TCS Coding Questions 2025 - Programming Logic & Hands-on Coding Problems with Solutions

TCS Placement Paper Coding Questions - Complete Guide

Practice with 25+ TCS placement paper coding questions covering Programming Logic MCQs and Hands-on Coding problems. These questions are representative of what you’ll encounter in the TCS NQT (National Qualifier Test) and TCS Digital hiring exams.

What’s Included:

• 10 Programming Logic Questions: Output prediction, loop analysis, pointer questions
• 15+ Hands-on Coding Problems: Easy and Medium level problems with solutions
• Multiple Language Solutions: Java, Python, and C++ solutions
• Time Complexity Analysis: Every solution includes complexity analysis

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TCS NQT Coding Section Overview

TCS NQT Coding Pattern 2025

TCS NQT coding section consists of two parts: Programming Logic (MCQs) testing code understanding and output prediction, and Hands-on Coding requiring you to write complete programs. Master one language thoroughly (Java/Python/C++) before the exam.

TCS NQT Coding Section Breakdown:

Section	Questions	Time	Difficulty	Focus Areas
Programming Logic	10-15	20 min	Easy-Medium	Output prediction, loops, arrays, pointers
Hands-on Coding	1-2	50 min	Medium	Arrays, strings, basic algorithms, recursion

Total Coding Questions: 11-17 questions in 70 minutes

Languages Allowed: C, C++, Java, Python

Programming Logic Questions (MCQs)

Output Prediction

Q1: What will be the output?

#include <stdio.h>
int main() {
    int x = 5;
    printf("%d", x++ + ++x);
    return 0;
}

Analysis:

• x++ is post-increment: uses current value (5), then increments x to 6
• ++x is pre-increment: increments x (6→7), then uses value 7
• Expression evaluates to: 5 + 7 = 12

Answer: 12

Q2: What will be the output?

#include <stdio.h>
int main() {
    int a = 10, b = 5;
    printf("%d", a > b ? a : b);
    return 0;
}

Analysis:

• Ternary operator: condition ? value_if_true : value_if_false
• 10 > 5 is true, so it returns ‘a’ which is 10

Answer: 10

Q3: What will be the output?

#include <stdio.h>
int main() {
    int i;
    for(i = 0; i < 5; i++);
    printf("%d", i);
    return 0;
}

Analysis:

• Note the semicolon after the for loop - it’s an empty loop body
• Loop runs 5 times (i: 0→1→2→3→4→5), then exits when i=5
• After loop, i = 5

Answer: 5

Q4: What will be the output?

public class Test {
    public static void main(String[] args) {
        String s1 = "Hello";
        String s2 = "Hello";
        System.out.println(s1 == s2);
    }
}

Analysis:

• String literals are stored in String pool
• Both s1 and s2 point to same object in pool
• == compares references, which are same here

Answer: true

Q5: What will be the output?

x = [1, 2, 3]
y = x
y.append(4)
print(len(x))

Analysis:

• y = x creates a reference, not a copy
• Both x and y point to the same list
• Appending to y also modifies x
• Length of x is now 4

Answer: 4

Loop Analysis

Q6: How many times will the loop execute?

int i = 0;
while (i < 10) {
    if (i % 2 == 0) {
        printf("%d ", i);
    }
    i++;
}

Analysis:

• Loop runs for i = 0 to 9 (10 iterations total)
• printf executes only when i is even (0, 2, 4, 6, 8)
• Output: 0 2 4 6 8

Answer: Loop executes 10 times, printf executes 5 times

Q7: What is the sum after the loop?

int sum = 0;
for (int i = 1; i <= 5; i++) {
    sum += i * i;
}
printf("%d", sum);

Analysis:

• i=1: sum = 0 + 1 = 1
• i=2: sum = 1 + 4 = 5
• i=3: sum = 5 + 9 = 14
• i=4: sum = 14 + 16 = 30
• i=5: sum = 30 + 25 = 55

