# Minimum Number of Platforms Required for a Railway/Bus Station

• Difficulty Level : Medium
• Last Updated : 04 May, 2022

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 9:40 to 12:00)

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

Naive 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.

• Run two nested loops, the outer loop from start to end and the inner loop from i+1 to end.
• For every iteration of the outer loop, find the count of intervals that intersect with the current interval.
• Update the answer with the maximum count of overlap in each iteration of the outer loop.

Implementation:

## C++14

 `// C++ program to implement the above approach``#include ``using` `namespace` `std;` `// Function to find the 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` `(max(arr[i], arr[j]) <= min(dep[i], dep[j]))``                ``plat_needed++;``        ``}` `        ``// update result``        ``result = max(result, plat_needed);``    ``}` `    ``return` `result;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 9775, 494, 252, 1680 };``    ``int` `dep[] = { 2052, 2254, 1395, 2130 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``cout << findPlatform(arr, dep, n);``    ``return` `0;``}`

## C

 `// C program to find minimum number of platforms required on``// a railway station` `// Importing the required header files``#include ` `// Creating MACRO for finding the maximum number``#define max(x, y) (((x) > (y)) ? (x) : (y))` `// Creating MACRO for finding the minimum number``#define min(x, y) (((x) < (y)) ? (x) : (y))` `// Function to returns the 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` `(max(arr[i], arr[j]) <= min(dep[i], dep[j]))``                ``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]);``    ``printf``(``"%d"``, findPlatform(arr, dep, n));``    ``return` `0;``}`

## Python3

 `# Program to find minimum number of platforms``# required on a railway station`  `def` `findPlatform(arr, dep, n):``    ``'''``    ``Accepts two arrays with arrival and departure time``    ``and the size of the array``    ``Returns minimum number of platforms required``    ``'''` `    ``# plat_needed indicates number of platforms``    ``# needed at a time``    ``plat_needed ``=` `1``    ``result ``=` `1` `    ``# run a nested loop to find overlap``    ``for` `i ``in` `range``(n):``        ``# minimum platform needed``        ``plat_needed ``=` `1` `        ``for` `j ``in` `range``(i``+``1``, n):``            ``# check for overlap``            ``if` `(``max``(arr[i], arr[j]) <``=` `min``(dep[i], dep[j])):``                ``plat_needed ``+``=` `1` `        ``# update result``        ``result ``=` `max``(result, plat_needed)` `    ``return` `result` `# Driver code`  `def` `main():``    ``arr ``=` `[``900``, ``940``, ``950``, ``1100``, ``1500``, ``1800``]``    ``dep ``=` `[``910``, ``1200``, ``1120``, ``1130``, ``1900``, ``2000``]` `    ``n ``=` `len``(arr)` `    ``print``(``"{}"``.``format``(``        ``findPlatform(arr, dep, n)))`  `if` `__name__ ``=``=` `'__main__'``:``    ``main()`

## Java

 `// Program to find minimum number of platforms``// required on a railway station``import` `java.io.*;` `class` `GFG {``    ``// Returns minimum number of platforms required``    ``public` `static` `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` `(i = ``0``; i < n; i++) {``            ``// minimum platform``            ``plat_needed = ``1``;` `            ``for` `(j = i + ``1``; j < n; j++) {``                ``// check for overlap``                ``if` `(Math.max(arr[i], arr[j])``                    ``<= Math.min(dep[i], dep[j]))``                    ``plat_needed++;``            ``}` `            ``// update result``            ``result = Math.max(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(findPlatform(arr, dep, n));``    ``}``}`

## C#

 `// Program to find minimum number of platforms``// required on a railway station` `using` `System;` `public` `class` `GFG {` `    ``// Returns minimum number of platforms required``    ``public` `static` `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` `(i = 0; i < n; i++) {``            ``// minimum platform``            ``plat_needed = 1;` `            ``for` `(j = i + 1; j < n; j++) {``                ``// check for overlap``                ``if` `(Math.Max(arr[i], arr[j])``                    ``<= Math.Min(dep[i], dep[j]))``                    ``plat_needed++;``            ``}` `            ``// update result``            ``result = Math.Max(result, plat_needed);``        ``}` `        ``return` `result;``    ``}` `    ``// Driver Code` `    ``static` `public` `void` `Main()``    ``{` `        ``int``[] arr = { 900, 940, 950, 1100, 1500, 1800 };``        ``int``[] dep = { 910, 1200, 1120, 1130, 1900, 2000 };``        ``int` `n = 6;``        ``Console.WriteLine(findPlatform(arr, dep, n));``    ``}``}`

## Javascript

 ``

Output

`3`

Time Complexity: O(n2), Two nested loops traverse the array.
Auxiliary space: O(1), As no extra space is required.

Efficient Approach: Store the arrival time and departure time and sort them based on arrival time then check if the departure time of the next train is smaller than the departure time of the previous train if it is smaller then increment the number of the platforms needed otherwise not.

• Store the arrival time and departure time in array arr and sort this array based on arrival time
• Declare a priority queue(min-heap) and store the departure time of the first train and also declare a counter cnt and initialize it with 1.
• Iterate over arr from 1 to n-1
• check if the arrival time of current train is less than or equals to the departure time of previous train which is kept on top of the priority queue
• If true, then push the new departure time and increment the counter cnt
• otherwise, we pop() the departure time
• push new departure time in the priority queue
• Finally, return the cnt.

