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TCS NQT Coding Questions with Solutions | PariksaHub

TCS NQT Coding Questions with Solutions

1. Write a program to check if a number is a palindrome. 2. Write a program to find the second largest number in an array. 3. Write a program to reverse a string.

10 questionsTCS NQTMixed

tcstsc nqtnqt examtcs exam

Questions & Code Solutions

Table of Contents10 questions

1Check if a Number is a Palindrome

Easy

Problem Statement

Given an integer, determine if it is a palindrome. A palindrome number reads the same backward as forward.

Example 1:

plain

Input: 121
Output: True

Example 2:

plain

Input: -121
Output: False

python

def isPalindrome(x):
    if x < 0:
        return False
    
    str_x = str(x)
    return str_x == str_x[::-1]


print(isPalindrome(121))
print(isPalindrome(-121))
print(isPalindrome(10))
print(isPalindrome(0))

python · Withou Built-in Functions

def isPalindrome(x):
    if x < 0:
        return False
    
    original = x
    reversed_num = 0
    
    while x > 0:

Why This Works?

Take the number 121. Extract the last digit (1), then the next (2), then the last (1). Build a new number from these digits in reverse order: 1 → 12 → 121. If the reversed number equals the original, it's a palindrome.

2Reverse a String (Without Built‑in Functions)

Medium

Problem Statement

You are given a string s.

Your task is to reverse the string without using built‑in reverse functions.

Return the reversed string.

Example

Example 1

Input: s = "hello"

Output: "olleh"

Example 2

Input: s = "TCSNQT"

Output: "TQNSCT"

Approach: Two Pointers

Convert the string into a character array (for mutability).
Use two pointers: one at the start, one at the end.
Swap characters until the pointers meet.
Return the new string formed from the reversed array.

Complexity

Time: O(n) (single pass)
Space: O(1) (in‑place reversal, ignoring output string storage)

java

public class Solution {
    public String reverse(String s) {
        char[] arr = s.toCharArray();
        int left = 0;
        int right = arr.length - 1;

        while (left < right) {
            char temp = arr[left];
            arr[left] = arr[right];
            arr[right] = temp;
            left++;
            right--;
        }

        return new String(arr);
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        System.out.println(sol.reverse("hello"));   // olleh
        System.out.println(sol.reverse("TCSNQT"));  // TQNSCT
    }
}

3Find GCD and LCM of two numbers

Medium

Problem Statement

You are given two integers a and b.

Your task is to compute:

The Greatest Common Divisor (GCD) of a and b.
The Least Common Multiple (LCM) of a and b.

Return both values.

Example 1

Input: a = 12, b = 18

Output: GCD = 6, LCM = 36

Explanation:

GCD of 12 and 18 is 6.
LCM is (a * b) / GCD = (12 * 18) / 6 = 36.

Example 2

Input: a = 7, b = 5

Output: GCD = 1, LCM = 35

Approach: Euclidean Algorithm

GCD: Use the Euclidean algorithm:
gcd(a, b) = gcd(b, a % b) until b = 0.
LCM: Once GCD is known, compute LCM using the formula:
LCM = (a * b) / GCD.

Complexity

Time: O(log(min(a, b))) (Euclidean algorithm)
Space: O(1) (constant extra space)

java

public class Solution {
    public static int gcd(int a, int b) {
        while (b != 0) {
            int temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }

    public static int lcm(int a, int b) {
        return (a * b) / gcd(a, b);
    }

    public static void main(String[] args) {
        int a = 12, b = 18;
        System.out.println("GCD: " + gcd(a, b)); // 6
        System.out.println("LCM: " + lcm(a, b)); // 36

        a = 7; b = 5;
        System.out.println("GCD: " + gcd(a, b)); // 1
        System.out.println("LCM: " + lcm(a, b)); // 35
    }
}

4Count vowels and consonants in a string

Medium

Problem Statement

You are given a string s consisting of lowercase and uppercase English letters.

Your task is to count the number of vowels and consonants in the string.

Return both counts.

Example 1

Input: s = "hello"

Output: Vowels = 2, Consonants = 3

Explanation:

Vowels: e, o → 2
Consonants: h, l, l → 3

Example 2

Input: s = "TCSNQT"

Output: Vowels = 1, Consonants = 5

Explanation:

Vowel: Q → none, only A, E, I, O, U are vowels.
Consonants: T, C, S, N, Q, T → 6 (but no vowels here).

Approach

Define a set of vowels: {a, e, i, o, u} (case‑insensitive).
Iterate through each character in the string.
If the character is a vowel, increment vowel count.
If it is an alphabet but not a vowel, increment consonant count.

Complexity

Time: O(n) (single pass through the string)
Space: O(1) (constant extra space)

java

public class Solution {
    public static void countVowelsAndConsonants(String s) {
        int vowels = 0, consonants = 0;
        s = s.toLowerCase();

        for (char c :

5Second Largest Number in An Array

Easy

Problem Statement

Given an array of integers, find the second largest number in the array.

