Minimum Platforms Required for Given Arrival and Departure Times
Last Updated :
26 Mar, 2025
Given two arrays, arr[] and dep[], that represent the arrival and departure times of trains respectively, the task is to find the minimum number of platforms required so that no train waits.
Examples:
Input: arr[] = [900, 940, 950, 1100, 1500, 1800], dep[] = [910, 1200, 1120, 1130, 1900, 2000]
Output: 3
Explanation: There are three trains during the time 9:40 to 12:00. So we need a minimum of 3 platforms.
Input: arr[] = [1, 5], dep[] = [3, 7]
Output: 1
Explanation: All train times are mutually exclusive. So we need only one platform
[Naive Approach] Using Two Nested Loops - O(n^2) time and O(1) space
The idea is to iterate through each train and for that train, check how many other trains have overlappingtimings with it - where current train's arrival time falls between the other train's arrival and departure times. We keep track of this count for each train and continuously update our answer with the maximum count found.
C++
// C++ program to find minimum Platforms Required
// for Given Arrival and Departure Times
#include <iostream>
#include <vector>
using namespace std;
// Function to find the minimum
// number of platforms required
int minPlatform(vector<int> &arr, vector<int>& dep) {
int n = arr.size();
int res = 0;
// Run a nested for-loop to find the overlap
for (int i = 0; i < n; i++) {
// Initially one platform is needed
int cnt = 1;
for (int j = 0; j < n; j++) {
if (i != j)
// Increment cnt if trains have overlapping
// time.
if (arr[i] >= arr[j] && dep[j] >= arr[i])
cnt++;
}
// Update the result
res = max(cnt, res);
}
return res;
}
int main() {
vector<int> arr = {900, 940, 950, 1100, 1500, 1800};
vector<int> dep = {910, 1200, 1120, 1130, 1900, 2000};
cout << minPlatform(arr, dep);
return 0;
}
// Java program to find minimum Platforms Required
// for Given Arrival and Departure Times
class GfG {
// Function to find the minimum
// number of platforms required
static int minPlatform(int[] arr, int[] dep) {
int n = arr.length;
int res = 0;
// Run a nested for-loop to find the overlap
for (int i = 0; i < n; i++) {
// Initially one platform is needed
int cnt = 1;
for (int j = 0; j < n; j++) {
if (i != j)
// Increment cnt if trains have overlapping
// time.
if (arr[i] >= arr[j] && dep[j] >= arr[i]) {
cnt++;
}
}
// Update the result
res = Math.max(cnt, res);
}
return res;
}
public static void main(String[] args) {
int[] arr = {900, 940, 950, 1100, 1500, 1800};
int[] dep = {910, 1200, 1120, 1130, 1900, 2000};
System.out.println(minPlatform(arr, dep));
}
}
# Python program to find minimum Platforms Required
# for Given Arrival and Departure Times
# Function to find the minimum
# number of platforms required
def minPlatform(arr, dep):
n = len(arr)
res = 0
# Run a nested for-loop to find the overlap
for i in range(n):
# Initially one platform is needed
cnt = 1
for j in range(n):
if i != j:
# Increment cnt if trains have overlapping
# time.
if arr[i] >= arr[j] and dep[j] >= arr[i]:
cnt += 1
# Update the result
res = max(cnt, res)
return res
if __name__ == "__main__":
arr = [900, 940, 950, 1100, 1500, 1800]
dep = [910, 1200, 1120, 1130, 1900, 2000]
print(minPlatform(arr, dep))
// C# program to find minimum Platforms Required
// for Given Arrival and Departure Times
using System;
class GfG {
// Function to find the minimum
// number of platforms required
static int minPlatform(int[] arr, int[] dep) {
int n = arr.Length;
int res = 0;
// Run a nested for-loop to find the overlap
for (int i = 0; i < n; i++) {
// Initially one platform is needed
int cnt = 1;
for (int j = 0; j < n; j++) {
if (i != j)
// Increment cnt if trains have overlapping
// time.