## C++

 `// C++ program to implement the above approach``#include ``using` `namespace` `std;` `// Function to find the minimum number``// of platforms required``int` `findPlatform(``int` `arr[], ``int` `dep[], ``int` `n)``{``    ``// Store the arrival and departure time``    ``vector > arr2(n);` `    ``for` `(``int` `i = 0; i < n; i++) {``        ``arr2[i] = { arr[i], dep[i] };``    ``}` `    ``// Sort arr2 based on arival time``    ``sort(arr2.begin(), arr2.end());` `    ``priority_queue<``int``, vector<``int``>, greater<``int``> > p;``    ``int` `count = 1;``    ``p.push(arr2[0].second);` `    ``for` `(``int` `i = 1; i < n; i++) {` `        ``// Check if arrival time of current train``        ``// is less than or equals to depature time``        ``// of previous train``        ``if` `(p.top() >= arr2[i].first) {``            ``count++;``        ``}``        ``else` `{``            ``p.pop();``        ``}``        ``p.push(arr2[i].second);``    ``}` `    ``// Return the number of train required``    ``return` `count;``}` `// 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 << findPlatform(arr, dep, n);``    ``return` `0;``}`

Output

`3`

Time Complexity: O(N*log(N))
Auxiliary Space: O(N)

Another efficient 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.

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 is greater than departure then one less platform is needed to 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.

## C++

 `// Program to find minimum number of platforms``// required on a railway station``#include ``#include ` `using` `namespace` `std;` `// Returns minimum number of platforms required``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 << findPlatform(arr, dep, n);``    ``return` `0;``}`

## C

 `// C program to find minimum number of platforms required on a railway station` `// Importing the required header files``#include ``#include ` `// Creating MACRO for finding the maximum number``#define max(x, y)(((x) > (y)) ? (x) : (y))``// Creating MACRO for finding the minimum number``#define min(x, y)(((x) < (y)) ? (x) : (y))` `// below method is needed for the sort function``// compare function, compares two elements``int` `compare (``const` `void` `* num1, ``const` `void` `* num2) {``   ``if``(*(``int``*)num1 > *(``int``*)num2)``    ``return` `1;``   ``else``    ``return` `-1;``}`  `// Returns minimum number of platforms required``int` `findPlatform(``int` `arr[], ``int` `dep[], ``int` `n)``{``    ``// Sort arrival and departure arrays``    ``qsort``(arr, n, ``sizeof``(``int``), compare);``    ``qsort``(dep, n, ``sizeof``(``int``), compare);` `    ``// 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]);``    ``printf``(``"%d"``, findPlatform(arr, dep, n));``    ``return` `0;``}`

## Java

 `// Program to find minimum number of platforms` `import` `java.util.*;` `class` `GFG {` `    ``// Returns minimum number of platforms required``    ``static` `int` `findPlatform(``int` `arr[], ``int` `dep[], ``int` `n)``    ``{``        ``// Sort arrival and departure arrays``        ``Arrays.sort(arr);``        ``Arrays.sort(dep);` `        ``// 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``    ``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 = "``                           ``+ findPlatform(arr, dep, n));``    ``}``}`

## Python3

 `# Program to find minimum``# number of platforms``# required on a railway``# station` `# Returns minimum number``# of platforms required`  `def` `findPlatform(arr, dep, n):` `    ``# Sort arrival and``    ``# departure arrays``    ``arr.sort()``    ``dep.sort()` `    ``# plat_needed indicates``    ``# number of platforms``    ``# needed at a time``    ``plat_needed ``=` `1``    ``result ``=` `1``    ``i ``=` `1``    ``j ``=` `0` `    ``# Similar to merge in``    ``# merge sort to process``    ``# all events in sorted order``    ``while` `(i < n ``and` `j < n):` `        ``# If next event in sorted``        ``# order is arrival,``        ``# increment count of``        ``# platforms needed``        ``if` `(arr[i] <``=` `dep[j]):` `            ``plat_needed ``+``=` `1``            ``i ``+``=` `1` `        ``# Else decrement count``        ``# of platforms needed``        ``elif` `(arr[i] > dep[j]):` `            ``plat_needed ``-``=` `1``            ``j ``+``=` `1` `        ``# Update result if needed``        ``if` `(plat_needed > result):``            ``result ``=` `plat_needed` `    ``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 Anant Agarwal.`

## C#

 `// C# program to find minimum number``// of platforms``using` `System;` `class` `GFG {` `    ``// Returns minimum number of platforms``    ``// required``    ``static` `int` `findPlatform(``int``[] arr, ``int``[] dep, ``int` `n)``    ``{` `        ``// Sort arrival and departure arrays``        ``Array.Sort(arr);``        ``Array.Sort(dep);` `        ``// 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``    ``public` `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 = "``                      ``+ findPlatform(arr, dep, n));``    ``}``}` `// This code os contributed by nitin mittal.`

## PHP

 ` ``\$dep``[``\$j``])``        ``{``            ``\$plat_needed``--;``            ``\$j``++;``        ``}` `        ``// Update result if needed``        ``if` `(``\$plat_needed` `> ``\$result``)``            ``\$result` `= ``\$plat_needed``;``    ``}``    ` `    ``return` `\$result``;``}` `    ``// Driver Code``    ``\$arr` `= ``array``(900, 940, 950, 1100, 1500, 1800);``    ``\$dep` `= ``array``(910, 1200, 1120, 1130, 1900, 2000);``    ``\$n` `= ``count``(``\$arr``);``    ``echo` `"Minimum Number of Platforms Required = "``, findPlatform(``\$arr``, ``\$dep``, ``\$n``);` `// This code is contributed by anuj_67.``?>`

## Javascript

 ``

Output

`3`

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

Note: There is one more approach to the problem, which uses O(n) extra space and O(n) time to solve the problem:
Minimum Number of Platforms Required for a Railway/Bus Station | Set 2 (Map-based approach)