Answer: 55

Q8: What will be the output?

int i = 1, j = 1;
while (i <= 3) {
    while (j <= 3) {
        printf("%d%d ", i, j);
        j++;
    }
    i++;
}

Analysis:

• Inner loop runs completely once (j: 1→2→3→4)
• j is not reset, so inner loop never runs again
• Output: 11 12 13

Answer: 11 12 13

Array & Pointer

Q9: What will be the output?

int arr[] = {10, 20, 30, 40, 50};
int *ptr = arr;
printf("%d", *(ptr + 3));

Analysis:

• ptr points to arr[0] = 10
• ptr + 3 points to arr[3] = 40
• *(ptr + 3) dereferences to get value

Answer: 40

Q10: What will be the output?

int arr[] = {1, 2, 3, 4, 5};
printf("%d", sizeof(arr)/sizeof(arr[0]));

Analysis:

• sizeof(arr) = 20 bytes (5 integers × 4 bytes each)
• sizeof(arr[0]) = 4 bytes (size of one integer)
• 20/4 = 5 (number of elements)

Answer: 5

Hands-on Coding Problems

Easy Level

Second Largest Element in Array

Problem: Find the second largest element in an array without sorting.

Example:

Input: [12, 35, 1, 10, 34, 1]
Output: 34

Solution (Java):

public int findSecondLargest(int[] arr) {
    if (arr.length < 2) return -1;

    int largest = Integer.MIN_VALUE;
    int secondLargest = Integer.MIN_VALUE;

    for (int num : arr) {
        if (num > largest) {
            secondLargest = largest;
            largest = num;
        } else if (num > secondLargest && num != largest) {
            secondLargest = num;
        }
    }

    return secondLargest == Integer.MIN_VALUE ? -1 : secondLargest;
}

Solution (Python):

def find_second_largest(arr):
    if len(arr) < 2:
        return -1

    largest = second_largest = float('-inf')

    for num in arr:
        if num > largest:
            second_largest = largest
            largest = num
        elif num > second_largest and num != largest:
            second_largest = num

    return second_largest if second_largest != float('-inf') else -1

Time Complexity: O(n) | Space Complexity: O(1)

Reverse Words in a String

Problem: Given a string, reverse the order of words.

Example:

Input: "the sky is blue"
Output: "blue is sky the"

Solution (Java):

public String reverseWords(String s) {
    String[] words = s.trim().split("\\s+");
    StringBuilder result = new StringBuilder();

    for (int i = words.length - 1; i >= 0; i--) {
        result.append(words[i]);
        if (i > 0) result.append(" ");
    }

    return result.toString();
}

Solution (Python):

def reverse_words(s):
    words = s.split()
    return ' '.join(words[::-1])

Time Complexity: O(n) | Space Complexity: O(n)

Prime Number Check

Problem: Determine if a given number is prime.

Example:

Input: 17
Output: true (17 is prime)

Input: 18
Output: false (18 is not prime)

Solution (Java):

public boolean isPrime(int n) {
    if (n <= 1) return false;
    if (n <= 3) return true;
    if (n % 2 == 0 || n % 3 == 0) return false;

    for (int i = 5; i * i <= n; i += 6) {
        if (n % i == 0 || n % (i + 2) == 0) {
            return false;
        }
    }
    return true;
}

Solution (Python):

def is_prime(n):
    if n <= 1:
        return False
    if n <= 3:
        return True
    if n % 2 == 0 or n % 3 == 0:
        return False

    i = 5
    while i * i <= n:
        if n % i == 0 or n % (i + 2) == 0:
            return False
        i += 6
    return True

Time Complexity: O(√n) | Space Complexity: O(1)

Armstrong Number

Problem: Check if a number is an Armstrong number (sum of cubes of digits equals the number).

