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 52 120 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 2 563-o'rinni va Hindiston mintaqasida 7 263-o'rinni egallagan.
📊 Auditoriya ko‘rsatkichlari va dinamika
невідомо sanasidan buyon loyiha tez o‘sib, 52 120 obunachiga ega bo‘ldi.
05 Iyun, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni 194 ga, so‘nggi 24 soatda esa 11 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya o‘rtacha 1.93% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 0.84% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 1 005 marta ko‘riladi; birinchi sutkada odatda 437 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 07 Iyun, 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.
SELECT MAX(salary)
FROM employee
WHERE salary < (SELECT MAX(salary) FROM employee);
2️⃣ Get the top 3 products by revenue from sales table.
SELECT product_id, SUM(revenue) AS total_revenue
FROM sales
GROUP BY product_id
ORDER BY total_revenue DESC
LIMIT 3;
3️⃣ Use JOIN to combine customer and order data.
SELECT c.customer_name, o.order_id, o.order_date
FROM customers c
JOIN orders o ON c.customer_id = o.customer_id;
(That's an INNER JOIN—use LEFT JOIN to include all customers, even without orders.)
4️⃣ Difference between WHERE and HAVING?
⦁ WHERE filters rows before aggregation (e.g., on individual records).
⦁ HAVING filters rows after aggregation (used with GROUP BY on aggregates).
Example:
SELECT department, COUNT(*)
FROM employee
GROUP BY department
HAVING COUNT(*) > 5;
5️⃣ Explain INDEX and how it improves performance.
An INDEX is a data structure that improves the speed of data retrieval.
It works like a lookup table and reduces the need to scan every row in a table.
Especially useful for large datasets and on columns used in WHERE, JOIN, or ORDER BY—think 10x faster queries, but it slows inserts/updates a bit.
💬 Tap ❤️ for more!def reverse_string(s):
return s[::-1]
C++:
string reverseString(string s) {
reverse(s.begin(), s.end());
return s;
}
Java:
String reverseString(String s) {
return new StringBuilder(s).reverse().toString();
}
2️⃣ Check for Palindrome
Q: Check if a string is a palindrome.
Python:
def is_palindrome(s):
s = s.lower().replace(" ", "")
return s == s[::-1]
C++:
bool isPalindrome(string s) {
transform(s.begin(), s.end(), s.begin(), ::tolower);
s.erase(remove(s.begin(), s.end(), ' '), s.end());
return s == string(s.rbegin(), s.rend());
}
Java:
boolean isPalindrome(String s) {
s = s.toLowerCase().replaceAll(" ", "");
return s.equals(new StringBuilder(s).reverse().toString());
}
3️⃣ Count Vowels in a String
Q: Count number of vowels in a string.
Python:
def count_vowels(s):
return sum(1 for c in s.lower() if c in "aeiou")
C++:
int countVowels(string s) {
int count = 0;
for (char c: s) {
c = tolower(c);
if (string("aeiou").find(c)!= string::npos)
count++;
}
return count;
}
Java:
int countVowels(String s) {
int count = 0;
s = s.toLowerCase();
for (char c : s.toCharArray()) {
if ("aeiou".indexOf(c) != -1)
count++;
}
return count;
}
4️⃣ Find Factorial (Recursion)
Q: Find factorial using recursion.
Python:
def factorial(n):
return 1 if n <= 1 else n * factorial(n - 1)
C++:
int factorial(int n) {
return (n <= 1) ? 1 : n * factorial(n - 1);
}
Java:
int factorial(int n) {
return (n <= 1) ? 1 : n * factorial(n - 1);
}
5️⃣ Find Duplicate Elements in List/Array
Q: Print all duplicates from a list.
Python:
from collections import Counter
def find_duplicates(lst):
return [k for k, v in Counter(lst).items() if v > 1]
C++:
vector<int> findDuplicates(vector<int>& nums) {
unordered_map<int, int> freq;
vector<int> res;
for (int n : nums) freq[n]++;
for (auto& p : freq)
if (p.second > 1) res.push_back(p.first);
return res;
}
Java:
List<Integer> findDuplicates(int[] nums) {
Map<Integer, Integer> map = new HashMap<>();
List<Integer> result = new ArrayList<>();
for (int n : nums) map.put(n, map.getOrDefault(n, 0) + 1);
for (Map.Entry<Integer, Integer> entry : map.entrySet())
if (entry.getValue() > 1) result.add(entry.getKey());
return result;
}
Double Tap ♥️ For Mores = "hello"
# Method 1: Slicing
reversed_s = s[::-1] # "olleh"
# Method 2: Two Pointers (In-place logic)
chars = list(s)
left, right = 0, len(chars) - 1
while left < right:
chars[left], chars[right] = chars[right], chars[left]
left += 1
right -= 1
reversed_s = ''.join(chars)
52. How do you check if a string is a palindrome?def is_palindrome(s):
# Clean string: lowercase and remove spaces
s = s.lower().replace(" ", "")
# Method 1: Slicing
return s == s[::-1]
# Method 2: Two Pointers
left, right = 0, len(s) - 1
while left < right:
if s[left] != s[right]:
return False
left += 1
right -= 1
return True
53. How do you find duplicates in an array?arr = [1, 2, 2, 3]
seen = set()
dups = set()
for num in arr:
if num in seen:
dups.add(num)
seen.add(num)
print(list(dups)) # Output: [2]
54. How do you find the missing number in a range from 1 to n?arr = [1, 2, 4] # Missing 3
n = len(arr) + 1 # Should be 4 elements total
expected_sum = n * (n + 1) // 2
actual_sum = sum(arr)
missing_number = expected_sum - actual_sum # 3
55. How do you merge two sorted arrays?arr1, arr2 = [1, 3], [2, 4]
i, j = 0, 0
result = []
while i < len(arr1) and j < len(arr2):
if arr1[i] < arr2[j]:
result.append(arr1[i])
i += 1
else:
result.append(arr2[j])
j += 1
# Append remaining elements
result.extend(arr1[i:])
result.extend(arr2[j:])
56. How do you find the nth Fibonacci number?def fib(n):
if n <= 1:
return n
a, b = 0, 1
for _ in range(2, n + 1):
a, b = b, a + b
return b
print(fib(6)) # Output: 8
57. How do you compute factorial? (Recursion vs Memoization)# Simple Recursion
def fact(n):
if n <= 1: return 1
return n * fact(n - 1)
# Recursive with Memoization (Optimization)
memo = {}
def fact_memo(n):
if n in memo: return memo[n]
if n <= 1: return 1
memo[n] = n * fact_memo(n - 1)
return memo[n]
print(fact(5)) # Output: 120
58. How do you remove duplicates from a sorted array in-place?arr = [1, 1, 2, 2, 3]
if not arr: return 0
slow = 0
for fast in range(1, len(arr)):
if arr[fast] != arr[slow]:
slow += 1
arr[slow] = arr[fast]
# Resulting array up to 'slow + 1' index
print(arr[:slow + 1]) # Output: [1, 2, 3]
59. How do you solve the Two Sum problem?nums, target = [2, 7, 11, 15], 9
mapping = {}
for i, num in enumerate(nums):
complement = target - num
if complement in mapping:
print([mapping[complement], i]) # Output: [0, 1]
mapping[num] = i
60. Interview tip you must remember
- Code Cleanly: Use meaningful variable names (e.g., current_sum instead of s).
- Test Immediately: Verbally walk through your code with a small test case before the interviewer asks you to.
- Discuss Optimization: Always mention Time and Space Complexity. Say: *"This is O(n) time and O(n) space. We could optimize space by..."*
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