TCS NQT CRUISE SHIP CODING PROBLEM WITH SOLUTION

TCS NQT CRUISE SHIP CODING PROBLEM WITH SOLUTION

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A party has been organised on cruise. The party is organised for a limited time(T). The number of guests entering (E[i]) and leaving (L[i]) the party at every hour is represented as elements of the array. The task is to find the maximum number of guests present on the cruise at any given instance within T hours.

Input format

– First line- T (Positive integer number)

– Second line- T number of values, where each value is separated by a new line.

– Third line- T number of values, where each value is separated by a new line.

Output format

– First line: A positive integer N denoting the maximum number of people on board anytime.

Examples

– Example 1:

– Input :

5

7 0 5 1 3

1 2 1 3 4

– Output :

8

– Explanation:

– 1st hour: Entry -> 7 ; Exit -> 1 ; No. of guests on ship : 6

– 2nd hour: Entry -> 0 ; Exit -> 2 ; No. of guests on ship : 6-2=4

– 3rd hour: Entry -> 5 ; Exit -> 1 ; No. of guests on ship : 4+5-1=8

– 4th hour: Entry -> 1 ; Exit -> 3 ; No. of guests on ship : 8+1-3=6

– 5th hour: Entry -> 3 ; Exit -> 4 ; No. of guests on ship: 6+3-4=5

– Hence, the maximum number of guests within 5 hours is 8.

– Example 2:

– Input:

4

3 5 2 0

0 2 4 4

– Output:

6

– Explanation:

– Hour 1: Entry -> 3 ; Exit -> 0 ; No. of guests on ship: 3

– Hour 2: Entry -> 5 ; Exit -> 2 ; No. of guest on ship: 3+5-2=6

– Hour 3: Entry -> 2 ; Exit -> 4 ; No. of guests on ship: 6+2-4= 4

– Hour 4: Entry -> 0 ; Exit -> 4 ; No. of guests on ship : 4+0-4=0

– Hence, the maximum number of guests within 5 hours is 6.

Constraints:

– 1 <= T <= 25

– 0 <= E[i] <= 500

– 0 <= L[i] <= 500

Program In C

``````#include <stdio.h>
#include <stdlib.h>

int main() {
unsigned short int T, i;
scanf("%hu", &T);
unsigned int * E = (unsigned int *) malloc (sizeof(unsigned int) * T);
unsigned int * L = (unsigned int *) malloc (sizeof(unsigned int) * T);
for (i = 0 ; i < T ; i++)
scanf("%u", &E[i]);
for (i = 0 ; i < T ; i++)
scanf("%u", &L[i]);
unsigned int current_presence = 0, max_presence = 0;
for (i = 0 ; i < T ; i++) {
current_presence += E[i] - L[i];
if (max_presence < current_presence)
max_presence = current_presence;
}
printf("%d", max_presence);
free(E);
free(L);
return 0;
}
``````

Program In C++

``````#include <iostream>
#include <vector>
using std::cin;
using std::cout;
int main() {
unsigned short int T, i;
cin >> T;
std::vector <unsigned int> E (T);
std::vector <unsigned int> L (T);
for (i = 0 ; i < T ; i++)
cin >> E.at(i);
for (i = 0 ; i < T ; i++)
cin >> L.at(i);
unsigned int current_presence {0}, max_presence {0};
for (i = 0 ; i < T ; i++) {
current_presence += E.at(i) - L.at(i);
if (max_presence < current_presence)
max_presence = current_presence;
}
cout << max_presence;
return 0;
}
``````

Program In Java

``````import java.util.Scanner;
class Main {
public static void main (String args[]) {
Scanner in = new Scanner(System.in);
int T = in.nextInt();
int E[] = new int[T];
int L[] = new int[T];
for (int i = 0 ; i < T ; i++)
E[i] = in.nextInt();
for (int i = 0 ; i < T ; i++)
L[i] = in.nextInt();
in.close();
int current_presence = 0, max_presence = 0;
for (int i = 0 ; i < T ; i++) {
current_presence += E[i] - L[i];
if (max_presence < current_presence)
max_presence = current_presence;
}
System.out.print(max_presence);
}
}
``````

Program In Python

``````if __name__ == '__main__':
T = int(input())
E = list(map(int, input().split()))
L = list(map(int, input().split()))
current_presence = 0
max_presence = 0
for i in range(T):
current_presence += E[i] - L[i]
if max_presence < current_presence:
max_presence = current_presence
print (max_presence, end = '')
``````