if (arr[i] >= arr[j] && dep[j] >= arr[i]) {
cnt++;
}
}
// Update the result
res = Math.Max(cnt, res);
}
return res;
}
static void Main(string[] args) {
int[] arr = {900, 940, 950, 1100, 1500, 1800};
int[] dep = {910, 1200, 1120, 1130, 1900, 2000};
Console.WriteLine(minPlatform(arr, dep));
}
}
// JavaScript program to find minimum Platforms Required
// for Given Arrival and Departure Times
// Function to find the minimum
// number of platforms required
function minPlatform(arr, dep) {
let n = arr.length;
let res = 0;
// Run a nested for-loop to find the overlap
for (let i = 0; i < n; i++) {
// Initially one platform is needed
let cnt = 1;
for (let j = 0; j < n; j++) {
if (i !== j)
// Increment cnt if trains have overlapping
// time.
if (arr[i] >= arr[j] && dep[j] >= arr[i]) {
cnt++;
}
}
// Update the result
res = Math.max(cnt, res);
}
return res;
}
// Driver Code
let arr = [900, 940, 950, 1100, 1500, 1800];
let dep = [910, 1200, 1120, 1130, 1900, 2000];
console.log(minPlatform(arr, dep));
[Expected Approach 1] Using Sorting and Two Pointers - O(n log(n)) time and O(1) space
This approach uses sorting and two-pointer to reduce the complexity. First, we sort the arrival and departure times of all trains. Then, using two pointers, we traverse through both arrays.
The idea is to maintain a count of platforms needed at any point in 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
Step by Step implementation:
- Sort the arrival and departure times so we can process train timings in order.
- Initialize two pointers:
- One for tracking arrivals (
i = 0). - One for tracking departures (
j = 0).
- Iterate through the arrival times:
- If the current train arrives before or at the departure of an earlier train, allocate a new platform (
cnt++). - Otherwise, if the arrival time is greater than the departure time, it means a train has left, freeing up a platform (
cnt--), and move the departure pointer forward (j++).
- Update the maximum number of platforms required after each step.
- Continue this process until all trains are processed.
C++
// C++ program to find minimum Platforms Required
// for Given Arrival and Departure Times
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
// Function to find the minimum
// number of platforms required
int minPlatform(vector<int> &arr, vector<int>& dep) {
int n = arr.size();
int res = 0;
// Sort the arrays
sort(arr.begin(), arr.end());
sort(dep.begin(), dep.end());
// Pointer to track the departure times
int j = 0;
// Tracks the number of platforms needed at any given time
int cnt = 0;
// Check for each train
for (int i=0; i<n; i++) {
// Decrement count if other
// trains have left
while (j<n && dep[j]<arr[i]) {
cnt--;
j++;
}
// one platform for current train
cnt++;
res = max(res, cnt);
}
return res;
}
int main() {
vector<int> arr = {900, 940, 950, 1100, 1500, 1800};
vector<int> dep = {910, 1200, 1120, 1130, 1900, 2000};
cout << minPlatform(arr, dep);
return 0;
}
// Java program to find minimum Platforms Required
// for Given Arrival and Departure Times
import java.util.Arrays;
class Main {
// Function to find the minimum
// number of platforms required
static int minPlatform(int[] arr, int[] dep) {
int n = arr.length;
int res = 0;
// Sort the arrays
Arrays.sort(arr);
Arrays.sort(dep);
// Pointer to track the departure times
int j = 0;
// Tracks the number of platforms needed at any given time
int cnt = 0;
// Check for each train
for (int i = 0; i < n; i++) {
// Decrement count if other
// trains have left
while (j < n && dep[j] < arr[i]) {
cnt--;
j++;
}
// one platform for current train
cnt++;
res = Math.max(res, cnt);
}
return res;
}
public static void main(String[] args) {
int[] arr = {900, 940, 950, 1100, 1500, 1800};
int[] dep = {910, 1200, 1120, 1130, 1900, 2000};
System.out.println(minPlatform(arr, dep));
}
}
# Python program to find minimum Platforms Required
# for Given Arrival and Departure Times
# Function to find the minimum
# number of platforms required
def minPlatform(arr, dep):
n = len(arr)
res = 0
# Sort the arrays
arr.sort()
dep.sort()
# Pointer to track the departure times
j = 0;
# Tracks the number of platforms needed at any given time
cnt = 0;
# Check for each train
for i in range(n):
# Decrement count if other
# trains have left
while j < n and dep[j] < arr[i]:
cnt -= 1
j += 1
# one platform for current train
cnt += 1
