Coding Projects
Channel specialized for advanced concepts and projects to master: * Python programming * Web development * Java programming * Artificial Intelligence * Machine Learning Managed by: @love_data
Show more📈 Analytical overview of Telegram channel Coding Projects
Channel Coding Projects (@programming_experts) in the English language segment is an active participant. Currently, the community unites 67 474 subscribers, ranking 1 884 in the Technologies & Applications category and 4 808 in the India region.
📊 Audience metrics and dynamics
Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 67 474 subscribers.
According to the latest data from 02 September, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by 427 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 2.84%. Within the first 24 hours after publication, content typically collects 1.11% reactions from the total number of subscribers.
- Post reach: On average, each post receives 1 917 views. Within the first day, a publication typically gains 747 views.
- Reactions and interaction: The audience actively supports content: the average number of reactions per post is 5.
- Thematic interests: Content is focused on key topics such as |--, algorithm, array, framework, javascript.
📝 Description and content policy
The author describes the resource as a platform for expressing subjective opinions:
“Channel specialized for advanced concepts and projects to master:
* Python programming
* Web development
* Java programming
* Artificial Intelligence
* Machine Learning
Managed by: @love_data”
Thanks to the high frequency of updates (latest data received on 03 September, 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.
str1 = "listen"
str2 = "silent"
if sorted(str1) == sorted(str2):
print("Anagrams")
else:
print("Not Anagrams")
Time Complexity: O(n log n)
A frequency-count approach can achieve O(n) average time.
1️⃣9️⃣0️⃣ How Do You Find the First Non-Repeating Character?
Answer:
Count the frequency of every character, then scan the string again and return the first character whose frequency is "1".
Example:
Input: "swiss"
Output: "w"
Python:
from collections import Counter
text = "swiss"
count = Counter(text)
for char in text:
if count[char] == 1:
print(char)
break
Time Complexity: O(n)
Space Complexity: O(k), where "k" is the number of distinct characters.
🔥 Double Tap ❤️ For Part-20
-----
2.47 ₽ · /balance_helptext = "hello"
reversed_text = text[::-1]
print(reversed_text)
Time Complexity: O(n)
Space Complexity: O(n)
1️⃣8️⃣2️⃣ How Do You Find the Largest Element in an Array?
Answer:
Traverse the array while keeping track of the largest value found so far.
Example:
Input: [10, 25, 7, 42, 18]
Output: 42
Python:
numbers = [10, 25, 7, 42, 18]
largest = numbers[0]
for num in numbers:
if num > largest:
largest = num
print(largest)
Time Complexity: O(n)
Space Complexity: O(1)
1️⃣8️⃣3️⃣ How Do You Find the Second Largest Element in an Array?
Answer:
Maintain two variables: one for the largest element and another for the second largest. Update them while traversing the array.
Example:
Input: [10, 25, 7, 42, 18]
Output: 25
Python:
numbers = [10, 25, 7, 42, 18]
largest = second = float('-inf')
for num in numbers:
if num > largest:
second = largest
largest = num
elif largest > num > second:
second = num
print(second)
Time Complexity: O(n)
Space Complexity: O(1)
1️⃣8️⃣4️⃣ How Do You Check Whether a String is a Palindrome?
Answer:
A palindrome is a string that reads the same forward and backward.
Examples:
"madam" → Palindrome
"level" → Palindrome
"hello" → Not a palindrome
Python:
text = "madam"
if text == text[::-1]:
print("Palindrome")
else:
print("Not a palindrome")
Time Complexity: O(n)
1️⃣8️⃣5️⃣ How Do You Find Duplicate Elements in an Array?
Answer:
Use a set to keep track of elements that have already appeared. If an element is already present in the set, it is a duplicate.
Example:
Input: [1, 2, 3, 2, 4, 1]
Output: [1, 2]
Python:
numbers = [1, 2, 3, 2, 4, 1]
seen = set()
duplicates = set()
for num in numbers:
if num in seen:
duplicates.add(num)
else:
seen.add(num)
print(duplicates)
Average Time Complexity: O(n)
Space Complexity: O(n)
1️⃣8️⃣6️⃣ How Do You Remove Duplicates from an Array?
Answer:
A common approach is to use a set, which stores only unique values.
Example:
Input: [1, 2, 2, 3, 3, 4]
Output: [1, 2, 3, 4]
Python:
numbers = [1, 2, 2, 3, 3, 4]
unique_numbers = list(set(numbers))
print(unique_numbers)
If the original order must be preserved:
unique_numbers = list(dict.fromkeys(numbers))
Average Time Complexity: O(n)
1️⃣8️⃣7️⃣ How Do You Find the Missing Number in an Array?
Answer:
If an array contains numbers from "1" to "n" with one number missing, calculate the expected sum and subtract the actual sum.
Example:
Input: [1, 2, 4, 5]
Output: 3
Python:
numbers = [1, 2, 4, 5]
n = 5
expected = n * (n + 1) // 2
missing = expected - sum(numbers)
print(missing)
Time Complexity: O(n)
Space Complexity: O(1)
1️⃣8️⃣8️⃣ How Do You Merge Two Sorted Arrays?
Answer:
Use two pointers to compare elements from both arrays and add the smaller element to the result.
Example:
Input:
[1, 3, 5]
[2, 4, 6]
Output:
[1, 2, 3, 4, 5, 6]
Python:
a = [1, 3, 5]
b = [2, 4, 6]
i = j = 0
result = []
while i < len(a) and j < len(b):
if a[i] < b[j]:
result.append(a[i])
i += 1
else:
result.append(b[j])
j += 1
while i < len(a):
result.append(a[i])
i += 1
while j < len(b):
result.append(b[j])
j += 1
print(result)