Coding Interview Resources
This channel contains the free resources and solution of coding problems which are usually asked in the interviews. Managed by: @love_data
Show more๐ Analytical overview of Telegram channel Coding Interview Resources
Channel Coding Interview Resources (@crackingthecodinginterview) in the English language segment is an active participant. Currently, the community unites 52 120 subscribers, ranking 2 563 in the Technologies & Applications category and 7 263 in the India region.
๐ Audience metrics and dynamics
Since its creation on ะฝะตะฒัะดะพะผะพ, the project has demonstrated rapid growth, gathering an audience of 52 120 subscribers.
According to the latest data from 05 June, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by 194 over the last 30 days and by 11 over the last 24 hours, overall reach remains high.
- Verification status: Not verified
- Engagement rate (ER): The average audience engagement rate is 1.93%. Within the first 24 hours after publication, content typically collects 0.84% reactions from the total number of subscribers.
- Post reach: On average, each post receives 1 005 views. Within the first day, a publication typically gains 437 views.
- Reactions and interaction: The audience actively supports content: the average number of reactions per post is 2.
- Thematic interests: Content is focused on key topics such as array, stack, algorithm, programming, sort.
๐ Description and content policy
The author describes the resource as a platform for expressing subjective opinions:
โThis channel contains the free resources and solution of coding problems which are usually asked in the interviews.
Managed by: @love_dataโ
Thanks to the high frequency of updates (latest data received on 07 June, 2026), the channel maintains relevance and a high level of publication reach. Analytics show that the audience actively interacts with content, making it an important point of influence in the Technologies & Applications category.
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I hope this message finds you well.
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I am confident that my skills and enthusiasm would make me a valuable addition to this role [Job ID / Link]. If possible, I would be incredibly grateful for your referral or any advice you could offer on how to best position myself for this opportunity.
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Best regards,
[Your Full Name]
[Your Email Address]def is_palindrome(s):
return s == s[::-1]
print(is_palindrome("madam")) # True
print(is_palindrome("hello")) # False
2. How to find the factorial of a number using recursion?
def factorial(n):
if n == 0 or n == 1:
return 1
return n * factorial(n - 1)
print(factorial(5)) # 120
3. How to merge two dictionaries in Python?
dict1 = {'a': 1, 'b': 2}
dict2 = {'c': 3, 'd': 4}
# Method 1 (Python 3.5+)
merged_dict = {**dict1, **dict2}
# Method 2 (Python 3.9+)
merged_dict = dict1 | dict2
print(merged_dict)
4. How to find the intersection of two lists?
list1 = [1, 2, 3, 4]
list2 = [3, 4, 5, 6]
intersection = list(set(list1) & set(list2))
print(intersection) # [3, 4]
5. How to generate a list of even numbers from 1 to 100?
even_numbers = [i for i in range(1, 101) if i % 2 == 0]
print(even_numbers)
6. How to find the longest word in a sentence?
def longest_word(sentence):
words = sentence.split()
return max(words, key=len)
print(longest_word("Python is a powerful language")) # "powerful"
7. How to count the frequency of elements in a list?
from collections import Counter
my_list = [1, 2, 2, 3, 3, 3, 4]
frequency = Counter(my_list)
print(frequency) # Counter({3: 3, 2: 2, 1: 1, 4: 1})
8. How to remove duplicates from a list while maintaining the order?
def remove_duplicates(lst):
return list(dict.fromkeys(lst))
my_list = [1, 2, 2, 3, 4, 4, 5]
print(remove_duplicates(my_list)) # [1, 2, 3, 4, 5]
9. How to reverse a linked list in Python?
class Node:
def __init__(self, data):
self.data = data
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
# Create linked list: 1 -> 2 -> 3
head = Node(1)
head.next = Node(2)
head.next.next = Node(3)
# Reverse and print the list
reversed_head = reverse_linked_list(head)
while reversed_head:
print(reversed_head.data, end=" -> ")
reversed_head = reversed_head.next
10. How to implement a simple binary search algorithm?
def binary_search(arr, target):
low, high = 0, len(arr) - 1
while low <= high:
mid = (low + high) // 2
if arr[mid] == target:
return mid
elif arr[mid] < target:
low = mid + 1
else:
high = mid - 1
return -1
print(binary_search([1, 2, 3, 4, 5, 6, 7], 4)) # 3
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