Example 1:

plain

Input: [10, 5, 20, 8, 15]
Output: 15

Example 2:

plain

Input: [10, 10, 10]
Output: None (or -1)

Example 3:

plain

Input: [5, 5, 10, 10, 8]
Output: 8

Constraints:

The array contains at least 1 element
Array elements are integers
If all elements are the same or only one unique element exists, return None or -1

The Solution

Traverse the array once and keep track of the largest and second largest numbers.

Steps:

Initialize largest and second_largest to negative infinity
Iterate through each number in the array
If number is greater than largest, update second_largest to largest, then update largest
Else if number is greater than second_largest and not equal to largest, update second_largest
Return second_largest if it exists, otherwise None

Time Complexity: O(n) - single pass through array

Space Complexity: O(1) - only use two variables

python

def secondLargest(arr):
    if len(

6Fibonacci Series up to N terms

Easy

Problem Statement

Given a number N, print the Fibonacci series up to N terms. The Fibonacci series is a sequence where each number is the sum of the two preceding ones, starting from 0 and 1.

Example 1:

plain

Input: N = 5
Output: 0 1 1 2 3

Example 2:

plain

Input: N = 8
Output: 0 1 1 2 3 5 8 13

Example 3:

plain

Input: N = 1
Output: 0

Constraints:

N >= 1
N is an integer

The Solution

Use two variables to keep track of the previous two numbers and generate the next number in the series.

Steps:

Handle edge case when N is 0 or less
Initialize first two numbers as 0 and 1
Print the first number
Loop N-1 times to generate remaining terms
Calculate next number as sum of previous two
Update the two variables for next iteration

Time Complexity: O(n) - generate n terms

Space Complexity: O(1) - only use two variables

python

def fibonacci(n):
    if n <= 0:
        return

7Prime Number

Easy

Problem Statement

Given an integer, determine if it is a prime number. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

Example 1:

plain

Input: 7
Output: True

Example 2:

plain

Input: 10
Output: False

Example 3:

plain

Input: 2
Output: True

Constraints:

The input is a positive integer
Number can be from 1 to 10^6

The Solution

Check divisibility only up to the square root, and optimize by handling 2 and 3 separately, then checking only numbers of form 6k±1.

Steps:

Handle numbers less than 2 (not prime)
Handle 2 and 3 specially (both prime)
Eliminate multiples of 2 and 3
Check only numbers of form 6k±1 up to sqrt(n)
Return True if no divisor found

Time Complexity: O(√n)

Space Complexity: O(1) - only use variables

python

def isPrime(n):
    if n <= 1:
        return

8Reverse a String

Medium

Problem Statement

Given a string, reverse it and return the reversed string.

Example 1:

plain

Input: "hello"
Output: "olleh"

Example 2:

plain

Input: "Python"
Output: "nohtyP"

Example 3:

plain

Input: "a"
Output: "a"

Constraints:

The string contains ASCII characters
String length is between 0 and 10^4

The Solution

Iterate through the string from end to start and build the reversed string.

Steps:

Initialize an empty result string
Loop through the input string from last index to first
Add each character to the result string
Return the result string

Time Complexity: O(n) - iterate through string once

Space Complexity: O(n) - store reversed string

python

def reverseString(s):
    result = ""
    
    i = len(s) -

9Rotate an Array by K Positions

Medium

Problem Statement

Given an array and an integer K, rotate the array to the right by K positions.

Example 1"

plain

Input: arr = [1, 2, 3, 4, 5], K = 2
Output: [4, 5, 1, 2, 3]

Example 2:

plain

Input: arr = [10, 20, 30, 40], K = 1
Output: [40, 10, 20, 30]

Example 3:

plain

Input: arr = [1, 2, 3], K = 5
Output: [2, 3, 1]

Constraints:

Array length is between 1 and 10^4
K can be greater than array length
Array elements are integers

The Solution

Use array slicing to split the array at the rotation point and recombine.

Steps:

Handle edge case when array is empty
Normalize K using modulo to handle K greater than array length
Split array into two parts at position (n - K)
Combine the second part with the first part
Return the rotated array

Time Complexity: O(n) - create new array with all elements

Space Complexity: O(n) - store rotated array

python

def rotateArray(arr, k):
    if len(arr) ==

10Find Missing Number in Array

Medium

Problem Statement

Given an array containing n distinct numbers taken from 1 to n+1, find the one number that is missing from the array.

Example 1:

plain

Input: [1, 2, 4, 5, 6]
Output: 3

Example 2:

plain

Input: [2, 3, 4, 5, 6, 7, 8]
Output: 1

Example 3:

plain

Input: [1, 2, 3, 4, 5, 7, 8]
Output: 6

Constraints:

Array length is n where numbers are from 1 to n+1
All numbers are distinct
Only one number is missing

The Solution

Calculate the expected sum of numbers from 1 to n+1, then subtract the actual sum of array elements.

Steps:

Find the length of array (n)
Calculate expected sum using formula: (n+1) * (n+2) / 2
Calculate actual sum of all elements in array
The difference is the missing number
Return the missing number

Time Complexity: O(n) - iterate through array once

Space Complexity: O(1) - only use variables

python

def findMissingNumber(arr):
    n = len(arr)
    
    expected_sum =