463. Island Perimeter | No BFS or DFS | With DFS | With BFS

[@YashJaiswal-lr7vi]

Great approach🔥. Another approach is we can find number of common sides of cell.
If there are x cells with 1 and y sides are common.
Then answer will be
4*x-2*y
Because single common side is attached to 2 cells.
Below is my python code:
class Solution:
def islandPerimeter(self, grid: List[List[int]]) -> int:
common=0
count=0
for i in range(len(grid)):
for j in range(len(grid[0])):
count+=grid[i][j]
if(i+1

[@SandeepT-p3o]

Hey bro! Start making videos of Codeforces Contests alsoo...

[@rahulsihara8946]

keep up the good work

[@yashwanthsai7808]

my solution : class Solution {
public:
int islandPerimeter(vector& grid) {
int n = grid.size();
int sum= 0 ;
int temp =0;
for(int i=0;i0 && grid[i-1][j]==1)
temp-=1;
if(i

[@mainakchanda7918]

Solve this
verifying an alien dictionary leedcode problem

[@harshal8781]

camera kaha gya aaj?😅