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.
Illustration:
Follow the steps mentioned below:
- Run two nested loops from start to end.
- For every iteration of the outer loop, find the count of intervals that intersect with the current interval except itself.
- Update the answer with the maximum count of overlap in each iteration of the outer loop.
- Print the answer.
Below is the implementation of the above approach:
C
#include <stdio.h>
#define max(x, y) (((x) > (y)) ? (x) : (y))
int findPlatform(int arr[], int dep[], int n)
{
int plat_needed = 1, result = 1;
for (int i = 0; i < n; i++) {
plat_needed = 1;
for (int j = 0; j < n; j++) {
if (i != j)
if (arr[i] >= arr[j] && dep[j] >= arr[i])
plat_needed++;
}
result = max(plat_needed, result);
}
return result;
}
int main()
{
int arr[] = { 100, 300, 500 };
int dep[] = { 900, 400, 600 };
int n = sizeof(arr) / sizeof(arr[0]);
printf("%d", findPlatform(arr, dep, n));
return 0;
}
|
C++14
#include <bits/stdc++.h>
using namespace std;
int findPlatform(int arr[], int dep[], int n)
{
int plat_needed = 1, result = 1;
for (int i = 0; i < n; i++) {
plat_needed = 1;
for (int j = 0; j < n; j++) {
if (i != j)
if (arr[i] >= arr[j] && dep[j] >= arr[i])
plat_needed++;
}
result = max(plat_needed, result);
}
return result;
}
int main()
{
int arr[] = { 100, 300, 500 };
int dep[] = { 900, 400, 600 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << findPlatform(arr, dep, n);
return 0;
}
|
Python3
def findPlatform(arr, dep, n):
plat_needed = 1
result = 1
for i in range(n):
plat_needed = 1
for j in range(n):
if i != j:
if (arr[i] >= arr[j] and dep[j] >= arr[i]):
plat_needed += 1
result = max(result, plat_needed)
return result
def main():
arr = [100, 300, 500]
dep = [900, 400, 600]
n = len(arr)
print("{}".format(
findPlatform(arr, dep, n)))
if __name__ == '__main__':
main()
|
Java
import java.io.*;
class GFG {
public static int findPlatform(int arr[], int dep[],
int n)
{
int plat_needed = 1, result = 1;
for (int i = 0; i < n; i++) {
plat_needed = 1;
for (int j = 0; j < n; j++) {
if (i != j)
if (arr[i] >= arr[j]
&& dep[j] >= arr[i])
plat_needed++;
}
result = Math.max(result, plat_needed);
}
return result;
}
public static void main(String[] args)
{
int arr[] = { 100, 300, 500 };
int dep[] = { 900, 400, 600 };
int n = 3;
System.out.println(findPlatform(arr, dep, n));
}
}
|
Javascript
<script>
function max(a,b)
{
if(a==b)
return a;
else{
if(a>b)
return a;
else
return b;
}
}
function findPlatform( arr, dep, n)
{
var plat_needed = 1, result = 1;
var i = 1, j = 0;
for (var i = 0; i < n; i++) {
plat_needed = 1;
for (var j = 0; j < n; j++) {
if(i != j)
if(arr[i] >= arr[j] && dep[j] >= arr[i])
plat_needed++;
}
result = max(result, plat_needed);
}
return result;
}
var arr = [100, 300, 500]
var dep = [900, 400, 600]
var n = 3;
document.write("Minimum Number of Platforms Required = "
+findPlatform(arr, dep, n));
</script>
|
C#
using System;
public class GFG {
public static int findPlatform(int[] arr, int[] dep,
int n)
{
int plat_needed = 1, result = 1;
int i = 0, j = 0;
for (i = 0; i < n; i++) {
plat_needed = 1;
for (j = 0; j < n; j++) {
if (i != j)
if (arr[i] >= arr[j]
&& dep[j] >= arr[i])
plat_needed++;
}
result = Math.Max(result, plat_needed);
}
return result;
}
static public void Main()
{
int[] arr = { 100, 300, 500 };
int[] dep = { 900, 400, 600 };
int n = 3;
Console.WriteLine(findPlatform(arr, dep, n));
}
}
|
Time Complexity: O(n2), Two nested loops traverse the array.
