Previously Asked Cognizant GenC Coding Questions – Set 3 (Beginner to Intermediate)
These coding questions are tailored for Cognizant GenC (Basic Track) preparation, mirroring the beginner to intermediate difficulty level seen in recent exams (2022-2024). Solutions are provided in C++ and Python with detailed explanations.
Question 1: Find Missing Number in Array
Given an array of n-1 integers in the range 1 to n with no duplicates, find the missing number.
Input: arr = [1, 2, 4, 6, 3, 7, 8], n = 8 Output: 5
Solution in C++
#includeusing namespace std; int findMissingNumber(vector & arr, int n) { int expectedSum = (n * (n + 1)) / 2; int actualSum = 0; for (int num : arr) actualSum += num; return expectedSum - actualSum; } int main() { vector arr = {1, 2, 4, 6, 3, 7, 8}; int n = 8; cout << findMissingNumber(arr, n) << endl; return 0; }
Solution in Python
def find_missing_number(arr, n):
expected_sum = (n * (n + 1)) // 2
actual_sum = sum(arr)
return expected_sum - actual_sum
arr = [1, 2, 4, 6, 3, 7, 8]
n = 8
print(find_missing_number(arr, n))
Explanation
Approach: Calculate the expected sum of numbers from 1 to n using the formula n*(n+1)/2. Subtract the actual sum of the array to find the missing number. Time Complexity: O(n), Space Complexity: O(1).
Relevance: Mathematical array problems are common in GenC to test logical reasoning.
Question 2: Swap Characters to Make Strings Equal
Given two strings of equal length, check if they can be made equal by swapping two characters in one string.
Input: s1 = "abcd", s2 = "adcb" Output: true (Swap 'b' and 'd' in s1)
Solution in C++
#includeusing namespace std; bool canBeEqual(string s1, string s2) { if (s1.size() != s2.size()) return false; vector diff; for (int i = 0; i < s1.size(); i++) { if (s1[i] != s2[i]) diff.push_back(i); if (diff.size() > 2) return false; } if (diff.size() == 0) return true; if (diff.size() != 2) return false; return s1[diff[0]] == s2[diff[1]] && s1[diff[1]] == s2[diff[0]]; } int main() { string s1 = "abcd", s2 = "adcb"; cout << (canBeEqual(s1, s2) ? "true" : "false") << endl; return 0; }
Solution in Python
def can_be_equal(s1, s2):
if len(s1) != len(s2):
return False
diff = []
for i in range(len(s1)):
if s1[i] != s2[i]:
diff.append(i)
if len(diff) > 2:
return False
if len(diff) == 0:
return True
if len(diff) != 2:
return False
return s1[diff[0]] == s2[diff[1]] and s1[diff[1]] == s2[diff[0]]
s1, s2 = "abcd", "adcb"
print(can_be_equal(s1, s2))
Explanation
Approach: Compare strings and collect positions of differences. If exactly two positions differ and swapping them aligns the strings, return true. Time Complexity: O(n), Space Complexity: O(1) or O(n) for diff array.
Relevance: String comparison with constraints is a typical GenC problem.
Question 3: Check Prime Number
Given an integer, check if it is a prime number (divisible only by 1 and itself).
Input: n = 17 Output: true
Solution in C++
#includeusing namespace std; bool isPrime(int n) { if (n <= 1) return false; if (n == 2) return true; if (n % 2 == 0) return false; for (int i = 3; i <= sqrt(n); i += 2) { if (n % i == 0) return false; } return true; } int main() { int n = 17; cout << (isPrime(n) ? "true" : "false") << endl; return 0; }
Solution in Python
def is_prime(n):
if n <= 1:
return False
if n == 2:
return True
if n % 2 == 0:
return False
for i in range(3, int(n ** 0.5) + 1, 2):
if n % i == 0:
return False
return True
n = 17
print(is_prime(n))
Explanation
Approach: Check divisibility up to the square root of n, optimizing by skipping even numbers after 2. Time Complexity: O(√n), Space Complexity: O(1).
Relevance: Prime number checks are a frequent GenC question testing number theory basics.
Question 4: Merge Two Sorted Arrays
Given two sorted arrays, merge them into a single sorted array.
Input: arr1 = [1, 3, 5], arr2 = [2, 4, 6] Output: [1, 2, 3, 4, 5, 6]
Solution in C++
#includeusing namespace std; vector mergeArrays(vector & arr1, vector & arr2) { int n1 = arr1.size(), n2 = arr2.size(); vector result(n1 + n2); int i = 0, j = 0, k = 0; while (i < n1 && j < n2) { if (arr1[i] <= arr2[j]) result[k++] = arr1[i++]; else result[k++] = arr2[j++]; } while (i < n1) result[k++] = arr1[i++]; while (j < n2) result[k++] = arr2[j++]; return result; } int main() { vector arr1 = {1, 3, 5}, arr2 = {2, 4, 6}; vector merged = mergeArrays(arr1, arr2); for (int x : merged) cout << x << " "; cout << endl; return 0; }
Solution in Python
def merge_arrays(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
arr1, arr2 = [1, 3, 5], [2, 4, 6]
print(merge_arrays(arr1, arr2))
Explanation
Approach: Use two pointers to compare elements from both arrays, adding the smaller one to the result. Append remaining elements. Time Complexity: O(n1 + n2), Space Complexity: O(n1 + n2).
Relevance: Merging sorted arrays tests pointer techniques, a GenC staple.
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