Example:

Input: 153
Output: true (1³ + 5³ + 3³ = 1 + 125 + 27 = 153)

Input: 123
Output: false

Solution (Java):

public boolean isArmstrong(int n) {
    int original = n;
    int digits = String.valueOf(n).length();
    int sum = 0;

    while (n > 0) {
        int digit = n % 10;
        sum += Math.pow(digit, digits);
        n /= 10;
    }

    return sum == original;
}

Solution (Python):

def is_armstrong(n):
    digits = len(str(n))
    original = n
    sum_val = 0

    while n > 0:
        digit = n % 10
        sum_val += digit ** digits
        n //= 10

    return sum_val == original

Time Complexity: O(d) where d is number of digits | Space Complexity: O(1)

Fibonacci Series

Problem: Generate Fibonacci series up to n terms.

Example:

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

Solution (Java):

public void printFibonacci(int n) {
    int a = 0, b = 1;

    for (int i = 0; i < n; i++) {
        System.out.print(a + " ");
        int sum = a + b;
        a = b;
        b = sum;
    }
}

Solution (Python):

def fibonacci(n):
    a, b = 0, 1
    result = []

    for _ in range(n):
        result.append(a)
        a, b = b, a + b

    return result

Time Complexity: O(n) | Space Complexity: O(1) or O(n) if storing

Palindrome Check (Number)

Problem: Check if a number reads the same forwards and backwards.

Example:

Input: 121
Output: true

Input: 123
Output: false

Solution (Java):

public boolean isPalindrome(int x) {
    if (x < 0) return false;

    int original = x;
    int reversed = 0;

    while (x > 0) {
        reversed = reversed * 10 + x % 10;
        x /= 10;
    }

    return original == reversed;
}

Solution (Python):

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

Time Complexity: O(log n) | Space Complexity: O(1)

Factorial Calculation

Problem: Calculate the factorial of a number.

Example:

Input: 5
Output: 120 (5! = 5×4×3×2×1 = 120)

Solution (Java):

public long factorial(int n) {
    if (n <= 1) return 1;

    long result = 1;
    for (int i = 2; i <= n; i++) {
        result *= i;
    }

    return result;
}

Solution (Python):

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

    result = 1
    for i in range(2, n + 1):
        result *= i
    return result

Time Complexity: O(n) | Space Complexity: O(1)

Sum of Digits

Problem: Find the sum of all digits in a number.

Example:

Input: 12345
Output: 15 (1+2+3+4+5 = 15)

Solution (Java):

public int sumOfDigits(int n) {
    int sum = 0;
    n = Math.abs(n);

    while (n > 0) {
        sum += n % 10;
        n /= 10;
    }

    return sum;
}

Solution (Python):

def sum_of_digits(n):
    return sum(int(d) for d in str(abs(n)))

Time Complexity: O(d) where d is number of digits | Space Complexity: O(1)

Medium Level

Array Rotation

Problem: Rotate an array to the right by k positions.

Example:

Input: nums = [1,2,3,4,5,6,7], k = 3
Output: [5,6,7,1,2,3,4]

Solution (Java):

public void rotate(int[] nums, int k) {
    int n = nums.length;
    k = k % n;

    reverse(nums, 0, n - 1);
    reverse(nums, 0, k - 1);
    reverse(nums, k, n - 1);
}

private void reverse(int[] nums, int start, int end) {
    while (start < end) {
        int temp = nums[start];
        nums[start] = nums[end];
        nums[end] = temp;
        start++;
        end--;
    }
}

Solution (Python):

def rotate(nums, k):
    n = len(nums)
    k = k % n
    nums[:] = nums[n-k:] + nums[:n-k]

Time Complexity: O(n) | Space Complexity: O(1)

Find Missing Number

Problem: Find the missing number in an array containing 1 to n.

