Programming
Aptitude
Syllabus
Interview Preparation
Interview Exp.
Off Campus
0
Notifications Mark All Read
- Login
- Get Prime
0
TCS Ninja Hands On Coding Questions with Solutions
Related Banners
Prepinsta Prime
Get Prepinsta Prime
Get all 200+ courses offered by Prepinsta
Check out our instagram
Instagram
Whatsapp
Telegram
LinkedinGet Hiring Updates
Amazon,Google,Delottie & 30+companies are hiring ! Get hiring Updates right in your inbox from PrepInsta
PrepInsta Prime
Get over 200+ course One Subscription
Courses like AI/ML, Cloud Computing, Ethical Hacking, C, C++, Java, Python, DSA (All Languages), Competitive Coding (All Languages), TCS, Infosys, Wipro, Amazon, DBMS, SQL and others.
PrepInsta Prime
Get over 200+ course One Subscription
Courses like AI/ML, Cloud Computing, Ethical Hacking, C, C++, Java, Python, DSA (All Languages), Competitive Coding (All Languages), TCS, Infosys, Wipro, Amazon, DBMS, SQL and others.
Get PrepInsta Prime Subscription
Get access to all the courses that PrepInsta offers, check the out below -
Companies
TCS, Cognizant, Delloite, Infosys, Wipro, CoCubes, KPMG, Amazone, ZS Associates, Accenture, Congnizant & other 50+ companies
Programming
Data Structures, Top 500 Codes, C, C++, Java Python & other 10+ subjects
Skills
Full Stack Web Development, Data Science, Machine Learning, AWS Cloud, & other 10+ skills and 20+ projects
Companies
TCS, Cognizant, Delloite, Infosys, Wipro, CoCubes, KPMG, Amazone, ZS Associates, Accenture, Congnizant & other 50+ companies
Programming
TCS, Cognizant, Delloite, Infosys, Wipro, CoCubes, KPMG, Amazone, ZS Associates, Accenture, Congnizant & other 50+ companies
Skills
TCS, Cognizant, Delloite, Infosys, Wipro, CoCubes, KPMG, Amazone, ZS Associates, Accenture, Congnizant & other 50+ companies
TCS Ninja 2022 Hands on Coding Questions
TCS Ninja 2022 hands-on coding questions answers with pattern has been announced as a part of TCS Test. TCS Ninja Hands on Coding Questions Paper with Solutions are given here on this page. Also, you will find TCS ninja coding questions with solved papers. There are 2 questions that you need to solve in 45 mins(30 + 15 mins)
- No Of Questions:- 2 Question(1 Easy + 1 Hard)
- Time – 45 mins(15 mins + 30 mins)
IMPORTANT NOTE:
- There will be no negative marking.
- TCS NQT is adaptive this year
- You will not get any extra rough paper in the exam as a calculator and Rough Paper will be available on your Desktop Screen. You are not allowed to move your eyes down while giving the examination.
| Coding Question For | TCS Ninja |
| Number of Questions | 2 |
| Time Allotted | 45 Mins |
| Negative Marking | No |
| Cut Off | Positive Testcases are Necessary |
Marking Scheme for TCS Hands on Coding Questions
|
Coding Marks
|
Marks
|
Type
|
|---|---|---|
|
0 Test Case
|
0 Marks
|
NA
|
|
1 Test Case
|
5 Marks
|
Shown
|
|
2 Test Case
|
8 Marks
|
Shown
|
|
3 Test Case
|
12 Marks
|
Hidden
|
|
4 Test Case
|
18 Marks
|
Hidden
|
|
5 Test Case
|
22 Marks
|
Hidden
|
Note
Over all marking in this section depends on how many of your test cases are running, the more the merrier
Word is the Key 💖 (Link to this question)
One programming language has the following keywords that cannot be used as identifiers:
break, case, continue, default, defer, else, for, func, goto, if, map, range, return, struct, type, var
Write a program to find if the given word is a keyword or not
Test cases
Case 1
- Input – defer
- Expected Output – defer is a keyword
Case 2
- Input – While
- Expected Output – while is not a keyword
C Program
C++
Python
Java
C Program