Auxiliary space: O(1), As no extra space is required.
Minimum Number of Platforms Required for a Railway/Bus Station using Heap:
Store the arrival time and departure time and sort them based on arrival time then check if the arrival 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.
Illustration:
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 the current train is less than or equal to the departure time of the 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.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int findPlatform(int arr[], int dep[], int n)
{
vector<pair<int, int> > arr2(n);
for (int i = 0; i < n; i++) {
arr2[i] = { arr[i], dep[i] };
}
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++) {
if (p.top() >= arr2[i].first) {
count++;
}
else {
p.pop();
}
p.push(arr2[i].second);
}
return count;
}
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;
}
|
Java
import java.io.*;
import java.util.Arrays;
import java.util.Comparator;
import java.util.PriorityQueue;
class GFG {
private static class TrainSchedule {
int arrivalTime, deptTime;
TrainSchedule(int arrivalTime, int deptTime)
{
this.arrivalTime = arrivalTime;
this.deptTime = deptTime;
}
public String toString()
{
return "(" + this.arrivalTime + ","
+ this.deptTime + ")";
}
}
private static class SortByArrival
implements Comparator<TrainSchedule> {
@Override
public int compare(TrainSchedule o1,
TrainSchedule o2)
{
return o1.arrivalTime - o2.arrivalTime;
}
}
public static int countPlatforms(int[] arr, int[] dep)
{
TrainSchedule[] trains
= new TrainSchedule[arr.length];
for (int i = 0; i < arr.length; i++) {
trains[i] = new TrainSchedule(arr[i], dep[i]);
}
Arrays.sort(trains, new SortByArrival());
PriorityQueue<Integer> pq = new PriorityQueue<>();
pq.add(trains[0].deptTime);
int count = 1;
for (int i = 1; i < arr.length; i++) {
TrainSchedule curr = trains[i];
if (curr.arrivalTime <= pq.peek()) {
count++;
}
else {
pq.poll();
}
pq.add(curr.deptTime);
}
return count;
}
public static void main(String[] args)
{
int[] arr = { 900, 940, 950, 1100, 1500, 1800 };
int[] dep = { 910, 1200, 1120, 1130, 1900, 2000 };
int res = countPlatforms(arr, dep);
System.out.println(res);
}
}
|
Python3
import heapq
def findPlatform(arr, dep, n):
arr2 = []
for i in range(n):
arr2.append([arr[i], dep[i]])
arr2.sort()
p = []
count = 1
heapq.heappush(p, arr2[0][1])
for i in range(1, n):
if p[0] >= arr2[i][0]:
count += 1
else:
heapq.heappop(p)
heapq.heappush(p, arr2[i][1])
return count
if __name__ == "__main__":
arr = [900, 940, 950, 1100, 1500, 1800]
dep = [910, 1200, 1120, 1130, 1900, 2000]
n = len(arr)
print(findPlatform(arr, dep, n))
|
Javascript
function findPlatform(arr, dep, n) {
const arr2 = new Array(n);
for (let i = 0; i < n; i++) {
arr2[i] = { arr: arr[i], dep: dep[i] };
}
arr2.sort((a, b) => a.arr - b.arr);
const p = [];
let count = 1;
p.push(arr2[0].dep);
for (let i = 1; i < n; i++) {
if (p[0] <= arr2[i].arr) {
p.shift();
} else {
count++;
}
p.push(arr2[i].dep);
p.sort();
}
return count;
}
const arr = [900, 940, 950, 1100, 1500, 1800];
const dep = [910, 1200, 1120, 1130, 1900, 2000];
const n = arr.length;
console.log(findPlatform(arr, dep, n));
|
C#
using System;
using System.Collections;
using System.Collections.Generic;
public class GFG {
public class TrainSchedule {
public int arrivalTime, deptTime;
public TrainSchedule(int arrivalTime, int deptTime)
