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
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Channel Coding Interview Resources (@crackingthecodinginterview) in the English language segment is an active participant. Currently, the community unites 52 236 subscribers, ranking 2 494 in the Technologies & Applications category and 6 880 in the India region.
📊 Audience metrics and dynamics
Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 52 236 subscribers.
According to the latest data from 26 August, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by 31 over the last 30 days and by -3 over the last 24 hours, overall reach remains high.
- Verification status: Not verified
- Engagement rate (ER): The average audience engagement rate is 1.85%. Within the first 24 hours after publication, content typically collects 0.76% reactions from the total number of subscribers.
- Post reach: On average, each post receives 966 views. Within the first day, a publication typically gains 398 views.
- Reactions and interaction: The audience actively supports content: the average number of reactions per post is 2.
- Thematic interests: Content is focused on key topics such as array, stack, algorithm, programming, sort.
📝 Description and content policy
The author describes the resource as a platform for expressing subjective opinions:
“This channel contains the free resources and solution of coding problems which are usually asked in the interviews.
Managed by: @love_data”
Thanks to the high frequency of updates (latest data received on 27 August, 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.
NULLs, constraints
🧠 Interview Tip: Be able to explain Primary vs Foreign Key.
2️⃣ Basic Queries
🔹 SELECT, FROM, WHERE, ORDER BY, LIMIT
🧠 Practice: Filter and sort data by multiple columns.
3️⃣ Joins – Very Frequently Asked!
🔹 INNER, LEFT, RIGHT, FULL OUTER JOIN
🧠 Interview Tip: Explain the difference with examples.
🧪 Practice: Write queries using joins across 2–3 tables.
4️⃣ Aggregations & GROUP BY
🔹 COUNT, SUM, AVG, MIN, MAX, HAVING
🧠 Common Question: Total sales per category where total > X.
5️⃣ Window Functions
🔹 ROW_NUMBER(), RANK(), DENSE_RANK(), LAG(), LEAD()
🧠 Interview Favorite: Top N per group, previous row comparison.
6️⃣ Subqueries & CTEs
🔹 Write queries inside WHERE, FROM, and using WITH
🧠 Use Case: Filtering on aggregated data, simplifying logic.
7️⃣ CASE Statements
🔹 Add logic directly in SELECT
🧠 Example: Categorize users based on spend or activity.
8️⃣ Data Cleaning & Transformation
🔹 Handle NULLs, format dates, string manipulation (TRIM, SUBSTRING)
🧠 Real-world Task: Clean user input data.
9️⃣ Query Optimization Basics
🔹 Understand indexing, query plan, performance tips
🧠 Interview Tip: Difference between WHERE and HAVING.
🔟 Real-World Scenarios
🧠 Must Practice:
• Sales funnel
• Retention cohort
• Churn rate
• Revenue by channel
• Daily active users
🧪 Practice Platforms
• LeetCode (Easy–Hard SQL)
• StrataScratch (Real business cases)
• Mode Analytics (SQL + Visualization)
• HackerRank SQL (MCQs + Coding)
💼 Final Tip:
Explain why your query works, not just what it does. Speak your logic clearly.
💬 Tap ❤️ for more!heapq.heappush(heap, (node.val, node))
Repeatedly:
• Pop smallest node
• Add next node from same list
🔹 Complexity
Complexity - Value
Time - O(n log k)
Space - O(k)
Where:
n = total nodes
k = number of lists
🔹 Interview Tip
Very common hard interview problem.
🚀 38. How do you implement LRU / LFU cache?
🔹 LRU Cache
LRU: Least Recently Used
Remove least recently accessed item.
🔹 Efficient Design
Use:
1. HashMap
2. Doubly Linked List
🔹 Python LRU Example
from collections import OrderedDict
class LRUCache:
def __init__(self, capacity):
self.cache = OrderedDict()
self.capacity = capacity
def get(self, key):
if key not in self.cache:
return -1
self.cache.move_to_end(key)
return self.cache[key]
def put(self, key, value):
if key in self.cache:
self.cache.move_to_end(key)
self.cache[key] = value
if len(self.cache) > self.capacity:
self.cache.popitem(last=False)
🔹 Complexity
Operation - Complexity
Get - O(1)
Put - O(1)
🔹 Interview Tip
LRU cache is a FAANG-favorite system design question.
🚀 39. How do you check for balanced parentheses?
Use a stack.
🔹 Idea
• Push opening brackets.
• When closing bracket appears: Check top of stack
🔹 Python Solution
def is_valid(s):
stack = []
mapping = {
')': '(',
'}': '{',
']': '['
}
for char in s:
if char in mapping.values():
stack.append(char)
elif char in mapping:
if not stack or stack.pop() != mapping[char]:
return False
return not stack
print(is_valid("({[]})"))
🔹 Output
True
🔹 Complexity
Complexity - Value
Time - O(n)
Space - O(n)
🔹 Uses
✅ Compilers
✅ Expression parsing
✅ Syntax validation
🚀 40. How do you implement a circular queue?
