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++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the minimum capacity requiredint 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 codeint 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;} |
Java
// Java implementation of the approachimport java.util.*;class GFG {// Function to return the minimum capacity requiredstatic 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 codepublic 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 requireddef 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 codeif __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 requiredstatic 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 codepublic 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
<script>// Javascript implementation of the approach// Function to return the minimum capacity requiredfunction minCapacity(enter, exit, n) { // To store the minimum capacity let minCap = 0; // To store the current capacity // of the train let currCap = 0; // For every station for (let 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 codelet enter = [3, 5, 2, 0];let exit = [0, 2, 4, 4];let n = enter.length;document.write(minCapacity(enter, exit, n));// This code is contributed by _saurabh_jaiswal.</script> |
Output:
6
Time Complexity: O(n)
Auxiliary Space: O(1), no extra space is required, so it is a constant
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