Campus Monk by Rachit Rastogi
📈 Analytical overview of Telegram channel Campus Monk by Rachit Rastogi
Channel Campus Monk by Rachit Rastogi (@rachityoutube) in the English language segment is an active participant. Currently, the community unites 10 731 subscribers, ranking 18 229 in the Education category and 36 224 in the India region.
📊 Audience metrics and dynamics
Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 10 731 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 -135 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 1.89%. Within the first 24 hours after publication, content typically collects 0.81% reactions from the total number of subscribers.
- Post reach: On average, each post receives 203 views. Within the first day, a publication typically gains 87 views.
- Reactions and interaction: The audience actively supports content: the average number of reactions per post is 0.
- Thematic interests: Content is focused on key topics such as tcs, sankalp, engineer, prep, intern.
📝 Description and content policy
The author describes the resource as a platform for expressing subjective opinions:
“Your one stop for knowledge in a better perspective !”
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 Education category.
'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? 👇
