Icertis Programming Questions: Advanced Q&A Guide to Top Your Icertis Technical Interview
by Aishwarya Bhatt
· February 5, 2025
Icertis is a global leader in contract intelligence. It leverages AI to transform how businesses manage contracts. With its innovative platform, Icertis empowers organizations to streamline operations, mitigate risks, and drive growth through smarter contract management.
Top 10 Icertis Programming Interview Questions and Answers
1. What is the significance of recursion in solving Icertis programming questions?
def factorial(n):
if n == 0 or n == 1:
return 1
return n * factorial(n - 1)
print(factorial(5)) # Output: 120
- Recursion simplifies problems by breaking them into smaller subproblems.
- It's ideal for tasks like tree traversals and dynamic programming.
- Base cases ensure termination and prevent infinite loops.
2. How do you reverse a string in Python for Icertis coding tests?
def reverse_string(s):
return s[::-1]
print(reverse_string("Icertis")) # Output: "sitreCi"
- Slicing (`[::-1]`) is a concise way to reverse strings in Python.
- This method is efficient and widely used in Icertis programming challenges.
- It demonstrates familiarity with Python's built-in features.
3. Explain how to implement a binary search algorithm for Icertis interview questions.
def binary_search(arr, target):
left, right = 0, len(arr) - 1
while left <= right:
mid = (left + right) // 2
if arr[mid] == target:
return mid
elif arr[mid] < target:
left = mid + 1
else:
right = mid - 1
return -1
print(binary_search([1, 3, 5, 7, 9], 5)) # Output: 2
- Binary search halves the search space at each step, improving efficiency.
- Requires a sorted array as input.
- Time complexity is O(log n), making it suitable for large datasets.
4. Write a function to check if two strings are anagrams for Icertis coding assessments.
def are_anagrams(str1, str2):
return sorted(str1) == sorted(str2)
print(are_anagrams("listen", "silent")) # Output: True
- Anagrams have identical characters but in different orders.
- Sorting both strings ensures they match if they're anagrams.
- This approach has a time complexity of O(n log n).
5. Solve the Fibonacci sequence problem using dynamic programming for Icertis technical interviews.
def fibonacci(n):
dp = [0] * (n + 1)
dp[1] = 1
for i in range(2, n + 1):
dp[i] = dp[i - 1] + dp[i - 2]
return dp[n]
print(fibonacci(6)) # Output: 8
- Dynamic programming avoids redundant calculations by storing intermediate results.
- The `dp` array stores Fibonacci numbers up to `n`.
- Time complexity is O(n), which is optimal for this problem.
6. How would you find the longest substring without repeating characters in Icertis coding challenges?
def longest_unique_substring(s):
char_set = set()
left = 0
max_len = 0
for right in range(len(s)):
while s[right] in char_set:
char_set.remove(s[left])
left += 1
char_set.add(s[right])
max_len = max(max_len, right - left + 1)
return max_len
print(longest_unique_substring("abcabcbb")) # Output: 3
- A sliding window technique efficiently tracks unique substrings.
- `char_set` ensures no duplicate characters are included.
- Time complexity is O(n), making it scalable for large inputs.
7. Implement a linked list reversal for Icertis data structure questions.
class Node:
def __init__(self, value):
self.value = value
self.next = None
def reverse_linked_list(head):
prev = None
current = head
while current:
next_node = current.next
current.next = prev
prev = current
current = next_node
return prev
# Example usage
head = Node(1)
head.next = Node(2)
head.next.next = Node(3)
new_head = reverse_linked_list(head)
while new_head:
print(new_head.value, end=" ") # Output: 3 2 1
new_head = new_head.next
- Linked list reversal modifies pointers to reverse the direction of nodes.
- `prev`, `current`, and `next_node` track node relationships during traversal.
- This solution runs in O(n) time with O(1) space complexity.
8. Design a function to merge two sorted arrays for Icertis algorithmic questions.
def merge_sorted_arrays(arr1, arr2):
merged = []
i, j = 0, 0
while i < len(arr1) and j < len(arr2):
if arr1[i] < arr2[j]:
merged.append(arr1[i])
i += 1
else:
merged.append(arr2[j])
j += 1
merged.extend(arr1[i:])
merged.extend(arr2[j:])
return merged
print(merge_sorted_arrays([1, 3, 5], [2, 4, 6])) # Output: [1, 2, 3, 4, 5, 6]
- Merging sorted arrays involves comparing elements from both arrays sequentially.
- The remaining elements from either array are appended after the main loop.
- This approach maintains sorted order with O(n + m) time complexity.
9. Solve the "Two Sum" problem for Icertis interview preparation.
def two_sum(nums, target):
num_map = {}
for i, num in enumerate(nums):
complement = target - num
if complement in num_map:
return [num_map[complement], i]
num_map[num] = i
print(two_sum([2, 7, 11, 15], 9)) # Output: [0, 1]
- A hash map (`num_map`) stores previously seen numbers and their indices.
- Checking for the complement reduces time complexity to O(n).
- This solution is efficient and commonly tested in Icertis programming questions.
10. How do you validate parentheses in a string for Icertis technical rounds?
def is_valid_parentheses(s):
stack = []
mapping = {')': '(', '}': '{', ']': '['}
for char in s:
if char in mapping.values():
stack.append(char)
elif char in mapping:
if not stack or stack.pop() != mapping[char]:
return False
return not stack
print(is_valid_parentheses("()[]{}")) # Output: True
- A stack is used to track opening parentheses.
- Each closing parenthesis must match the top of the stack.
- If the stack is empty at the end, the string is valid.