Artificial Intelligence
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Ko'proq ko'rsatish๐ Telegram kanali Artificial Intelligence analitikasi
Artificial Intelligence (@machinelearning_deeplearning) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 53 094 obunachidan iborat bo'lib, Taสผlim toifasida 3 252-o'rinni va Hindiston mintaqasida 7 063-o'rinni egallagan.
๐ Auditoriya koโrsatkichlari va dinamika
ะฝะตะฒัะดะพะผะพ sanasidan buyon loyiha tez oโsib, 53 094 obunachiga ega boโldi.
06 Iyun, 2026 dagi oxirgi maโlumotlarga koโra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni 1 082 ga, soโnggi 24 soatda esa 17 ga oโzgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya oโrtacha 5.70% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining N/A% ini tashkil etuvchi reaksiyalarni toโplaydi.
- Post qamrovi: Har bir post oโrtacha 3 027 marta koโriladi; birinchi sutkada odatda 0 ta koโrish yigโiladi.
- Reaksiyalar va oโzaro taโsir: Auditoriya faol: har bir postga oโrtacha 11 ta reaksiya keladi.
- Tematik yoโnalishlar: Kontent learning, classification, layer, pattern, chatbot kabi asosiy mavzularga jamlangan.
๐ Tavsif va kontent siyosati
Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida taโriflaydi:
โ๐ฐ Machine Learning & Artificial Intelligence Free Resources
๐ฐ Learn Data Science, Deep Learning, Python with Tensorflow, Keras & many more
For Promotions: @love_dataโ
Yuqori yangilanish chastotasi (oxirgi maโlumot 08 Iyun, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli boโlib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Taสผlim toifasidagi muhim taโsir nuqtasiga aylantirishini koโrsatadi.
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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