{
this.arrivalTime = arrivalTime;
this.deptTime = deptTime;
}
public override string ToString()
{
return "(" + this.arrivalTime + ","
+ this.deptTime + ")";
}
}
public static int countPlatforms(int[] arr, int[] dep)
{
TrainSchedule[] trains
= new TrainSchedule[arr.Length];
for (int i = 0; i < arr.Length; i++) {
trains[i] = new TrainSchedule(arr[i], dep[i]);
}
Array.Sort(trains, (a, b) = > a.arrivalTime
- b.arrivalTime);
var pq = new Queue<int>();
pq.Enqueue(trains[0].deptTime);
int count = 1;
for (int i = 1; i < arr.Length; i++) {
TrainSchedule curr = trains[i];
if (curr.arrivalTime <= pq.Peek()) {
count++;
}
else {
pq.Dequeue();
}
pq.Enqueue(curr.deptTime);
}
return count;
}
public static void Main(string[] args)
{
int[] arr = { 900, 940, 950, 1100, 1500, 1800 };
int[] dep = { 910, 1200, 1120, 1130, 1900, 2000 };
int res = countPlatforms(arr, dep);
Console.WriteLine(res);
}
}
|
Time Complexity: O(N*log(N)), Heaps take log(n) time for pushing element and there are n elements.
Auxiliary Space: O(N), Space required by heap to store the element.
Minimum Number of Platforms Required for a Railway/Bus Station using Sorting:
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.
Illustration:
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 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.
Follow the steps mentioned below:
- Sort the arrival and departure times of trains.
- Create two pointers i=1, and j=0, and a variable to store ans and current count plat
- Run a loop while i<n and j<n and compare the ith element of arrival array and jth element of departure array.
- 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
- 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
- Update the ans, i.e. ans = max(ans, plat).
Below is the implementation of the above approach:
C
#include <stdio.h>
#include <stdlib.h>
#define max(x, y) (((x) > (y)) ? (x) : (y))
#define min(x, y) (((x) < (y)) ? (x) : (y))
int compare(const void* num1, const void* num2)
{
if (*(int*)num1 > *(int*)num2)
return 1;
else
return -1;
}
int findPlatform(int arr[], int dep[], int n)
{
qsort(arr, n, sizeof(int), compare);
qsort(dep, n, sizeof(int), compare);
int plat_needed = 1, result = 1;
int i = 1, j = 0;
while (i < n && j < n) {
if (arr[i] <= dep[j]) {
plat_needed++;
i++;
}
else if (arr[i] > dep[j]) {
plat_needed--;
j++;
}
if (plat_needed > result)
result = plat_needed;
}
return result;
}
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;
}
|
C++
#include <algorithm>
#include <iostream>
using namespace std;
int findPlatform(int arr[], int dep[], int n)
{
sort(arr, arr + n);
sort(dep, dep + n);
int plat_needed = 1, result = 1;
int i = 1, j = 0;
while (i < n && j < n) {
if (arr[i] <= dep[j]) {
plat_needed++;
i++;
}
else if (arr[i] > dep[j]) {
plat_needed--;
j++;
}
if (plat_needed > result)
result = plat_needed;
}
return result;
}
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;
}
|
Java
import java.util.*;
class GFG {
static int findPlatform(int arr[], int dep[], int n)
{
Arrays.sort(arr);
Arrays.sort(dep);
int plat_needed = 1, result = 1;
int i = 1, j = 0;
while (i < n && j < n) {
if (arr[i] <= dep[j]) {
plat_needed++;
i++;
}
else if (arr[i] > dep[j]) {
plat_needed--;
j++;
}
if (plat_needed > result)
result = plat_needed;
}
return result;
}
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
