Python Interviews
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Ko'proq ko'rsatish๐ Telegram kanali Python Interviews analitikasi
Python Interviews (@pythoninterviews) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 28 757 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 4 793-o'rinni va Hindiston mintaqasida 15 226-o'rinni egallagan.
๐ Auditoriya koโrsatkichlari va dinamika
ะฝะตะฒัะดะพะผะพ sanasidan buyon loyiha tez oโsib, 28 757 obunachiga ega boโldi.
04 Iyun, 2026 dagi oxirgi maโlumotlarga koโra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni 95 ga, soโnggi 24 soatda esa 2 ga oโzgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya oโrtacha 0.63% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 0.85% ini tashkil etuvchi reaksiyalarni toโplaydi.
- Post qamrovi: Har bir post oโrtacha 181 marta koโriladi; birinchi sutkada odatda 243 ta koโrish yigโiladi.
- Reaksiyalar va oโzaro taโsir: Auditoriya faol: har bir postga oโrtacha 1 ta reaksiya keladi.
- Tematik yoโnalishlar: Kontent |--, link:-, learning, sql, analytic kabi asosiy mavzularga jamlangan.
๐ Tavsif va kontent siyosati
Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida taโriflaydi:
โJoin this channel to learn python for web development, data science, artificial intelligence and machine learning with quizzes, projects and amazing resources for free
For collaborations: @coderfunโ
Yuqori yangilanish chastotasi (oxirgi maโlumot 05 Iyun, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli boโlib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Texnologiyalar & Aralashmalar 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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