TCS NQT – October 3rd, 2024 (Shift 1)
Question 1: Weekly Exercise Summary
Problem Statement:
You have to design a weekly exercise summary by taking the number of minutes of daily exercise for 7 consecutive days.
The exercise duration for all days will be entered by the user.
Your task is to:
- Calculate the total exercise duration for the week
- Calculate the average daily workout duration (rounded to nearest integer)
Input Format:
Day 1 exercise duration: <minutes> Day 2 exercise duration: <minutes> ... Day 7 exercise duration: <minutes>
Output Format:
Exercise summary Total exercise duration : <total> (minutes) Average daily exercise duration: <average> minutes
Example Input:
Day 1 exercise duration: 45 Day 2 exercise duration: 15 Day 3 exercise duration: 30 Day 4 exercise duration: 15 Day 5 exercise duration: 5 Day 6 exercise duration: 10 Day 7 exercise duration: 20
Example Output:
Exercise summary Total exercise duration : 140 (minutes) Average daily exercise duration: 20 minutes
Solution (Java):
import java.util.Scanner;
public class ExerciseSummary {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int[] exerciseDuration = new int[7];
int total = 0;
for (int i = 0; i < 7; i++) {
System.out.print(“Day ” + (i + 1) + ” exercise duration: “);
exerciseDuration[i] = scanner.nextInt();
total += exerciseDuration[i];
}
double average = total / 7.0;
average = Math.round(average * 100.0) / 100.0;
System.out.println(“Exercise Summary”);
System.out.println(“Total exercise duration: ” + total + ” minutes”);
System.out.println(“Average daily exercise duration: ” + average + ” minutes”);
scanner.close();
}
}
Question 2: Problem Statement:
You are given a range of integers from M to N (inclusive), where:
Mis the lower limitNis the upper limit
Your task is to find all palindrome numbers in the range and return the count of such numbers.
Input Format:
Enter M : <lower limit integer> Enter N : <upper limit integer>
Output Format:
<number of palindrome numbers>
Definition:
A palindrome number is a number that reads the same backward as forward.
Example: 121, 131, 11, 9
Example Input:
Enter M : 10 Enter N : 20
Example Output:
1
Explanation:
Only one palindrome exists between 10 and 20 → 11
Solution (C++):
#include <iostream>
using namespace std;
bool isPalindrome(int number) {
if (number < 0) return false;
int original = number, reversed = 0;
while (number > 0) {
reversed = reversed * 10 + number % 10;
number /= 10;
}
return original == reversed;
}
int main() {
int M, N;
cout << “Enter M: “;
cin >> M;
cout << “Enter N: “;
cin >> N;
int count = 0;
for (int i = M; i <= N; i++) {
if (isPalindrome(i)) count++;
}
cout << count << endl;
return 0;
}
TCS NQT – October 3rd, 2024 (Shift 2)
Question 3: Train Travel Time
Problem: A train covers 800 meters (400m track + 400m bridge) at a given speed (km/hr). Calculate the time taken in seconds using the formula: (Total distance / Speed) * (18/5).
Solution (C++):
#include <iostream>
using namespace std;
int main() {
double speed;
cin >> speed;
int totalDistance = 800;
double speedMs = speed * 5.0 / 18.0; // Convert km/hr to m/s
int time = totalDistance / speedMs;
cout << time << endl;
return 0;
}
Question 4: Split Array with Equal Averages
Problem: Given an array of integers, check if it can be split into two subarrays with equal averages. Output true or false.
Solution (Java):
import java.util.Scanner;
public class ArraySplit {
public static boolean canSplit(int[] arr, int n) {
int totalSum = 0;
for (int i = 0; i < n; i++) totalSum += arr[i];
int leftSum = 0;
for (int i = 0; i < n – 1; i++) {
leftSum += arr[i];
int rightSum = totalSum – leftSum;
if (leftSum * (n – i – 1) == rightSum * (i + 1)) return true;
}
return false;
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] arr = new int[n];
for (int i = 0; i < n; i++) arr[i] = sc.nextInt();
System.out.println(canSplit(arr, n) ? “true” : “false”);
sc.close();
}
}
TCS NQT – October 4th, 2024 (Shift 1)
Question 5: Perfect Donation Amount
Problem: Check if a donation amount is a perfect number (sum of its proper divisors equals the number). Return true if perfect, false otherwise.
Solution (Java):
import java.util.Scanner;
public class PerfectNumber {
public static boolean isPerfect(int num) {
int sum = 0;
for (int i = 1; i <= Math.sqrt(num); i++) {
if (num % i == 0) {
sum += i;
if (num / i != i && num / i != num) sum += num / i;
}
}
return sum == num;
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int num = sc.nextInt();
System.out.println(isPerfect(num));
sc.close();
}
}
Question 6: Inventory Frequency Counter
Problem: Given a string of space-separated item names, count the frequency of each item. If the input contains digits, output “Invalid input.”
Solution (Java):
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
public class InventoryManager {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String s = sc.nextLine();
Map<String, Integer> freq = new HashMap<>();
String word = “”;
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) >= ‘0’ && s.charAt(i) <= ‘9’) {
System.out.println(“Invalid input”);
return;
} else if (s.charAt(i) == ‘ ‘) {
if (!word.isEmpty()) freq.put(word, freq.getOrDefault(word, 0) + 1);
word = “”;
} else {
word += s.charAt(i);
}
}
if (!word.isEmpty()) freq.put(word, freq.getOrDefault(word, 0) + 1);
for (Map.Entry<String, Integer> entry : freq.entrySet()) {
System.out.println(entry.getKey() + ” ” + entry.getValue());
}
sc.close();
}
}
TCS NQT – October 4th, 2024 (Shift 2)
Question 7: Calculate Speed in km/hr
Problem: Given a distance of 1000 meters and time in seconds, calculate the speed in km/hr.
Solution (C++):
#include <iostream>
using namespace std;
int main() {
double time;
cin >> time;
double distance = 1000.0; // meters
double speedMs = distance / time; // m/s
double speedKmh = speedMs * 18.0 / 5.0; // Convert m/s to km/hr
cout << speedKmh << endl;
return 0;
}
Question 8: Jump Game
Problem: Given an array where each element is the maximum jump length from that index, check if you can reach the last index from the first index.
Solution (Java):
import java.util.Scanner;
public class JumpGame {
public static boolean canJump(int[] nums) {
int maxReach = 0;
for (int i = 0; i < nums.length && i <= maxReach; i++) {
maxReach = Math.max(maxReach, i + nums[i]);
if (maxReach >= nums.length – 1) return true;
}
return false;
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String input = sc.nextLine();
String[] s = input.split(“,”);
int[] nums = new int[s.length];
for (int i = 0; i < s.length; i++) {
nums[i] = Integer.parseInt(s[i].trim());
}
System.out.println(canJump(nums));
sc.close();
}
}
Tips for TCS NQT Success
- Practice Input Handling: Many questions use Scanner (Java) or cin (C++). Test with sample inputs to avoid errors.
- Master Arrays and Strings: Most problems involve arrays or string manipulation.
- Understand Math Basics: Questions like Train Travel Time and Speed Calculation require unit conversions. Brush up on formulas!
- Learn HashMaps: The Inventory Frequency Counter uses a HashMap, which is super common in coding tests.
- Test Edge Cases: For example, in the Jump Game, check if an array with one element or zeros works correctly.
- Time Management: TCS NQT has tight time limits. Solve easier problems first, then tackle complex ones like Jump Game.