def findPlatform(arr, dep, n):
arr.sort()
dep.sort()
plat_needed = 1
result = 1
i = 1
j = 0
while (i < n and j < n):
if (arr[i] <= dep[j]):
plat_needed += 1
i += 1
elif (arr[i] > dep[j]):
plat_needed -= 1
j += 1
if (plat_needed > result):
result = plat_needed
return result
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))
|
Javascript
<script>
function findPlatform(arr, dep, n)
{
arr = arr.sort((a,b) => a-b));
dep = dep.sort((a,b) => a-b));
let plat_needed = 1;
let result = 1;
let i = 1;
let j = 0;
while (i < n && j < n)
{
if (arr[i] <= dep[j])
{
plat_needed++;
i++;
}
else if (arr[i] > dep[j])
{
plat_needed--;
j++;
}
if (plat_needed > result)
result = plat_needed;
}
return result;
}
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));
</script>
|
C#
using System;
class GFG {
static int findPlatform(int[] arr, int[] dep, int n)
{
Array.Sort(arr);
Array.Sort(dep);
int plat_needed = 1, result = 1;
int i = 1, j = 0;
while (i < n && j < n) {
if (arr[i] <= dep[j]) {
plat_needed++;
i++;
}
else if (arr[i] > dep[j]) {
plat_needed--;
j++;
}
if (plat_needed > result)
result = plat_needed;
}
return result;
}
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));
}
}
|
PHP
<?php
function findPlatform($arr, $dep, $n)
{
sort($arr);
sort($dep);
$plat_needed = 1;
$result = 1;
$i = 1;
$j = 0;
while ($i < $n and $j < $n)
{
if ($arr[$i] <= $dep[$j])
{
$plat_needed++;
$i++;
}
elseif ($arr[$i] > $dep[$j])
{
$plat_needed--;
$j++;
}
if ($plat_needed > $result)
$result = $plat_needed;
}
return $result;
}
$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);
?>
|
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.
Minimum Number of Platforms Required for a Railway/Bus Station using Sweep Line Algorithm:
The sweep line algorithm is an efficient method for solving problems involving intervals or segments, and can be used to solve the problem of finding the minimum number of platforms needed at a train station based on the arrival and departure times of trains. The algorithm maintains a count of the number of platforms needed at each time, which is used to determine the minimum number of platforms needed at the station.
By using the idea behind this article: Constant time range add operation on an array
When a train arrives at the station, we increase the count by 1 because it occupies a platform. Similarly, when a train departs from the station, we decrease the count by 1 because it frees up a platform.
We first create a vector of size maxDepartureTime+2 and initialize all its values to 0. We then iterate over the input arrays and increment the count at the arrival time and decrement the count at the departure time in the vector. At any point in time, the sum of values (cumulative sum) in the vector will give us the number of trains present at the station. We can then find the maximum of this sum, which will give us the minimum number of platforms required.
Illustration:
Follow the steps mentioned below:
- Initialize a variable count to 0.
- Create a vector v of size maxDepartureTime+2 and initialize all its values to 0.
- Iterate over the input arrays and increment the value at the arrival time in v and decrement the value at the departure time+1 in v.
- Iterate over the vector v and keep track of the maximum sum seen so far. This maximum sum will give us the minimum number of platforms required.
