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
Ko'proq ko'rsatish📈 Telegram kanali Coding Projects analitikasi
Coding Projects (@programming_experts) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 67 482 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 1 868-o'rinni va Hindiston mintaqasida 4 791-o'rinni egallagan.
📊 Auditoriya ko‘rsatkichlari va dinamika
невідомо sanasidan buyon loyiha tez o‘sib, 67 482 obunachiga ega bo‘ldi.
31 Avgust, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni 468 ga, so‘nggi 24 soatda esa 28 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya o‘rtacha 2.81% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.12% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 1 895 marta ko‘riladi; birinchi sutkada odatda 755 ta ko‘rish yig‘iladi.
- Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 4 ta reaksiya keladi.
- Tematik yo‘nalishlar: Kontent |--, algorithm, array, framework, javascript kabi asosiy mavzularga jamlangan.
📝 Tavsif va kontent siyosati
Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida ta’riflaydi:
“Channel specialized for advanced concepts and projects to master:
* Python programming
* Web development
* Java programming
* Artificial Intelligence
* Machine Learning
Managed by: @love_data”
Yuqori yangilanish chastotasi (oxirgi ma’lumot 01 Sentabr, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Texnologiyalar & Aralashmalar toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.
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)