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

Given the 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 that 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)```
Recommended Practice

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.

Implementation:

## 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[0]);``    ``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[``1``])``    ``times ``=` `sorted``(times, key ``=` `lambda` `x: x[``0``])``    ` `    ``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][``1``] ``=``=` `'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`

## C#

 `// C# program to find minimum number``// of platforms required on a railway station``using` `System;``using` `System.Collections.Generic;` `public` `class` `pair {``  ``public` `int` `first;``  ``public` `char` `second;` `  ``public` `pair(``int` `key1, ``char` `key2) {``    ``this``.first = key1;``    ``this``.second = key2;``  ``}``}` `public` `class` `GFG {` `  ``public` `static` `int` `findPlatform(``int` `[]arr, ``int` `[]dep, ``int` `n) {` `    ``// Insert all the times (arr. and dep.)``    ``// in the List of pairs.``    ``List order = ``new` `List();``    ``for` `(``int` `i = 0; i < n; i++) {``      ``order.Add(``new` `pair(arr[i], ``'a'``));``      ``order.Add(``new` `pair(dep[i], ``'d'``));``    ``}` `    ``// Custom sort order List, first``    ``// by time than by arrival or departure``    ``order.Sort((p1,p2)=> p1.first == p2.first? p1.second -``               ``p2.second:  p1.first - p2.first);` `    ``int` `result = 1;``    ``int` `plat_needed = 0;` `    ``for` `(``int` `i = 0; i < order.Count; i++) {``      ``pair p = order[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;` `    ``Console.WriteLine(``"Minimum Number of "` `+``                      ``"Platforms Required = "` `+``                      ``findPlatform(arr, dep, n));``  ``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``

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 .

### Using constant space and O(n) time complexity

All the trains will arrive and depart in only one day, i.e. a 24-hour time frame, hence we can use an array and for every arrival increase the value at that time (time in 24-hour format), and for every departure decrement the value at the index. In the array, every index denotes time in 24hr format and the sum of traversal up to that index is the number of platforms required, at that time.

Illustration:

arrivals – { 1200 , 1230 , 1240 , 1245}
departures – { 1210 , 1245 , 1250 , 1250}
make an array of size 2361 having 0 as the default value .
for the above example the array will have the following values – 1 ,-1 , 1 , -1 , 0 , -2
at indices – 1200 , 1210 , 1230 ,1240 ,1245,1250
Rest all indices have the default value of 0,in the above situation the number of platforms required at different times are –
platforms     time
1             1200
2             1210
3             1230
2             1240
2             1245
1             1250
3 is the max so 3 is the answer

Approach:

Follow the below steps to solve the problem:

1. Make an array for 24 hrs i.e. – 2361 size , named time.
2. Now for every arrival increment the element at the index in the array time.
3. For every departure decrement the element (default integer value in time is 0).
4. Then traverse the entire array and keep adding the value at current index to a variable count.
5. The max value of the variable count is the answer.

Implementation:

## 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[2361] = {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 for 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[0]);``    ``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``[``2361``];``    ``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 for 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 ``=` `[``0``] ``*` `2361``    ``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``, ``2361``):` `        ``# Taking cumulative sum of``        ``# platform give us required``        ``# number of platform for 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``[2361];``        ``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 for 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).

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