- Return the maximum sum.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int findPlatformOptimized(int arr[], int dep[], int n)
{
int count = 0, maxPlatforms = 0;
int maxDepartureTime = dep[0];
for (int i = 1; i < n; i++) {
maxDepartureTime = max(maxDepartureTime, dep[i]);
}
vector<int> v(maxDepartureTime + 2, 0);
for (int i = 0; i < n; i++) {
v[arr[i]]++;
v[dep[i] + 1]--;
}
for (int i = 0; i <= maxDepartureTime + 1; i++) {
count += v[i];
maxPlatforms = max(maxPlatforms, count);
}
return maxPlatforms;
}
int main()
{
int arr[] = { 100, 300, 600 };
int dep[] = { 900, 400, 500 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << findPlatformOptimized(arr, dep, n);
return 0;
}
|
Java
import java.util.*;
public class Main {
public static int findPlatformOptimized(int[] arr, int[] dep, int n)
{
int count = 0, maxPlatforms = 0;
int maxDepartureTime = dep[0];
for (int i = 1; i < n; i++) {
maxDepartureTime = Math.max(maxDepartureTime, dep[i]);
}
List<Integer> v = new ArrayList<>(maxDepartureTime + 2);
for (int i = 0; i < maxDepartureTime + 2; i++) {
v.add(0);
}
for (int i = 0; i < n; i++) {
v.set(arr[i], v.get(arr[i]) + 1);
v.set(dep[i] + 1, v.get(dep[i] + 1) - 1);
}
for (int i = 0; i <= maxDepartureTime + 1; i++) {
count += v.get(i);
maxPlatforms = Math.max(maxPlatforms, count);
}
return maxPlatforms;
}
public static void main(String[] args)
{
int[] arr = { 100, 300, 600 };
int[] dep = { 900, 400, 500 };
int n = arr.length;
System.out.println(findPlatformOptimized(arr, dep, n));
}
}
|
Python3
from typing import List
def find_platform_optimized(arr: List[int], dep: List[int], n: int) -> int:
count = 0
max_platforms = 0
max_departure_time = max(dep)
v = [0] * (max_departure_time + 2)
for i in range(n):
v[arr[i]] += 1
v[dep[i] + 1] -= 1
for i in range(max_departure_time + 2):
count += v[i]
max_platforms = max(max_platforms, count)
return max_platforms
if __name__ == '__main__':
arr = [100, 300, 600]
dep = [900, 400, 500]
n = len(arr)
print(find_platform_optimized(arr, dep, n))
|
Javascript
function findPlatformOptimized(arr, dep, n) {
let count = 0, maxPlatforms = 0;
let maxDepartureTime = dep[0];
for (let i = 1; i < n; i++) {
maxDepartureTime = Math.max(maxDepartureTime, dep[i]);
}
const v = new Array(maxDepartureTime + 2).fill(0);
for (let i = 0; i < n; i++) {
v[arr[i]]++;
v[dep[i] + 1]--;
}
for (let i = 0; i <= maxDepartureTime + 1; i++) {
count += v[i];
maxPlatforms = Math.max(maxPlatforms, count);
}
return maxPlatforms;
}
const arr = [100, 300, 600];
const dep = [900, 400, 500];
const n = arr.length;
console.log(findPlatformOptimized(arr, dep, n));
|
C#
using System;
using System.Collections.Generic;
class Program {
static int FindPlatformOptimized(int[] arr, int[] dep,
int n)
{
int count = 0, maxPlatforms = 0;
int maxDepartureTime = dep[0];
for (int i = 1; i < n; i++) {
maxDepartureTime
= Math.Max(maxDepartureTime, dep[i]);
}
List<int> v = new List<int>(maxDepartureTime + 2);
for (int i = 0; i <= maxDepartureTime + 1; i++) {
v.Add(0);
}
for (int i = 0; i < n; i++) {
v[arr[i]]++;
v[dep[i] + 1]--;
}
for (int i = 0; i <= maxDepartureTime + 1; i++) {
count += v[i];
maxPlatforms = Math.Max(maxPlatforms, count);
}
return maxPlatforms;
}
static void Main(string[] args)
{
int[] arr = { 100, 300, 600 };
int[] dep = { 900, 400, 500 };
int n = arr.Length;
Console.WriteLine(
FindPlatformOptimized(arr, dep, n));
}
}
|
Complexity Analysis:
- Time Complexity: O(n), where n is the number of trains. The algorithm iterates over the input arrays once to update the vector, and then iterates over the vector once to calculate the cumulative sum. Both iterations take O(n) time. The space complexity of the algorithm is O(1) as the vector used has a fixed size.
- Auxiliary space: O(maxDepartureTime), as we create a vector of size maxDepartureTime+2 to store the count of trains at each time.
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)
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Last Updated :
13 Apr, 2023
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