Circular queue reuses empty spaces efficiently.
🔹 Visualization
Front → [1,2,3,_,_]
After dequeue + enqueue:
[,2,3,4,]
🔹 Python Implementation
class CircularQueue:
def __init__(self, size):
self.queue = [None] * size
self.front = 0
self.rear = 0
self.size = size
self.count = 0
def enqueue(self, value):
if self.count == self.size:
return "Full"
self.queue[self.rear] = value
self.rear = (self.rear + 1) % self.size
self.count += 1
def dequeue(self):
if self.count == 0:
return "Empty"
value = self.queue[self.front]
self.front = (self.front + 1) % self.size
self.count -= 1
return value
🔹 Complexity
Operation - Complexity
Enqueue - O(1)
Dequeue - O(1)
🔹 Real-World Uses
✅ CPU scheduling
✅ Streaming systems
✅ Buffers
✅ Embedded systems
🔥 Double Tap ❤️ For Part-5class MaxStack:
def __init__(self):
self.stack = []
self.max_stack = []
def push(self, value):
self.stack.append(value)
if not self.max_stack or value >= self.max_stack[-1]:
self.max_stack.append(value)
def pop(self):
if self.stack[-1] == self.max_stack[-1]:
self.max_stack.pop()
return self.stack.pop()
def get_max(self):
return self.max_stack[-1]
🔹 Complexity
Operation - Complexity
Push - O(1)
Pop - O(1)
Get Max - O(1)
🔹 Interview Tip
Very common design-based stack question.
🚀 32. How do you implement a queue using two stacks?
Queues are FIFO. Stacks are LIFO.
We can combine two stacks.
🔹 Idea
Stack1 → enqueue
Stack2 → dequeue
🔹 Python Solution
class Queue:
def __init__(self):
self.s1 = []
self.s2 = []
def enqueue(self, value):
self.s1.append(value)
def dequeue(self):
if not self.s2:
while self.s1:
self.s2.append(self.s1.pop())
return self.s2.pop()
🔹 Complexity
Operation - Complexity
Enqueue - O(1)
Dequeue - Amortized O(1)
🔹 Interview Tip
Interviewers love this because it tests understanding of stack behavior.
🚀 33. How do you design a stack that supports getMin() in O(1)?
Very similar to Max Stack.
🔹 Idea
Maintain:
• Main stack
• Min stack
🔹 Python Solution
class MinStack:
def __init__(self):
self.stack = []
self.min_stack = []
def push(self, value):
self.stack.append(value)
if not self.min_stack or value <= self.min_stack[-1]:
self.min_stack.append(value)
def pop(self):
if self.stack[-1] == self.min_stack[-1]:
self.min_stack.pop()
return self.stack.pop()
def get_min(self):
return self.min_stack[-1]
🔹 Complexity
Operation - Complexity
Push - O(1)
Pop - O(1)
Get Min - O(1)
🔹 Interview Tip
This is one of the highest-frequency interview problems.
🚀 34. What is a monotonic stack and when is it useful?
A monotonic stack maintains elements in:
• Increasing order OR
• Decreasing order
🔹 Uses
✅ Next Greater Element
✅ Largest Rectangle in Histogram
✅ Stock Span Problem
✅ Daily Temperatures
🔹 Example
arr = [2, 1, 3]
stack = []
for num in arr:
while stack and stack[-1] > num:
stack.pop()
stack.append(num)
🔹 Complexity
Most monotonic stack problems:
O(n)
because every element is pushed and popped once.
🔹 Interview Tip
Extremely important pattern for medium/hard problems.
🚀 35. How do you implement a priority queue / heap?
A heap is a complete binary tree.
Types:
• Min Heap
• Max Heap
🔹 Python Min Heap
import heapq
heap = []
heapq.heappush(heap, 10)
heapq.heappush(heap, 5)
heapq.heappush(heap, 20)
print(heapq.heappop(heap))
🔹 Output
5
🔹 Complexity
Operation - Complexity
Insert - O(log n)
Delete - O(log n)
Peek - O(1)
🔹 Uses
✅ Task scheduling
✅ Dijkstra’s algorithm
✅ Top K problems
✅ Priority processing
🚀 36. How do you find the top K frequent elements?
🔹 Approach
1. Count frequency using hashmap
2. Use heap
🔹 Python Solution
from collections import Counter
import heapq
def top_k(nums, k):
freq = Counter(nums)
return heapq.nlargest(k, freq.keys(), key=freq.get)
print(top_k([1, 1, 1, 2, 2, 3], 2))
🔹 Output
[1, 2]
🔹 Complexity
Complexity - Value
Time - O(n log k)
Space - O(n)
🔹 Interview Tip
Heap + hashmap combination is frequently tested.
🚀 37. How do you merge K sorted lists?
🔹 Efficient Approach
Use a Min Heap.
Heap stores:
smallest current node
🔹 Python Idea
import heapq