res = max(res, cnt)
return res
if __name__ == "__main__":
arr = [900, 940, 950, 1100, 1500, 1800]
dep = [910, 1200, 1120, 1130, 1900, 2000]
print(minPlatform(arr, dep))
// C# program to find minimum Platforms Required
// for Given Arrival and Departure Times
using System;
class GfG {
// Function to find the minimum
// number of platforms required
static int minPlatform(int[] arr, int[] dep) {
int n = arr.Length;
int res = 0;
// Sort the arrays
Array.Sort(arr);
Array.Sort(dep);
// Pointer to track the departure times
int j = 0;
// Tracks the number of platforms needed at any given time
int cnt = 0;
// Check for each train
for (int i = 0; i < n; i++) {
// Decrement count if other
// trains have left
while (j < n && dep[j] < arr[i]) {
cnt--;
j++;
}
// one platform for current train
cnt++;
res = Math.Max(res, cnt);
}
return res;
}
static void Main(string[] args) {
int[] arr = {900, 940, 950, 1100, 1500, 1800};
int[] dep = {910, 1200, 1120, 1130, 1900, 2000};
Console.WriteLine(minPlatform(arr, dep));
}
}
// JavaScript program to find minimum Platforms Required
// for Given Arrival and Departure Times
// Function to find the minimum
// number of platforms required
function minPlatform(arr, dep) {
let n = arr.length;
let res = 0;
// Sort the arrays
arr.sort((a, b) => a - b);
dep.sort((a, b) => a - b);
// Pointer to track the departure times
let j = 0;
// Tracks the number of platforms needed at any given time
let cnt = 0;
// Check for each train
for (let i = 0; i < n; i++) {
// Decrement count if other
// trains have left
while (j < n && dep[j] < arr[i]) {
cnt--;
j++;
}
// one platform for current train
cnt++;
res = Math.max(res, cnt);
}
return res;
}
// Driver Code
let arr = [900, 940, 950, 1100, 1500, 1800];
let dep = [910, 1200, 1120, 1130, 1900, 2000];
console.log(minPlatform(arr, dep));
[Expected Approach 2] Using Sweep line algorithm
The Sweep Line Algorithm is an efficient technique for solving interval-based problems. It works by treating each train's arrival and departure times as events on a timeline. By processing these events in chronological order, we can track the number of trains at the station at any moment, which directly indicates the number of platforms required at that time. The maximum number of overlapping trains during this process determines the minimum number of platforms needed.
Step by Step implementation:
- Create an array v[] of size greater than the maximum departure time. This array will help track the number of platforms needed at each time.
- Mark arrivals and departures:
- For each arrival time, increment v[arrival_time] by 1, indicating that a platform is needed.
- For each departure time, decrement v[departure_time + 1] by 1, indicating that a platform is freed as the train has left.
- Iterate through
v[] and compute the cumulative sum. - The running sum keeps track of the number of trains present at any given time.
- The maximum value encountered represents the minimum number of platforms required.
C++
// C++ program to find minimum Platforms Required
// for Given Arrival and Departure Times
#include <iostream>
#include <vector>
using namespace std;
// Function to find the minimum
// number of platforms required
int minPlatform(vector<int> &arr, vector<int>& dep) {
int n = arr.size();
int res = 0;
// Find the max Departure time
int maxDep = dep[0];
for (int i=1; i<n; i++) {
maxDep = max(maxDep, dep[i]);
}
// Create a vector to store the count of trains at each
// time
vector<int> v(maxDep + 2, 0);
// Increment the count at the arrival time and decrement
// at the departure time
for (int i = 0; i < n; i++) {
v[arr[i]]++;
v[dep[i] + 1]--;
}
int count = 0;
// Iterate over the vector and keep track of the maximum
// sum seen so far
for (int i = 0; i <= maxDep + 1; i++) {
count += v[i];
res = max(res, count);
}
return res;
}
int main() {
vector<int> arr = {900, 940, 950, 1100, 1500, 1800};
vector<int> dep = {910, 1200, 1120, 1130, 1900, 2000};
cout << minPlatform(arr, dep);
return 0;
}
// Java program to find minimum Platforms Required
// for Given Arrival and Departure Times
import java.util.Arrays;
class GfG {
// Function to find the minimum
// number of platforms required
static int minPlatform(int[] arr, int[] dep) {
int n = arr.length;
int res = 0;
// Find the max Departure time
int maxDep = dep[0];
