# Find the minimum capacity of the train required to hold the passengers

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: 6
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: 5

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 `

Output:

```6
```

Time Complexity: O(n)

My Personal Notes arrow_drop_up Competitive Programmer, Full Stack Developer, Technical Content Writer, Machine Learner

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

2

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.