Example:

Input: [1, 2, 4, 5, 6] (n = 6)
Output: 3

Solution (Java):

public int findMissing(int[] nums, int n) {
    int expectedSum = n * (n + 1) / 2;
    int actualSum = 0;

    for (int num : nums) {
        actualSum += num;
    }

    return expectedSum - actualSum;
}

Solution (Python):

def find_missing(nums, n):
    expected_sum = n * (n + 1) // 2
    actual_sum = sum(nums)
    return expected_sum - actual_sum

Time Complexity: O(n) | Space Complexity: O(1)

Remove Duplicates from Sorted Array

Problem: Remove duplicates in-place from a sorted array.

Example:

Input: [1, 1, 2, 2, 3, 4, 4]
Output: 4 (array becomes [1, 2, 3, 4, ...])

Solution (Java):

public int removeDuplicates(int[] nums) {
    if (nums.length == 0) return 0;

    int j = 0;
    for (int i = 1; i < nums.length; i++) {
        if (nums[i] != nums[j]) {
            j++;
            nums[j] = nums[i];
        }
    }

    return j + 1;
}

Solution (Python):

def remove_duplicates(nums):
    if not nums:
        return 0

    j = 0
    for i in range(1, len(nums)):
        if nums[i] != nums[j]:
            j += 1
            nums[j] = nums[i]

    return j + 1

Time Complexity: O(n) | Space Complexity: O(1)

Count Vowels and Consonants

Problem: Count the number of vowels and consonants in a string.

Example:

Input: "Hello World"
Output: Vowels: 3, Consonants: 7

Solution (Java):

public int[] countVowelsConsonants(String s) {
    int vowels = 0, consonants = 0;
    String vowelStr = "aeiouAEIOU";

    for (char c : s.toCharArray()) {
        if (Character.isLetter(c)) {
            if (vowelStr.indexOf(c) != -1) {
                vowels++;
            } else {
                consonants++;
            }
        }
    }

    return new int[]{vowels, consonants};
}

Solution (Python):

def count_vowels_consonants(s):
    vowels = set('aeiouAEIOU')
    vowel_count = 0
    consonant_count = 0

    for char in s:
        if char.isalpha():
            if char in vowels:
                vowel_count += 1
            else:
                consonant_count += 1

    return vowel_count, consonant_count

Time Complexity: O(n) | Space Complexity: O(1)

Two Sum

Problem: Find two numbers in an array that add up to a target sum.

Example:

Input: nums = [2, 7, 11, 15], target = 9
Output: [0, 1] (nums[0] + nums[1] = 2 + 7 = 9)

Solution (Java):

public int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> map = new HashMap<>();

    for (int i = 0; i < nums.length; i++) {
        int complement = target - nums[i];
        if (map.containsKey(complement)) {
            return new int[]{map.get(complement), i};
        }
        map.put(nums[i], i);
    }

    return new int[]{};
}

Solution (Python):

def two_sum(nums, target):
    seen = {}

    for i, num in enumerate(nums):
        complement = target - num
        if complement in seen:
            return [seen[complement], i]
        seen[num] = i

    return []

Time Complexity: O(n) | Space Complexity: O(n)

GCD of Two Numbers

Problem: Find the Greatest Common Divisor of two numbers.

Example:

Input: a = 48, b = 18
Output: 6

Solution (Java):

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

Solution (Python):

def gcd(a, b):
    while b:
        a, b = b, a % b
    return a

Time Complexity: O(log(min(a,b))) | Space Complexity: O(1)

Check Anagram

Problem: Check if two strings are anagrams of each other.

Example:

Input: s1 = "listen", s2 = "silent"
Output: true

Solution (Java):

public boolean isAnagram(String s1, String s2) {
    if (s1.length() != s2.length()) return false;

    int[] count = new int[26];

    for (int i = 0; i < s1.length(); i++) {
        count[s1.charAt(i) - 'a']++;
        count[s2.charAt(i) - 'a']--;
    }

    for (int c : count) {
        if (c != 0) return false;
    }

    return true;
}

Solution (Python):

def is_anagram(s1, s2):
    return sorted(s1.lower()) == sorted(s2.lower())

Time Complexity: O(n) or O(n log n) for sorting | Space Complexity: O(1) or O(n)

Merge Two Sorted Arrays

Problem: Merge two sorted arrays into one sorted array.

