Previously Asked Cognizant GenC Coding Questions – Set 1 (Beginner to Intermediate)
These coding questions are designed for Cognizant GenC (Basic Track) preparation, reflecting the beginner to intermediate difficulty level seen in recent exams (2022-2024). Solutions are provided in C++ and Python with detailed explanations.
Question 1: Reverse an Array
Given an array of integers, reverse it in-place without using extra space.
Input: arr = [1, 2, 3, 4, 5] Output: [5, 4, 3, 2, 1]
Solution in C++
#includeusing namespace std; void reverseArray(vector & arr) { int left = 0, right = arr.size() - 1; while (left < right) { swap(arr[left], arr[right]); left++; right--; } } int main() { vector arr = {1, 2, 3, 4, 5}; reverseArray(arr); for (int x : arr) cout << x << " "; cout << endl; return 0; }
Solution in Python
def reverse_array(arr):
left, right = 0, len(arr) - 1
while left < right:
arr[left], arr[right] = arr[right], arr[left]
left += 1
right -= 1
arr = [1, 2, 3, 4, 5]
reverse_array(arr)
print(arr)
Explanation
Approach: Use two pointers (left and right) starting from the ends of the array. Swap elements and move pointers inward until they meet. Time Complexity: O(n/2) = O(n), Space Complexity: O(1).
Relevance: Frequently asked in GenC to test array manipulation and in-place operations.
Question 2: Check Palindrome String
Given a string, check if it is a palindrome (reads the same forward and backward). Ignore case and non-alphanumeric characters.
Input: s = "A man, a plan, a canal: Panama" Output: true
Solution in C++
#includeusing namespace std; bool isPalindrome(string s) { string clean = ""; for (char c : s) { if (isalnum(c)) clean += tolower(c); } int left = 0, right = clean.size() - 1; while (left < right) { if (clean[left] != clean[right]) return false; left++; right--; } return true; } int main() { string s = "A man, a plan, a canal: Panama"; cout << (isPalindrome(s) ? "true" : "false") << endl; return 0; }
Solution in Python
def is_palindrome(s):
clean = ''.join(c.lower() for c in s if c.isalnum())
return clean == clean[::-1]
s = "A man, a plan, a canal: Panama"
print(is_palindrome(s))
Explanation
Approach: Clean the string by keeping only alphanumeric characters and converting to lowercase. Compare characters from both ends (C++) or reverse the string (Python). Time Complexity: O(n), Space Complexity: O(n) for the cleaned string.
Relevance: Common in GenC to test string handling and two-pointer technique.
Question 3: Find Second Largest Element
Given an array of integers, find the second largest element without sorting the entire array.
Input: arr = [10, 5, 8, 12, 3] Output: 10
Solution in C++
#includeusing namespace std; int secondLargest(vector & arr) { if (arr.size() < 2) return -1; int first = INT_MIN, second = INT_MIN; for (int num : arr) { if (num > first) { second = first; first = num; } else if (num > second && num != first) { second = num; } } return second == INT_MIN ? -1 : second; } int main() { vector arr = {10, 5, 8, 12, 3}; cout << secondLargest(arr) << endl; return 0; }
Solution in Python
def second_largest(arr):
if len(arr) < 2:
return -1
first = second = float('-inf')
for num in arr:
if num > first:
second = first
first = num
elif num > second and num != first:
second = num
return second if second != float('-inf') else -1
arr = [10, 5, 8, 12, 3]
print(second_largest(arr))
Explanation
Approach: Track the largest (`first`) and second largest (`second`) elements in one pass. Update `second` when a number is less than `first` but greater than current `second`. Time Complexity: O(n), Space Complexity: O(1).
Relevance: Tests array traversal and conditional logic, a staple in GenC coding rounds.
Question 4: Sum of Digits Until Single Digit
Given a number, repeatedly sum its digits until the result is a single digit.
Input: n = 1234 Output: 1 (1+2+3+4 = 10, 1+0 = 1)
Solution in C++
#includeusing namespace std; int digitalRoot(int n) { while (n > 9) { int sum = 0; while (n > 0) { sum += n % 10; n /= 10; } n = sum; } return n; } int main() { int n = 1234; cout << digitalRoot(n) << endl; return 0; }
Solution in Python
def digital_root(n):
while n > 9:
n = sum(int(digit) for digit in str(n))
return n
n = 1234
print(digital_root(n))
Explanation
Approach: Extract digits by modulo/division (C++) or string conversion (Python), sum them, and repeat until a single digit remains. Alternatively, use the formula: digital root = 1 + ((n-1) % 9) if n > 0, but iterative is more common in GenC. Time Complexity: O(log n) per digit extraction, Space Complexity: O(1).
Relevance: Tests number manipulation and iteration, recently seen in GenC exams.
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