for (int i = 1; i < n; i++) {
maxDep = Math.max(maxDep, dep[i]);
}
// Create an array to store the count of trains at each
// time
int[] v = new int[maxDep + 2];
// Increment the count at the arrival time and decrement
// at the departure time
for (int i = 0; i < n; i++) {
v[arr[i]]++;
v[dep[i] + 1]--;
}
int count = 0;
// Iterate over the array and keep track of the maximum
// sum seen so far
for (int i = 0; i <= maxDep + 1; i++) {
count += v[i];
res = Math.max(res, count);
}
return res;
}
public static void main(String[] args) {
int[] arr = {900, 940, 950, 1100, 1500, 1800};
int[] dep = {910, 1200, 1120, 1130, 1900, 2000};
System.out.println(minPlatform(arr, dep));
}
}
# Python program to find minimum Platforms Required
# for Given Arrival and Departure Times
# Function to find the minimum
# number of platforms required
def minPlatform(arr, dep):
n = len(arr)
res = 0
# Find the max Departure time
maxDep = max(dep)
# Create a list to store the count of trains at each
# time
v = [0] * (maxDep + 2)
# Increment the count at the arrival time and decrement
# at the departure time
for i in range(n):
v[arr[i]] += 1
v[dep[i] + 1] -= 1
count = 0
# Iterate over the list and keep track of the maximum
# sum seen so far
for i in range(maxDep + 2):
count += v[i]
res = max(res, count)
return res
if __name__ == "__main__":
arr = [900, 940, 950, 1100, 1500, 1800]
dep = [910, 1200, 1120, 1130, 1900, 2000]
print(minPlatform(arr, dep))
// C# program to find minimum Platforms Required
// for Given Arrival and Departure Times
using System;
class GfG {
// Function to find the minimum
// number of platforms required
static int minPlatform(int[] arr, int[] dep) {
int n = arr.Length;
int res = 0;
// Find the max Departure time
int maxDep = dep[0];
for (int i = 1; i < n; i++) {
maxDep = Math.Max(maxDep, dep[i]);
}
// Create an array to store the count of trains at each
// time
int[] v = new int[maxDep + 2];
// Increment the count at the arrival time and decrement
// at the departure time
for (int i = 0; i < n; i++) {
v[arr[i]]++;
v[dep[i] + 1]--;
}
int count = 0;
// Iterate over the array and keep track of the maximum
// sum seen so far
for (int i = 0; i <= maxDep + 1; i++) {
count += v[i];
res = Math.Max(res, count);
}
return res;
}
static void Main(string[] args) {
int[] arr = {900, 940, 950, 1100, 1500, 1800};
int[] dep = {910, 1200, 1120, 1130, 1900, 2000};
Console.WriteLine(minPlatform(arr, dep));
}
}
// JavaScript program to find minimum Platforms Required
// for Given Arrival and Departure Times
// Function to find the minimum
// number of platforms required
function minPlatform(arr, dep) {
let n = arr.length;
let res = 0;
// Find the max Departure time
let maxDep = Math.max(...dep);
// Create an array to store the count of trains at each
// time
let v = new Array(maxDep + 2).fill(0);
// Increment the count at the arrival time and decrement
// at the departure time
for (let i = 0; i < n; i++) {
v[arr[i]]++;
v[dep[i] + 1]--;
}
let count = 0;
// Iterate over the array and keep track of the maximum
// sum seen so far
for (let i = 0; i <= maxDep + 1; i++) {
count += v[i];
res = Math.max(res, count);
}
return res;
}
// Driver Code
let arr = [900, 940, 950, 1100, 1500, 1800];
let dep = [910, 1200, 1120, 1130, 1900, 2000];
console.log(minPlatform(arr, dep));
Time Complexity: O(n + k), where n is the number of trains and k is the maximum value present in the arrays.
Auxiliary space: O(k), where k is the maximum value present in both the arrays.
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Job Scheduling with two jobs allowed at a timeGiven a 2d array jobs[][] of order n * 2, where each element jobs[i], contains two integers, representing the start and end time of the job. Your task is to check if it is possible to complete all the jobs, provided that two jobs can be done simultaneously at a particular moment. Note: If a job star
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Optimal Page Replacement AlgorithmIn operating systems, whenever a new page is referred and not present in memory, page fault occurs, and Operating System replaces one of the existing pages with newly needed page. Different page replacement algorithms suggest different ways to decide which page to replace. The target for all algorit
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Greedy algorithm on Graph