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Find the minimum capacity of the train required to hold the passengers
• Difficulty Level : Easy
• Last Updated : 24 May, 2021

Given the number of passengers entering and exiting the train, the task is to find the minimum capacity of the train to keep all the passengers in throughout the journey.
Examples:

Input: enter[] = {3, 5, 2, 0}, exit[] = {0, 2, 4, 4}
Output:
Station 1: Train capacity = 3
Station 2: Train capacity = 3 + 5 – 2 = 6
Station 3: Train capacity = 6 + 2 – 4 = 4
Station 4: Train capacity = 4 – 4 = 0
The maximum passengers that can be in the
train at any instance of time is 6.
Input: enter[] = {5, 2, 2, 0}, exit[] = {0, 2, 2, 5}
Output:

Approach: The current capacity of the train at a particular station can be calculated by adding the number of people entering the train and subtracting the number of people exiting the train. The minimum capacity required will be the maximum of all the values of current capacities at all the stations.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the minimum capacity required``int` `minCapacity(``int` `enter[], ``int` `exit``[], ``int` `n)``{` `    ``// To store the minimum capacity``    ``int` `minCap = 0;` `    ``// To store the current capacity``    ``// of the train``    ``int` `currCap = 0;` `    ``// For every station``    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// Add the number of people entering the``        ``// train and subtract the number of people``        ``// exiting the train to get the``        ``// current capacity of the train``        ``currCap = currCap + enter[i] - ``exit``[i];` `        ``// Update the minimum capacity``        ``minCap = max(minCap, currCap);``    ``}` `    ``return` `minCap;``}` `// Driver code``int` `main()``{``    ``int` `enter[] = { 3, 5, 2, 0 };``    ``int` `exit``[] = { 0, 2, 4, 4 };``    ``int` `n = ``sizeof``(enter) / ``sizeof``(enter);` `    ``cout << minCapacity(enter, ``exit``, n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG``{` `// Function to return the minimum capacity required``static` `int` `minCapacity(``int` `enter[],``                       ``int` `exit[], ``int` `n)``{` `    ``// To store the minimum capacity``    ``int` `minCap = ``0``;` `    ``// To store the current capacity``    ``// of the train``    ``int` `currCap = ``0``;` `    ``// For every station``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{` `        ``// Add the number of people entering the``        ``// train and subtract the number of people``        ``// exiting the train to get the``        ``// current capacity of the train``        ``currCap = currCap + enter[i] - exit[i];` `        ``// Update the minimum capacity``        ``minCap = Math.max(minCap, currCap);``    ``}``    ``return` `minCap;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `enter[] = { ``3``, ``5``, ``2``, ``0` `};``    ``int` `exit[] = { ``0``, ``2``, ``4``, ``4` `};``    ``int` `n = enter.length;` `    ``System.out.println(minCapacity(enter, exit, n));``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 implementation of the approach` `# Function to return the``# minimum capacity required``def` `minCapacity(enter, exit, n):``    ` `    ``# To store the minimum capacity``    ``minCap ``=` `0``;` `    ``# To store the current capacity``    ``# of the train``    ``currCap ``=` `0``;` `    ``# For every station``    ``for` `i ``in` `range``(n):``        ` `        ``# Add the number of people entering the``        ``# train and subtract the number of people``        ``# exiting the train to get the``        ``# current capacity of the train``        ``currCap ``=` `currCap ``+` `enter[i] ``-` `exit[i];` `        ``# Update the minimum capacity``        ``minCap ``=` `max``(minCap, currCap);``    ``return` `minCap;` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``enter ``=` `[``3``, ``5``, ``2``, ``0``];``    ``exit ``=` `[``0``, ``2``, ``4``, ``4``];``    ``n ``=` `len``(enter);` `    ``print``(minCapacity(enter, exit, n));` `# This code is contributed by Princi Singh`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{` `// Function to return the minimum``// capacity required``static` `int` `minCapacity(``int` `[]enter,``                       ``int` `[]exit, ``int` `n)``{` `    ``// To store the minimum capacity``    ``int` `minCap = 0;` `    ``// To store the current capacity``    ``// of the train``    ``int` `currCap = 0;` `    ``// For every station``    ``for` `(``int` `i = 0; i < n; i++)``    ``{` `        ``// Add the number of people entering the``        ``// train and subtract the number of people``        ``// exiting the train to get the``        ``// current capacity of the train``        ``currCap = currCap + enter[i] - exit[i];` `        ``// Update the minimum capacity``        ``minCap = Math.Max(minCap, currCap);``    ``}``    ``return` `minCap;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]enter = { 3, 5, 2, 0 };``    ``int` `[]exit = { 0, 2, 4, 4 };``    ``int` `n = enter.Length;` `    ``Console.WriteLine(minCapacity(enter, exit, n));``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``
Output:

`6`

Time Complexity: O(n)

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