[i]];                     // Once train departs we decrease              // the platform count             --platform[departure[i] + 1];         }                    // We are running loop till 2361 because         // maximum time value in a day can be 23:60         for(int i = 1; i < 2361; i++)         {                            // Taking cumulative sum of platform             // give us required number of             // platform fro every minute             platform[i] = platform[i] +                           platform[i - 1];             requiredPlatform = Math.Max(requiredPlatform,                                         platform[i]);         }         return requiredPlatform;     }     static void Main() {     int[] arr = { 900, 940, 950, 1100, 1500, 1800 };     int[] dep = { 910, 1200, 1120, 1130, 1900, 2000 };     int n = arr.Length;            Console.Write("Minimum Number of " +                        "Platforms Required = " +                        minPlatform(arr, dep, n));   } }   // This code is contributed by divyesh072019.

Javascript



Output:

Minimum Number of Platforms Required = 3

Complexity Analysis:

• Time Complexity: O(N).
• Space Complexity: O(1).

This article is contributed by Jatin Goyal and Abhinav Kumar Singh. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.