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

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)
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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 multimap. The first value of element in multimap 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 multimap and no need to store the times in the array.

 `// 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 multimap. ` `    ``multimap<``int``, ``char``> 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; ` ` `  `    ``multimap<``int``, ``char``>::iterator it = order.begin(); ` ` `  `    ``// Start iterating the multimap. ` `    ``for` `(; it != order.end(); it++) { ` ` `  `        ``// 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; ` `} `

Output:

```3
```

Method 2: If all the arrival and departure occur on the same day then we can use and auxiliary array to compute the required number of the platform in O(n).

Whenever an arrival occurs we increase the count of the required platform when a departure occurs we decrease the required platform at that point of time, after that, we take the cumulative sum, which would provide the required number of platform for all point of time, out of these values maximum value is our answer.

 `// 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 = {}; ` `    ``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; ` `} `

Output:

```3
```

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