TCS Coding Questions 2022 Day 2 Slot 1

September 20, 2021

Coding Question 1 for 2022 (September slot)

In this article, we will discuss about the TCS Coding Question which is asked in the TCS placement test. This type of Coding Questions will help you to crack your upcoming TCS exam as well as during your inteview process.

TCS Coding Question 1 Day 1 Slot 1

TCS Coding Question Day 2 Slot 1 – Question 1

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.

Example 1:

Input :

  • 5    -> Value of T
  • [7,0,5,1,3]  -> E[], Element of E[0] to E[N-1], where input each element is separated by new line 
  • [1,2,1,3,4]   -> L[], Element of L[0] to L[N-1], while input each element is separate by new line.

Output :

8     -> Maximum number of guests on cruise at an instance.

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

Hour 3:

Entry: 5 Exit: 1

No. of guests on ship : 4+5-1=8

Hour 4:

Entry : 1 Exit : 3

No. of guests on ship : 8+1-3=6

Hour 5:

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  -> Value of T

[3,5,2,0]   -> E[], Element of E[0] to E[N-1], where input each element is separated by new line.

[0,2,4,4]    -> L[], Element of L[0] to L[N-1], while input each element in separated by new line

Output:

6

Cruise at an instance

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.

The input format for testing

The candidate has to write the code to accept 3 input.

First input- Accept  value for number of T(Positive integer number)

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

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

The output format for testing

The output should be a positive integer number or a message as given in the problem statement(Check the output in Example 1 and Example 2)

Constraints:

  • 1<=T<=25
  • 0<= E[i] <=500
  • 0<= L[i] <=500

29 comments on “TCS Coding Questions 2022 Day 2 Slot 1”


  • Priyadarshan Gahukar

    ent=[3,5,2,0]
    leav=[0,2,4,4]
    lis=[]

    a=ent[0]-leav[0]
    lis.append(a)
    for i in range(1,len(ent)):
    a = a+ent[i]-leav[i]
    lis.append(a)
    print(max(lis))


  • Tanya Nagpal

    // CPP SOLUTION WITH COMMENTS
    #include
    using namespace std;

    int main() {
    int t; // GIVEN TIME
    cin >> t;
    vector entry(t); // Vector to store values of ‘guests entering the cruise’
    vector exit(t); // Vector to store values of ‘guests leaving the cruise’

    // Input ‘entry’ values
    for (int i = 0; i > entry[i];

    // Input ‘exit’ values
    for (int i = 0; i > exit[i];

    int max = 0; // Variable to store the maximum difference
    int sum = 0; // Variable to store the current cumulative difference

    for (int i = 0; i < t; i++) {
    sum += entry[i] – exit[i]; // Calculate the difference between 'entry' and 'exit' for the current index
    max = max(sum, max); // Update 'max' with the maximum cumulative difference
    }

    cout << max << endl; // Output the maximum cumulative difference

    return 0;
    }


    • Tanya Nagpal

      #include
      using namespace std;

      int main() {
      int t; // GIVEN TIME
      cin >> t;

      vector entry(t); // Vector to store values of ‘guests entering the cruise’
      vector exit(t); // Vector to store values of ‘guests leaving the cruise’
      for (int i = 0; i> entry[i]; // Input ‘entry’ values

      for (int i = 0; i exit[i]; // Input ‘entry’ values

      int max = 0; // Variable to store the maximum difference
      int sum = 0; // Variable to store the current cumulative difference between entry and exit
      for (int i = 0; i < t; i++) {
      sum += entry[i] – exit[i]; // Stores cumulative sum
      max = max(sum, max); // Update 'max' with the maximum cumulative difference
      }
      cout << max << endl; // Output the maximum cumulative difference return 0;
      }


  • D.YAMINI

    t=int(input())
    e=[]
    l=[]
    for i in range(t):
    e.append(int(input()))
    for i in range(t):
    l.append(int(input()))
    m=[]
    a=0
    rem=0
    for i in range(t):
    rem=rem+e[i]
    rem=e[i]-l[i]
    if rem>a:
    a=rem

    print(a)


  • kimaya

    python solution

    time = 5
    people_in = [7,0,5,1,3]
    people_out = [1,2,1,3,4]
    universal_count = 0
    universal_count_list = list()

    for duration in range(0,time):
    universal_count = (universal_count+people_in[duration])-people_out[duration]
    universal_count_list.append(universal_count)

    print(max(universal_count_list))


  • sumanth

    Java Code by Sumanth HD
    This is Prefect code Without Complexity It is Simple to Understand

    package TCS;

    import java.util.Iterator;

    public class P5
    {

    public static void main(String[] args)
    {
    int E[]= {7,0,5,1,3};
    int L[]= {1,2,1,3,4};
    int Ans=0;
    int max=0;

    for (int i = 0; i max)
    {
    max=Ans;
    }

    }
    System.out.println(“\nMaximum number of guests on cruise at an instance is “+max);
    }

    }


  • sumanth

    Java Code by Sumanth HD
    This is Prefect code Without Complexity It is Simple to Understand
    package TCS;

    import java.util.Iterator;

    public class P5
    {

    public static void main(String[] args)
    {
    int E[]= {7,0,5,1,3};
    int L[]= {1,2,1,3,4};
    int Ans=0;
    int max=0;

    // int arr[]=new int[5];
    for (int i = 0; i max)
    {
    max=Ans;
    }

    }
    System.out.println(“\nMaximum number of guests on cruise at an instance is “+max);
    }

    }


  • adepuharidwathi

    t=int(input())
    a=list(map(int,input().split()))
    b=list(map(int,input().split()))
    d=[]
    c=0
    for i in range(t):
    c+=a[i]-b[i]
    d.append(c)
    print(max(d))


  • AJAHARUDDIN

    Python Solution:
    T = int(input(‘Limited time: ‘))
    E,L = [],[]
    for i in range(T):
    E.append(int(input(‘entry value:’)))
    for i in range(T):
    L.append(int(input(‘exit value:’)))
    Guests = 0
    maxm = 0
    for i in range(T):
    Guests+=E[i]-L[i]
    if Guests>maxm:
    maxm=Guests
    print(maxm)


  • nuthalapatidinesh098

    E=[]
    L=[]
    T=int(input(“range T:”))
    for i in range(T):
    e=int(input(“E:”))
    E.append(e)
    for i in range(T):
    l=int(input(“L:”))
    L.append(l)
    Sum=0
    Max=0
    for i in range(T):
    Sum+=E[i]-L[i]
    Max=max(Sum,Max)
    print(“output”,Max)


  • Suhag

    def maxPeople():
    T = int(input())
    E = list(map(int,input().split()))
    L = list(map(int,input().split()))

    people =0
    m=0
    for i in range(T):
    people = people + E[i]-L[i]
    m = max(people,m)

    return m

    maxPeople()


  • Suhag

    def maxPeople():
    T = int(input())
    E = list(map(int,input().split()))
    L = list(map(int,input().split()))

    people =0
    m=0
    for i in range(T):
    people = people + E[i]-L[i]
    m = max(people,m)

    return m

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