Cognizant GenC Coding Questions 2025 – Set 4

Cognizant GenC Coding Questions – Set 4 (Beginner to Intermediate)

Previously asked Cognizant GenC Coding Questions – Set 4 (Beginner to Intermediate)

These coding questions are curated 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: Find the Frequency of Elements in an Array

Given an array of integers, print the frequency of each element.

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

Solution in C++

#include 
using namespace std;

void printFrequencies(vector& arr) {
    unordered_map freq;
    for (int num : arr) freq[num]++;
    for (auto& pair : freq) {
        cout << pair.first << ": " << pair.second << endl;
    }
}

int main() {
    vector arr = {1, 2, 2, 3, 1, 4};
    printFrequencies(arr);
    return 0;
}

Solution in Python

from collections import Counter

def print_frequencies(arr):
    freq = Counter(arr)
    for num, count in freq.items():
        print(f"{num}: {count}")

arr = [1, 2, 2, 3, 1, 4]
print_frequencies(arr)

Explanation

Approach: Use a hash map (unordered_map in C++, Counter in Python) to count occurrences of each element. Iterate and print the results. Time Complexity: O(n), Space Complexity: O(k) where k is the number of unique elements.

Relevance: Frequency counting tests basic data structure usage, a common GenC topic.

Question 2: Check if Two Strings are Anagrams

Given two strings, check if they are anagrams (contain the same characters with the same frequencies).

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

Solution in C++

#include 
using namespace std;

bool areAnagrams(string s1, string s2) {
    if (s1.size() != s2.size()) return false;
    int count[26] = {0};
    for (int i = 0; i < s1.size(); i++) {
        count[s1[i] - 'a']++;
        count[s2[i] - 'a']--;
    }
    for (int i = 0; i < 26; i++) {
        if (count[i] != 0) return false;
    }
    return true;
}

int main() {
    string s1 = "listen", s2 = "silent";
    cout << (areAnagrams(s1, s2) ? "true" : "false") << endl;
    return 0;
}

Solution in Python

def are_anagrams(s1, s2):
    if len(s1) != len(s2):
        return False
    return sorted(s1) == sorted(s2)

s1, s2 = "listen", "silent"
print(are_anagrams(s1, s2))

Explanation

Approach: In C++, use a frequency array to increment for s1 and decrement for s2; if all counts are 0, they’re anagrams. In Python, sort and compare (simpler but less efficient). Time Complexity: O(n) for C++, O(n log n) for Python, Space Complexity: O(1) for C++, O(n) for Python.

Relevance: Anagram checks test string manipulation, a frequent GenC question.

Question 3: Find Largest Element in an Array

Given an array of integers, find the largest element without using built-in functions like max().

Input: arr = [3, 7, 2, 9, 1]
Output: 9

Solution in C++

#include 
using namespace std;

int findLargest(vector& arr) {
    if (arr.empty()) return -1;
    int largest = arr[0];
    for (int i = 1; i < arr.size(); i++) {
        if (arr[i] > largest) largest = arr[i];
    }
    return largest;
}

int main() {
    vector arr = {3, 7, 2, 9, 1};
    cout << findLargest(arr) << endl;
    return 0;
}

Solution in Python

def find_largest(arr):
    if not arr:
        return -1
    largest = arr[0]
    for num in arr[1:]:
        if num > largest:
            largest = num
    return largest

arr = [3, 7, 2, 9, 1]
print(find_largest(arr))

Explanation

Approach: Initialize the largest as the first element, then iterate to update it if a larger element is found. Time Complexity: O(n), Space Complexity: O(1).

Relevance: Finding extremes in arrays is a basic GenC problem testing iteration skills.

Question 4: Factorial of a Number

Given a non-negative integer, compute its factorial.

Input: n = 5
Output: 120 (5 * 4 * 3 * 2 * 1 = 120)

Solution in C++

#include 
using namespace std;

long long factorial(int n) {
    if (n < 0) return -1;
    long long result = 1;
    for (int i = 1; i <= n; i++) {
        result *= i;
    }
    return result;
}

int main() {
    int n = 5;
    cout << factorial(n) << endl;
    return 0;
}

Solution in Python

def factorial(n):
    if n < 0:
        return -1
    result = 1
    for i in range(1, n + 1):
        result *= i
    return result

n = 5
print(factorial(n))

Explanation

Approach: Multiply numbers from 1 to n iteratively. Use long long in C++ to handle larger values. Time Complexity: O(n), Space Complexity: O(1).

Relevance: Factorial computation tests loops and basic math, a classic GenC question.

Master your Cognizant GenC prep with additional coding problems and expert guidance! Keep learning and stay ahead.

Unlock More Placement Prep Resources at JobsAddaa

Discover more from Jobs Addaa

Subscribe to get the latest posts sent to your email.

Cognizant GenC Coding Questions 2025 – Set 4

Previously asked Cognizant GenC Coding Questions – Set 4 (Beginner to Intermediate)

Question 1: Find the Frequency of Elements in an Array

Solution in C++

Solution in Python

Explanation

Question 2: Check if Two Strings are Anagrams

Solution in C++

Solution in Python

Explanation

Question 3: Find Largest Element in an Array

Solution in C++

Solution in Python

Explanation

Question 4: Factorial of a Number

Solution in C++

Solution in Python

Explanation

Share to your friends:

Related

Discover more from Jobs Addaa

Discover more from Jobs Addaa