Cognizant GenC Coding Questions 2025 – Set 2

Previously Asked Cognizant GenC Coding Questions – Set 2 (Beginner to Intermediate) These coding questions are crafted for Cognizant GenC (Basic Track) preparation, reflecting the beginner to intermediate difficulty level observed in recent exams (2022-2024). Solutions are provided in C++ and Python with detailed explanations. Question 1: Rotate Array by K Positions Given an array of integers and a number k, rotate the array to the right by k positions. Input: arr = [1, 2, 3, 4, 5], k = 2 Output: [4, 5, 1, 2, 3] Solution in C++ #include using namespace std; void rotateArray(vector& arr, int k) { int n = arr.size(); k = k % n; // Handle k > n if (k == 0) return; reverse(arr.begin(), arr.begin() + n - k); reverse(arr.begin() + n - k, arr.end()); reverse(arr.begin(), arr.end()); } int main() { vector arr = {1, 2, 3, 4, 5}; int k = 2; rotateArray(arr, k); for (int x : arr) cout << x << " "; cout << endl; return 0; } Solution in Python def rotate_array(arr, k): n = len(arr) k = k % n # Handle k > n if k == 0: return arr[:] = arr[n-k:] + arr[:n-k] arr = [1, 2, 3, 4, 5] k = 2 rotate_array(arr, k) print(arr) Explanation Approach: In C++, use the reversal method: reverse the last k elements, then the first n-k, then the whole array. In Python, use slicing for simplicity. Time Complexity: O(n) for both, Space Complexity: O(1) for C++, O(n) for Python due to slicing. Relevance: Array rotation is a common GenC question testing manipulation skills. Question 2: Remove Duplicates from Sorted Array Given a sorted array, remove duplicates in-place and return the new length. Input: arr = [1, 1, 2, 3, 3, 4] Output: 4 (arr becomes [1, 2, 3, 4]) Solution in C++ #include using namespace std; int removeDuplicates(vector& arr) { if (arr.empty()) return 0; int writeIndex = 1; for (int i = 1; i < arr.size(); i++) { if (arr[i] != arr[i-1]) { arr[writeIndex] = arr[i]; writeIndex++; } } return writeIndex; } int main() { vector arr = {1, 1, 2, 3, 3, 4}; int newLength = removeDuplicates(arr); for (int i = 0; i < newLength; i++) cout << arr[i] << " "; cout << endl; return 0; } Solution in Python def remove_duplicates(arr): if not arr: return 0 write_index = 1 for i in range(1, len(arr)): if arr[i] != arr[i-1]: arr[write_index] = arr[i] write_index += 1 return write_index arr = [1, 1, 2, 3, 3, 4] new_length = remove_duplicates(arr) print(arr[:new_length]) Explanation Approach: Use a pointer to track the position for unique elements. Since the array is sorted, compare adjacent elements and move unique ones forward. Time Complexity: O(n), Space Complexity: O(1). Relevance: In-place array modification is a frequent GenC topic. Question 3: Count Vowels in a String Given a string, count the number of vowels (a, e, i, o, u) in it. Ignore case. Input: s = "Hello World" Output: 3 (e, o, o) Solution in C++ #include using namespace std; int countVowels(string s) { int count = 0; for (char c : s) { c = tolower(c); if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') { count++; } } return count; } int main() { string s = "Hello World"; cout << countVowels(s) << endl; return 0; } Solution in Python def count_vowels(s): vowels = set('aeiouAEIOU') return sum(1 for char in s if char in vowels) s = "Hello World" print(count_vowels(s)) Explanation Approach: Iterate through the string, check each character against vowels (case-insensitive). C++ uses a condition, Python uses a set for efficiency. Time Complexity: O(n), Space Complexity: O(1). Relevance: String traversal and character checking are staples in GenC exams. Question 4: Find GCD of Two Numbers Given two positive integers, find their Greatest Common Divisor (GCD) using the Euclidean algorithm. Input: a = 48, b = 18 Output: 6 Solution in C++ #include using namespace std; int gcd(int a, int b) { while (b != 0) { int temp = b; b = a % b; a = temp; } return a; } int main() { int a = 48, b = 18; cout << gcd(a, b) << endl; return 0; } Solution in Python def gcd(a, b): while b: a, b = b, a % b return a a, b = 48, 18 print(gcd(a, b)) Explanation Approach: Use the Euclidean algorithm: repeatedly divide and take the remainder until one number becomes 0. The other is the GCD. Time Complexity: O(log(min(a,b))), Space Complexity: O(1). Relevance: Mathematical logic questions like GCD are common in GenC coding rounds.