#include<stdio.h> #include<string.h> int main(){ char str[16][10] = {"break", "case", "continue", "default", "defer", "else","for", "func", "goto", "if", "map", "range", "return", "struct", "type", "var"}; char input[20]; int flag = 0; scanf("%s",input); for(int i = 0; i<16;i++){ if(strcmp(input,str[i]) == 0){ flag = 1; break; } } if(flag==1){ printf("%s is a keyword",input); } else{ printf("%s is not a keyword",input); } return 0; }
C++
#include<iostream> #include<string.h> using namespace std; int main(){ char str[16][10] = {"break", "case", "continue", "default", "defer", "else","for", "func", "goto", "if", "map", "range", "return", "struct", "type", "var"}; char input[20]; int flag = 0; cin >> input; for(int i = 0; i<16;i++){ if(strcmp(input,str[i]) == 0){ flag = 1; break; } } if(flag==1){ cout << input << " is a keyword"; } else{ cout << input << " is not a keyword"; } return 0; }
Python
keyword = {"break", "case", "continue", "default", "defer", "else", "for",
"func", "goto", "if", "map", "range", "return", "struct", "type", "var"}
input_var = input()
if input_var in keyword:
print(input_var+ " is a keyword")
else:
print(input_var+ " is a not keyword")Java
import java.util.Scanner; public class Prep { public static voidmain(String args[]) { String str[]= {"break", "case", "continue", "default", "defer", "else","for", "func", "goto", "if", "map", "range", "return", "struct", "type", "var"}; int flag = 0; Scanner sc = new Scanner(System.in); String input=sc.nextLine(); for(int i = 0; i<16;i++){ if(str[i].equals(input)){ flag = 1; break; } } if(flag==1){ System.out.println(input+" is a keyword"); } else{ System.out.println(input+" is not a keyword"); } } } //The above is code is contributed by Jonty Chakraborty (PrepInsta Placement Cell Student)
Program 2
Sweet Seventeen
Sweet Seventeen 💖 (Link to this question)
Given a maximum of four digit to the base 17 (10 – A, 11 – B, 12 – C, 13 – D … 16 – G} as input, output its decimal value.
Test Cases
Case 1
- Input – 1A
- Expected Output – 27
Case 2
- Input – 23GF
- Expected Output – 10980
C Program
C++
Python
Java
C Program
#include <stdio.h> #include <math.h> #include <string.h> int main(){ char hex[17]; long long decimal, place; int i = 0, val, len; decimal = 0; place = 1; scanf("%s",hex); len = strlen(hex); len--; for(i = 0;hex[i]!='\0';i++) { if(hex[i]>='0'&& hex[i]<='9'){ //48 to 57 are ascii values of 0 - 9 //say value is 8 its ascii will be 56 //val = hex[i] - 48 => 56 - 48 => val = 8 val = hex[i] - 48; } else if(hex[i]>='a'&& hex[i]<='g'){ //97 to 103 are ascii values of a - g //say value is g its ascii will be 103 //val = hex[i] - 97 + 10 => 103 - 97 + 10=> val = 16 //10 is added as g value is 16 not 6 or a value is 10 not 0 val = hex[i] - 97 + 10; } else if(hex[i]>='A'&& hex[i]<='G'){ //similarly, 65 to 71 are values of A - G val = hex[i] - 65 + 10; } decimal = decimal + val * pow(17,len); len--; } printf("%lld",decimal); return 0; }
C++
#include <iostream> #include <math.h> #include <string.h> using namespace std; int main(){ char hex[17]; long long decimal, place; int i = 0, val, len; decimal = 0; place = 1; cin>> hex; len = strlen(hex); len--; for(i = 0;hex[i]!='\0';i++) { if(hex[i]>='0'&& hex[i]<='9'){ //48 to 57 are ascii values of 0 - 9 //say value is 8 its ascii will be 56 //val = hex[i] - 48 => 56 - 48 => val = 8 val = hex[i] - 48; } else if(hex[i]>='a'&& hex[i]<='g'){ //97 to 103 are ascii values of a - g //say value is g its ascii will be 103 //val = hex[i] - 97 + 10 => 103 - 97 + 10=> val = 16 //10 is added as g value is 16 not 6 or a value is 10 not 0 val = hex[i] - 97 + 10; } else if(hex[i]>='A'&& hex[i]<='G'){ //similarly, 65 to 71 are values of A - G val = hex[i] - 65 + 10; } decimal = decimal + val * pow(17,len); len--; } cout<< decimal; return 0; }
Python
