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Minimum Number of Platforms Required for a Railway/Bus Station

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  • 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.

Follow the steps mentioned below:

  • 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.
  • Print the answer.

Implementation: 

C++14




// C++ program to implement the above approach
#include <bits/stdc++.h>
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 <stdio.h>
 
// 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




<script>
// Program to find minimum number of platforms
// required on a railway station
 
function max(a,b)
{
    if(a==b)
         return a;
    else{
        if(a>b)
            return a;
        else
            return b;
       }
}
 
// Returns minimum number of platforms required
function findPlatform( arr, dep, n)
{
 
    // plat_needed indicates number of platforms
    // needed at a time
    var plat_needed = 1, result = 1;
    var i = 1, j = 0;
 
    // run a nested loop to find overlap
    for (var i = 0; i < n; i++) {
        // minimum platform
        plat_needed = 1;
 
        for (var 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;
}
 
var arr = [ 900, 940, 950, 1100, 1500, 1800 ];
    var dep = [ 910, 1200, 1120, 1130, 1900, 2000 ];
    var n =6;
    document.write("Minimum Number of Platforms Required = "
        +findPlatform(arr, dep, n));
 
 
</script>


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.

Follow the steps mentioned below:

  • 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 <bits/stdc++.h>
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<pair<int, int> > 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 <algorithm>
#include <iostream>
 
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 <stdio.h>
#include <stdlib.h>
 
// 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




<?php
// PHP Program to find minimum number
// of platforms  required on a railway
// station
 
// Returns minimum number of
// platforms required
function findPlatform($arr, $dep, $n)
{
     
    // Sort arrival and
    // departure arrays
    sort($arr);
    sort($dep);
     
    // 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++;
            $i++;
        }
     
        // Else decrement count
        // of platforms needed
        elseif ($arr[$i] > $dep[$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




<script>
// Javascript Program to find minimum number
// of platforms  required on a railway
// station
 
// Returns minimum number of
// platforms required
function findPlatform(arr, dep, n)
{
     
    // Sort arrival and
    // departure arrays
    arr = arr.sort((a,b) => a-b));
    dep = dep.sort((a,b) => a-b));
     
    // plat_needed indicates
    // number of platforms
    // needed at a time
    let plat_needed = 1;
    let result = 1;
    let i = 1;
    let 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
    let arr = new Array(900, 940, 950, 1100, 1500, 1800);
    let dep = new Array(910, 1200, 1120, 1130, 1900, 2000);
    let n = arr.length;
    document.write("Minimum Number of Platforms Required = " + findPlatform(arr, dep, n));
 
// This code is contributed by Saurabh Jaiswal.
</script>


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)
 

This article is contributed by Shivam. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above


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