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Cognizant GenC Coding Questions 2025 – Set 1

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Cognizant GenC Coding Questions – Set 1 (Beginner to Intermediate)

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++

#include 
using 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++

#include 
using 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++

#include 
using 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++

#include 
using 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.