Campus Monk by Rachit Rastogi
📈 Telegram kanali Campus Monk by Rachit Rastogi analitikasi
Campus Monk by Rachit Rastogi (@rachityoutube) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 10 731 obunachidan iborat bo'lib, Taʼlim toifasida 18 229-o'rinni va Hindiston mintaqasida 36 224-o'rinni egallagan.
📊 Auditoriya ko‘rsatkichlari va dinamika
невідомо sanasidan buyon loyiha tez o‘sib, 10 731 obunachiga ega bo‘ldi.
26 Avgust, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -135 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 1.89% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 0.81% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 203 marta ko‘riladi; birinchi sutkada odatda 87 ta ko‘rish yig‘iladi.
- Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 0 ta reaksiya keladi.
- Tematik yo‘nalishlar: Kontent tcs, sankalp, engineer, prep, intern kabi asosiy mavzularga jamlangan.
📝 Tavsif va kontent siyosati
Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida ta’riflaydi:
“Your one stop for knowledge in a better perspective !”
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 Taʼlim toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.
'1' (land) and '0' (water), count the number of islands (connected groups of land, horizontally/vertically).
Input: 11000 11000 00100 00011 Output: 3💡 Hint: This is graph traversal on an implicit grid graph - each land cell is a node, adjacent land cells are connected edges. DFS or BFS, "sinking" each island as you find it so you don't count it twice. Solution:
python
def num_islands(grid):
if not grid:
return 0
rows, cols = len(grid), len(grid[0])
count = 0
def sink(r, c):
if r < 0 or r >= rows or c < 0 or c >= cols or grid[r][c] != '1':
return
grid[r][c] = '0' # mark as visited by sinking it
sink(r+1, c)
sink(r-1, c)
sink(r, c+1)
sink(r, c-1)
for r in range(rows):
for c in range(cols):
if grid[r][c] == '1':
count += 1
sink(r, c)
return count
Complexity: O(rows × cols) time - every cell is visited a constant number of times. Space is O(rows × cols) worst case for the recursion stack, if the entire grid is one giant island. Common mistake: Modifying the grid in place without realizing that mutates the input the caller passed in - perfectly fine for most interview settings, but worth mentioning out loud: "I'm mutating the grid directly to track visited cells - if we need to preserve the original input, I'd use a separate visited set instead." This exact pattern - grid + DFS/BFS + "sinking"/marking visited - solves a huge family of "connected regions" problems. Worth having memorized cold. Would you use DFS or BFS here, and does it actually matter for this particular problem? 👇
