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Find the minimum capacity of the train required to hold the passengers
  • Difficulty Level : Easy
  • Last Updated : 09 Oct, 2019

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

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++

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// C++ implementation of the approach
#include <bits/stdc++.h>
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[0]);
  
    cout << minCapacity(enter, exit, n);
  
    return 0;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

6

Time Complexity: O(n)

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