TCS NQT Chain Marketing Coding Question
We at gradjobopenings.com provide free job alerts of freshers job drives. In this website we list on campus job openings for freshers and off campus job openings for freshers and also work from home job openings. This is the best website to apply for off campus drive in India. Visit our website for government job alerts and private job alerts. We also list free interview notes and study materials, one of the best interview study website.comfortable to face the interviews:
Problem statement
Chain Marketing Sales Organization has a scheme for income generation through which its members generate income for themselves. The scheme is such that suppose A joins the scheme and makes R and V join this scheme then A is Parent Member of R and V who are child Members. When any member joins the scheme then the parent gets a total commission of 10% from each of its child members. Child members receive a commission of 5% respectively. If a Parent member does not have any member joined under him, then he/she gets a commission of 5%.
Take the Name of the Members Joining the Scheme as input. Display how many members joined the scheme including parent members, Calculate the Total commission gained by each member in the scheme. The fixed amount for joining the scheme is Rs.5000 on which commission will be generated.
Scheme Amount=5000
Examples
1. When there are more than one child members
– **Input:** (Do not give Input Prompts. Accept values as follows.)
“`c
Amit //Enter Parent Member Name as this
Y //Enter Y if Parent Member have child members otherwise enter N
Rajesh, Virat //Enter Name of child members of Amit in comma separated
“`
– Output:* (Final output must be in the format given below.)
“`
TOTAL MEMBERS: 3
COMMISSION DETAILS
Amit: 1000 INR
Rajesh: 250 INR
Virat: 250INR
“`
2. When there is only one child member in the hierarchy
Input:
“`c
Amit
Y
Rajesh
“`
– Output:
“`
TOTAL MEMBERS: 2
COMMISSION DETAILS
Amit: 500 INR
Rajesh: 250 INR
“`
Algorithm
“`
1. Read head
2. Read whether head has child or not (y/n)
3. if head has child (y), do
Read children
count = number of commas in children
Print TOTAL MEMBERS: (count + 2)
Print COMMISSION DETAILS
Print (head) ((commas + 1) * 500) “INR”
children = split children with ‘, ‘ as delimiter
for every child in children, do
Print (child): 250 INR
else do,
print TOTAL MEMBERS: 1
print COMMISSION DETAILS
print (head): 250 INR
Program In C Language
#include <stdio.h>
int count_commas(char children[])
{
int count = 0, i = 0;
while(children[i] != '\0')
{
if (children[i] == ',')
count++;
i++;
}
return count;
}
int main()
{
char head[100], has_child;
scanf("%s", head);
scanf(" %c", &has_child);
char children[100];
if (has_child == 'Y' || has_child == 'y')
{
scanf(" %[^\n]s", children);
int commas = count_commas(children);
printf("TOTAL MEMBERS: %d\n", commas + 2);
printf("COMMISSION DETAILS\n");
printf("%s: %d INR\n", head, (commas + 1) * 500);
commas = 0;
while(1)
{
if (children[commas] == '\0')
{
printf(": 250 INR");
break;
}
if (children[commas] == ',')
{
printf(": 250 INR\n");
commas += 2;
continue;
}
printf("%c", children[commas]);
commas++;
}
}
else
{
printf("TOTAL MEMBERS: 1\n");
printf("COMMISSION DETAILS\n");
printf("%s: 250 INR\n", head);
}
return 0;
}
Program In C++
#include <iostream>
#include <algorithm>
#include <string>
using std::string;
using std::cin;
using std::cout;
using std::count;
using std::endl;
int main() {
string head;
char has_child;
cin >> head >> has_child;
if ( has_child == 'Y' || has_child == 'y' )
{
string children;
cin.ignore();
getline(cin, children);
size_t commas = count (children.begin(), children.end(), ',');
cout << "TOTAL MEMBERS: " << commas + 2;
cout << endl << "COMMISSION DETAILS" << endl;
cout << head << ": " << (commas + 1) * 500 << " INR" << endl;
int start;
while(1)
{
start = children.find(", ");
if (start == -1)
{
cout << children << ": 250 INR" << endl;
break;
}
cout << children.substr(0, start) << ": 250 INR" << endl;
children.erase ( 0, start + 2 );
}
}
else
{
cout << "TOTAL MEMBERS: 1" << endl;
cout << "COMMISSION DETAILS" << endl;
cout << head << ": 250 INR" << endl;
}
return 0;
}
Program in Java
import java.util.Scanner;
class Main
{
public static void main(String[] args)
{
Scanner in = new Scanner(System.in);
String head = in.next();
char has_child = in.next().charAt(0);
in.nextLine();
String children="";
int comm = 1;
String[] children_arr = {};
if (has_child == 'Y' || has_child == 'y')
{
children = in.nextLine();
children_arr = children.split(", ");
comm += children_arr.length;
}
System.out.println("TOTAL MEMBERS: " + comm);
System.out.println("COMMISSION DETAILS");
if (has_child == 'Y' || has_child == 'y')
{
System.out.println(head + ": " + (comm - 1) * 500 + " INR");
for (String child : children_arr)
System.out.println(child + ": 250 INR");
}
else
System.out.println(head + ": 250 INR");
}
}
Program in Python
if __name__ == "__main__":
head = input()
has_child = input()
children = []
if has_child is 'Y' or has_child is 'y':
children = input().split(', ')
print ("TOTAL MEMBERS:", len(children) + 1)
print ("COMMISSION DETAILS")
if has_child is 'Y' or has_child is 'y':
print (head + ": " + str(len(children) * 500) + " INR")
for i in children:
print(i + ": 250 INR")
else:
print(head + ": 250 INR")
Pingback: TCS NQT CODING QUESTIONS WITH SOLUTIONS - GRAD JOB OPENINGS