'''The int() function converts the specified value into an integer number. We are using the same int() method to convert the given input. int() accepts two arguments, number and base. Base is optional and the default value is 10. In the following program we are converting to base 17''' num = str(input()) print(int(num,17))
Java
import java.util.*; public class Main { public static void main(String[] args) { HashMap<Character,Integer> hmap = new HashMap<Character,Integer>(); hmap.put('A',10); hmap.put('B',11); hmap.put('C',12); hmap.put('D',13); hmap.put('E',14); hmap.put('F',15); hmap.put('G',16); hmap.put('a',10); hmap.put('b',11); hmap.put('c',12); hmap.put('d',13); hmap.put('e',14); hmap.put('f',15); hmap.put('g',16); Scanner sin = new Scanner(System.in); String s = sin.nextLine(); long num=0; int k=0; for(int i=s.length()-1;i>=0;i--) { if((s.charAt(i)>='A'&&s.charAt(i)<='Z')||(s.charAt(i)>='a' &&s.charAt(i)<='z')) { num = num + hmap.get(s.charAt(i))*(int)Math.pow(17,k++); } else { num = num+((s.charAt(i)-'0')*(int)Math.pow(17,k++)); } } System.out.println(num); } }
If you want to prepare for more TCS Ninja Coding Round Questions with Solutions, click on the button.
If you want to prepare for TCS Digital Advanced Coding Round Question, click on the button.
Program 3
Oddly Even
The Oddly even 💖 (Link to this question)
Problem
Given a maximum of 100 digit numbers as input, find the difference between the sum of odd and even position digits
Test Cases
Case 1
- Input: 4567
- Expected Output: 2
Explanation : Odd positions are 4 and 6 as they are pos: 1 and pos: 3, both have sum 10. Similarly, 5 and 7 are at even positions pos: 2 and pos: 4 with sum 12. Thus, difference is 12 – 10 = 2
Case 2
- Input: 5476
- Expected Output: 2
Case 3
- Input: 9834698765123
- Expected Output: 1
C
C++
Python
Java
C
#include <stdio.h> #include <string.h> #include <stdlib.h> int main() { int a = 0,b = 0,i = 0, n; char num[100]; printf("Enter the number:"); scanf("%s",num); //get the input up to 100 digit n = strlen(num); while(n>0) { if(i==0) //add even digits when no of digit is even and vise versa { a+=num[n-1]-48; n--; i=1; } else //add odd digits when no of digit is even and vice versa { b+=num[n-1]-48; n--; i=0; } } printf("%d",abs(a-b)); //print the difference of odd and even return 0; }
C++
#include <iostream> #include <string.h> #include <stdlib.h>
using namespace std; int main() { int a = 0,b = 0,i = 0, n; char num[100]; cout<< "Enter the number:"; cin>> num; //get the input up to 100 digit n = strlen(num); while(n>0) { if(i==0) //add even digits when no of digit is even and vise versa { a+=num[n-1]-48; n--; i=1; } else //add odd digits when no of digit is even and vice versa { b+=num[n-1]-48; n--; i=0; } } cout<< abs(a-b); //print the difference of odd and even return 0; }
Python
num = [int(d) for d in str(input("Enter the number:"))] even,odd = 0,0 for i in range(0,len(num)): if i % 2 ==0: even = even + num[i] else: odd = odd + num[i] print(abs(odd-even)) # code contributed by Shubhanshu Arya PrepInsta Placement Cell Student
Java
import java.util.*; public class Main { public static void main(String[] args) { Scanner sin = new Scanner(System.in); String s=sin.nextLine(); long num = 0, num1 = 0; num=num + s.charAt(0)-'0'; for(int i=1;i<s.length();i++) { if(i%2==0) num = num + s.charAt(i)-'0'; else num1 = num1 + s.charAt(i)-'0'; } System.out.println(Math.abs(num-num1)); } }
Program 4
Prime Number with a Twist
Ques. Write a code to check whether no is prime or not. Condition use function check() to find whether entered no is positive or negative ,if negative then enter the no, And if yes pas no as a parameter to prime() and check whether no is prime or not?
