TCS Coding Questions 2022 Day 2 Slot 1

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
import java.util.*;
class Solution
{
public static void main (String[]args)
{
Scanner sc = new Scanner (System.in);
int t = sc.nextInt ();
int e[] = new int[t];
int l[] = new int[t];

for (int i = 0; i < t; i++)
e[i] = sc.nextInt ();

for (int i = 0; i < t; i++)
l[i] = sc.nextInt ();

int max = 0, sum = 0;
for (int i = 0; i < t; i++)
{
sum += e[i] - l[i];
max = Math.max (sum, max);
}
System.out.println (max);
}
}

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


  • Suresh

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


  • Ankan

    //Using C
    #include
    int main()
    {
    int t,i,total=0,count=0;
    printf(“Enter no. of hours:”);
    scanf(“%d”,&t);

    int a[t],b[t];

    printf(“\nEnter the val of a array:”);
    for(i=0;i<t;i++)
    {
    scanf("%d",&a[i]);
    }

    printf("\nEnter the val of b array:");
    for(i=0;i<t;i++)
    {
    scanf("%d",&b[i]);
    }

    for(i=0;i<t;i++)
    {
    total += a[i]-b[i];
    if(totalcount)
    {
    count=total;
    }
    }

    printf(“%d”,count);
    return 0;
    }


  • 072

    Using Python
    time=int(input())
    entry=[int(x) for x in input().split()]
    exit=[int(x) for x in input().split()]
    count=0
    guests=[]
    for i in range(len(entry)):
    count=count+entry[i]-exit[i]
    guests.append(count)
    print(max(guests))


  • RATUL

    java code
    ———————————————————————————————————————————————————————————————————————————————————————- public static void main(String[] args) {
    // TODO code application logic here
    int t = 0;
    Scanner sc = new Scanner(System.in);
    t=sc.nextInt();
    int e[]=new int[t];
    int l[]=new int [t];
    for(int i=0;i<t;i++)
    {
    e[i]=sc.nextInt();
    }
    for(int i=0;i<t;i++)
    {
    l[i]=sc.nextInt();
    }
    int v[]=new int[t];
    v[0]=e[0]-l[0];
    for(int i=1;i<t;i++)
    {
    v[i]=v[i-1]+e[i]-l[i];
    }

    int m=v[0];
    for(int i=0;i<t;i++)
    {
    if(m<v[i])
    m=v[i];
    }

    System.out.println(m);
    }


  • suraj

    using python

    n = int(input(“Enter number of elements : “))

    a = list(map(int,input(“\nEnter the numbers : “).strip().split()))[:n]
    b = list(map(int,input(“\nEnter the numbers : “).strip().split()))[:n]
    temp=[]
    for i in range (0,n):
    s=a[i]+b[i]
    temp.append(s)
    print(max(temp))


  • Ansh

    C CODE FOR THIS QUESTION

    #include
    #include
    void main()
    {
    int max = INT_MIN;
    int sum = 0;
    int t =4;
    int E[5] = {3,5,2,0};
    int n = sizeof(E)/sizeof(E[0]);
    int L[5] = {0,2,4,4};
    int m = sizeof(L)/sizeof(L[0]);
    int j =0,k=0;
    for(int i=0;i<t;i++)
    {
    while(j<n && kmax)
    {
    max = sum;
    }
    }
    }

    printf(“%d”,max);
    }


  • goutham

    T=int(input(“value of number T: “))
    E=0
    L=0
    N=0
    Mox=0
    for i in range(1,T+1):
    E=int(input(“value of entry: “))
    L=int(input(“value of exit: “))
    N=N+E-L
    Mox=max(N,Mox)
    print(“maximum”,Mox)
    print(“Hour”,i,”:”)
    print(“Entry:”,E,” “,”Exit:”,L)
    print(“No. of guests on ship:”,N)
    print(“Hence,the maximum number of guests within {0} hours is {1}”.format(T+1,Mox))


    • Tamradhwaj

      java
      import java.util.Scanner;

      class GFG {
      public static void main(String[] arg) {
      Scanner sc = new Scanner(System.in);
      int a = sc.nextInt();
      int e = 0;
      int y[] = new int[a];
      int z[] = new int[a];
      for (int i = 0; i < a; i++) {
      y[i] = sc.nextInt();
      }
      for (int i = 0; i < a; i++)
      z[i] = sc.nextInt();

      for (int j = 0; j <= 0; j++) {
      e = y[j] – z[j];
      }
      for (int j = 1; j < a; j++) {
      e = e + y[j] – z[j];
      }
      System.out.println(e);
      }
      }