@anshu mishra:
Hi, kindly clarify on this small doubt of mine.
Your algo is
while (!q.empty())
{
x = q.top();
q.pop();
if(node notvisited already && adjacent to x && value = x)
add to queue
}
For the graph,
1 1 2
1 3 1
2 3 4
first,
queue = 1 (at 0,0)
then you would add the 2 1s at (0,1) and (1,0) to the queue.
So far, i can understand.
now, if you start to pop elements, you ll see that there s no element
with
*equal value as 1* adjacent to the popped elements and the queue
would be
empty.
. How does the algo proceed from here and keep track of count?
.
On May 30, 10:22 am, anshu mishra <[email protected]> wrote:
> @ross no, a particular element has to read only 5 times maximum.
>
> 1 reading (i,j) (if its flag is already false skip)
> 2 read by top element
> 3. read by bottom element
> 4 read by left element
> 5 read by right element
>
> coz atleast one of the this operation its flag will be unset(false), then we
> have to just skip it.
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