Example:

Input: arr1 = [1, 3, 5], arr2 = [2, 4, 6]
Output: [1, 2, 3, 4, 5, 6]

Solution (Java):

public int[] mergeSorted(int[] arr1, int[] arr2) {
    int[] result = new int[arr1.length + arr2.length];
    int i = 0, j = 0, k = 0;

    while (i < arr1.length && j < arr2.length) {
        if (arr1[i] <= arr2[j]) {
            result[k++] = arr1[i++];
        } else {
            result[k++] = arr2[j++];
        }
    }

    while (i < arr1.length) {
        result[k++] = arr1[i++];
    }

    while (j < arr2.length) {
        result[k++] = arr2[j++];
    }

    return result;
}

Solution (Python):

def merge_sorted(arr1, arr2):
    result = []
    i = j = 0

    while i < len(arr1) and j < len(arr2):
        if arr1[i] <= arr2[j]:
            result.append(arr1[i])
            i += 1
        else:
            result.append(arr2[j])
            j += 1

    result.extend(arr1[i:])
    result.extend(arr2[j:])
    return result

Time Complexity: O(n + m) | Space Complexity: O(n + m)

Pattern Printing - Right Triangle

Problem: Print a right triangle pattern of stars.

Example:

Input: n = 5
Output:
*
**
***
****
*****

Solution (Java):

public void printRightTriangle(int n) {
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= i; j++) {
            System.out.print("*");
        }
        System.out.println();
    }
}

Solution (Python):

def print_right_triangle(n):
    for i in range(1, n + 1):
        print('*' * i)

Time Complexity: O(n²) | Space Complexity: O(1)

Pattern Printing - Pyramid

Problem: Print a pyramid pattern of stars.

Example:

Input: n = 5
Output:
    *
   ***
  *****
 *******
*********

Solution (Java):

public void printPyramid(int n) {
    for (int i = 1; i <= n; i++) {
        // Print spaces
        for (int j = 1; j <= n - i; j++) {
            System.out.print(" ");
        }
        // Print stars
        for (int j = 1; j <= 2 * i - 1; j++) {
            System.out.print("*");
        }
        System.out.println();
    }
}

Solution (Python):

def print_pyramid(n):
    for i in range(1, n + 1):
        spaces = ' ' * (n - i)
        stars = '*' * (2 * i - 1)
        print(spaces + stars)

Time Complexity: O(n²) | Space Complexity: O(1)

String Problems

Palindrome String Check

Problem: Check if a string is a palindrome (ignoring case and non-alphanumeric characters).

Example:

Input: "A man, a plan, a canal: Panama"
Output: true

Solution (Java):

public boolean isPalindrome(String s) {
    int left = 0, right = s.length() - 1;

    while (left < right) {
        while (left < right && !Character.isLetterOrDigit(s.charAt(left))) {
            left++;
        }
        while (left < right && !Character.isLetterOrDigit(s.charAt(right))) {
            right--;
        }

        if (Character.toLowerCase(s.charAt(left)) !=
            Character.toLowerCase(s.charAt(right))) {
            return false;
        }
        left++;
        right--;
    }

    return true;
}

Solution (Python):

def is_palindrome(s):
    cleaned = ''.join(c.lower() for c in s if c.isalnum())
    return cleaned == cleaned[::-1]

Time Complexity: O(n) | Space Complexity: O(1) or O(n)

First Non-Repeating Character

Problem: Find the first non-repeating character in a string.