C
C++
Java
C
#include<bits/stdc++.h>
using namespace std;
void prime(int num){
int count=0;
for(int i=2;i<num;i++){
if(num%i==0){
count++;
break;
}
}
if(count==0){
cout<<"prime"<<endl;
}
else{
cout<<"Not Prime"<<endl;
}
}
int main(){
int n;
cout<<"Enter the number: ";
cin>>n;
if(n>0){
prime(n);
}
else{
cout<<"negative number.Please enter a postive number"<<endl;
}
return 0;
}
C++
#include
using namespace std;
void enter();
void check(int);
void prime(int);
int main()
{
enter();
return 0;
}
void check(int num)
{
if(num<0)
{
cout<<"invalid input enter value again"<<endl;
enter();
}
else
{
prime(num);
}
}
void enter()
{
int num;
cout<<"Enter number:";
cin>>num;
check(num);
}
void prime(int num)
{
int i,div=0;
for(i=1;i<=num;i++)
{
if(num%i==0)
{
div++;
}
}
if(div==2)
{
cout<<num<<" is a prime number";
}
else
{
cout<<" is not a prime number";
}
}
Java
/* Write a progam in java and check if a number which is entered is prime number or not. However if the number
happens to be negative then we must ask the user to enter a positive number again*/
import java.util.Scanner;
class Main
{
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
//First we will ask the user to enter a number
System.out.println("Enter value to be evaluated : ");
int n = sc.nextInt();
//create object of class CheckPrime
Main prime=new Main();
//calling function with value n, as parameter
prime.verify(n);
}
//function for checking number is positive or negative
void verify(int n)
{
if(n<0)
System.out.println("Negative number detected enter positive number");
else
calc(n);
}
//creating function for checking prime or not
void calc(int n)
{
int x=0;
for(int i=2;i<n;i++)
{
if(n%i==0)
++x;
}
if(x>=1)
System.out.println("The number that you have entered is not prime");
else
System.out.println("The number that you have entered is prime");
}
}
Program 5
Leap Year or not
Program to check if a year is Leap Year or not 💖
C Program
C++
Java
Perl
Python
C Program
#include<stdio.h>
int main()
{
int year;
printf("Enter a year to check if it is a leap year\n");
scanf("%d", &year);
if (year%400 == 0) // Exactly divisible by 400 e.g. 1600, 2000
printf("%d is a leap year.\n", year);
else if (year%100 == 0) // Exactly divisible by 100 and not by 400 e.g. 1900, 2100
printf("%d isn't a leap year.\n", year);
else if (year%4 == 0) // Exactly divisible by 4 and neither by 100 nor 400 e.g. 2016, 2020
printf("%d is a leap year.\n", year);
else // Not divisible by 4 or 100 or 400 e.g. 2017, 2018, 2019
printf("%d isn't a leap year.\n", year);
return 0
}
C++
#include using namespace std; int main() { int year = 2016; if (((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0)) cout<<year<<" is a leap year"; else cout<<year<<" is not a leap year"; return 0; }
Output
2016 is a leap year
In the above program, if a year is divisible by 4 and not divisible by 100, then it is a leap year. Also, if a year is divisible by 400, it is a leap year.
This is demonstrated by the following code snippet.
if (((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0)) cout<<year<<" is a leap year"; else cout<<year<<" is not a leap year";
The program to check if a year is leap year or not can also be written using nested if statements. This is given as follows −
Java
import java.util.Scanner; public class Demo { public static void main(String[] args) { int year; Scanner scan = new Scanner(System.in); System.out.println("Enter any Year:"); year = scan.nextInt(); scan.close(); boolean isLeap = false; if(year % 4 == 0) { if( year % 100 == 0) { if ( year % 400 == 0) isLeap = true; else isLeap = false; } else isLeap = true; } else { isLeap = false; } if(isLeap==true) System.out.println(year + " is a Leap Year."); else System.out.println(year + " is not a Leap Year."); } }
Perl
# perl Script
# leap year
print "Enter Year: ";
$year=;
# condition to check for leap year
if( (0 == $year % 4) && (0 != $year % 100) || (0 == $year % 400) )
{
print "Leap year";
}
else
{
print "Not a leap year";
}
Output
Enter Year: 2016
Leap year
Python
# Python Program to Check Leap Year using If Statement year = int(input("Please Enter the Year Number you wish: ")) if (( year%400 == 0)or (( year%4 == 0 ) and ( year%100 != 0))): print("%d is a Leap Year" %year) else: print("%d is Not the Leap Year" %year)
Login/Signup to comment