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
Ko'proq ko'rsatish📈 Telegram kanali Coding Interview Resources analitikasi
Coding Interview Resources (@crackingthecodinginterview) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 52 248 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 2 474-o'rinni va Hindiston mintaqasida 6 815-o'rinni egallagan.
📊 Auditoriya ko‘rsatkichlari va dinamika
невідомо sanasidan buyon loyiha tez o‘sib, 52 248 obunachiga ega bo‘ldi.
26 Avgust, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni 31 ga, so‘nggi 24 soatda esa -3 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya o‘rtacha 1.85% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 0.76% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 966 marta ko‘riladi; birinchi sutkada odatda 398 ta ko‘rish yig‘iladi.
- Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 2 ta reaksiya keladi.
- Tematik yo‘nalishlar: Kontent array, stack, algorithm, programming, sort kabi asosiy mavzularga jamlangan.
📝 Tavsif va kontent siyosati
Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida ta’riflaydi:
“This channel contains the free resources and solution of coding problems which are usually asked in the interviews.
Managed by: @love_data”
Yuqori yangilanish chastotasi (oxirgi ma’lumot 27 Avgust, 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.
class Node:
def __init__(self, data):
self.data = data
self.next = None
def reverse(head):
prev = None
current = head
while current:
next_node = current.next
current.next = prev
prev = current
current = next_node
return prev
🔹 Complexity
Time → O(n)
Space → O(1)
🔹 Interview Tip
This is one of the most important linked-list questions.
🚀 22. How do you detect a cycle in a linked list?
Use Floyd’s Cycle Detection Algorithm.
Also called: Tortoise and Hare Algorithm
🔹 Idea
• Slow pointer moves 1 step
• Fast pointer moves 2 steps
• If they meet → cycle exists
🔹 Python Solution
def has_cycle(head):
slow = fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
if slow == fast:
return True
return False
🔹 Complexity
Time → O(n)
Space → O(1)
🔹 Interview Tip
Very common interview question.
🚀 23. How do you find the middle node of a linked list?
Use two pointers.
🔹 Approach
• Slow pointer → moves 1 step
• Fast pointer → moves 2 steps
When fast reaches end:
slow = middle
🔹 Python Solution
def middle_node(head):
slow = fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
return slow
🔹 Complexity
Time → O(n)
Space → O(1)
🔹 Interview Tip
Two-pointer technique is heavily used in linked lists.
🚀 24. How do you merge two sorted linked lists?
🔹 Example
1 → 3 → 5
2 → 4 → 6
Merged:
1 → 2 → 3 → 4 → 5 → 6
🔹 Python Solution
def merge_lists(l1, l2):
dummy = Node(0)
current = dummy
while l1 and l2:
if l1.data < l2.data:
current.next = l1
l1 = l1.next
else:
current.next = l2
l2 = l2.next
current = current.next
current.next = l1 or l2
return dummy.next
🔹 Complexity
Time → O(n + m)
Space → O(1)
🔹 Interview Tip
This problem is the base concept behind merge sort on linked lists.
🚀 25. How do you find and remove a duplicate in a list?
🔹 Using HashSet
def remove_duplicates(head):
seen = set()
current = head
prev = None
while current:
if current.data in seen:
prev.next = current.next
else:
seen.add(current.data)
prev = current
current = current.next
return head
🔹 Complexity
Time → O(n)
Space → O(n)
🔹 Without Extra Space
Can also be solved using nested loops: O(n²)
🔹 Interview Tip
Interviewers may ask: Can you solve it without extra memory?
🚀 26. How do you implement a dummy head in linked-list problems?
A dummy node simplifies edge cases.
🔹 Why Useful?
Without dummy node: Handling head insertion/deletion becomes complex
With dummy node: Logic becomes cleaner
🔹 Example
dummy = Node(0)
dummy.next = head
🔹 Use Cases
✅ Remove nodes
✅ Merge lists
✅ Partition lists
✅ Reverse sublists
🔹 Interview Tip
Using dummy nodes often makes solutions cleaner and bug-free.
🚀 27. How do you delete a node given only that node (no head)?
Important constraint: No access to head pointer
🔹 Trick
Copy next node value into current node.
🔹 Python Solution
def delete_node(node):
node.data = node.next.data
node.next = node.next.next
🔹 Limitation
Cannot delete last node because no next node exists.
🔹 Interview Tip
Classic interview trick question.
🚀 28. How do you implement a circular linked list?
In a circular linked list: Last node → points to head instead of NULL.
🔹 Visualization
1 → 2 → 3
↑ ↓
← ← ← ←
🔹 Python Example
class Node:
def init(self, data):
self.data = data
self.next = None