Example:

Input: "leetcode"
Output: 'l' (index 0)

Input: "loveleetcode"
Output: 'v' (index 2)

Solution (Java):

public char firstUniqChar(String s) {
    int[] count = new int[26];

    for (char c : s.toCharArray()) {
        count[c - 'a']++;
    }

    for (char c : s.toCharArray()) {
        if (count[c - 'a'] == 1) {
            return c;
        }
    }

    return '\0';
}

Solution (Python):

def first_unique_char(s):
    from collections import Counter
    count = Counter(s)

    for i, char in enumerate(s):
        if count[char] == 1:
            return char
    return None

Time Complexity: O(n) | Space Complexity: O(1) (fixed 26 chars)

Longest Common Prefix

Problem: Find the longest common prefix among an array of strings.

Example:

Input: ["flower", "flow", "flight"]
Output: "fl"

Solution (Java):

public String longestCommonPrefix(String[] strs) {
    if (strs == null || strs.length == 0) return "";

    String prefix = strs[0];

    for (int i = 1; i < strs.length; i++) {
        while (strs[i].indexOf(prefix) != 0) {
            prefix = prefix.substring(0, prefix.length() - 1);
            if (prefix.isEmpty()) return "";
        }
    }

    return prefix;
}

Solution (Python):

def longest_common_prefix(strs):
    if not strs:
        return ""

    prefix = strs[0]

    for s in strs[1:]:
        while not s.startswith(prefix):
            prefix = prefix[:-1]
            if not prefix:
                return ""

    return prefix

Time Complexity: O(S) where S is sum of all characters | Space Complexity: O(1)

String Compression

Problem: Compress a string using counts of repeated characters.

Example:

Input: "aabcccccaaa"
Output: "a2b1c5a3"

Solution (Java):

public String compress(String s) {
    if (s == null || s.length() == 0) return s;

    StringBuilder result = new StringBuilder();
    int count = 1;

    for (int i = 1; i <= s.length(); i++) {
        if (i < s.length() && s.charAt(i) == s.charAt(i - 1)) {
            count++;
        } else {
            result.append(s.charAt(i - 1)).append(count);
            count = 1;
        }
    }

    return result.length() < s.length() ? result.toString() : s;
}

Solution (Python):

def compress(s):
    if not s:
        return s

    result = []
    count = 1

    for i in range(1, len(s) + 1):
        if i < len(s) and s[i] == s[i - 1]:
            count += 1
        else:
            result.append(s[i - 1] + str(count))
            count = 1

    compressed = ''.join(result)
    return compressed if len(compressed) < len(s) else s

Time Complexity: O(n) | Space Complexity: O(n)

Practice Tips for TCS NQT Coding

Master One Language

• Choose C, C++, Java, or Python
• Master syntax, data types, and control structures
• Know standard library functions
• Practice writing code without IDE autocomplete

Focus on Fundamentals

• Arrays, strings, loops, conditionals
• Basic algorithms: sorting, searching
• Pattern printing problems
• Mathematical operations (GCD, factorial, prime)

Time Management

• Programming Logic: 1-2 min per MCQ
• Hands-on Coding: 25-30 min per problem
• Always test with sample inputs
• Handle edge cases (empty input, single element)

Common Mistakes to Avoid

• Off-by-one errors in loops
• Integer overflow for large numbers
• Not handling null/empty inputs
• Forgetting to return correct data type

TCS NQT Coding Section Strategy

Important Tips

TCS NQT coding section requires both speed and accuracy. Practice these problems until you can solve them quickly without errors.

1. Read the problem carefully - Understand input/output format
2. Identify the pattern - Most TCS problems follow standard patterns
3. Write clean code - Use meaningful variable names
4. Test with examples - Verify with given test cases
5. Handle edge cases - Empty arrays, negative numbers, large inputs

Practice More TCS Coding Questions →

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Complete TCS placement guide with eligibility, process, and salary details

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Practice TCS coding questions regularly to master programming fundamentals! Focus on array manipulation, string processing, and mathematical problems. These are the most common problem types in TCS NQT.

Pro Tip: TCS values clean, working code over complex solutions. Start with a brute force approach, then optimize if time permits.

Last updated: February 2026