The code can be reduced one step further in the inner loop.


for (int i = 2; i <= k; i++) {
for (int j = 2; j <= n; j++) {
int best = M[j][i-1];
for (int p = 1; p <= j-1; p++) {
[…]

Oh, okay. I got it. Thanks. I face the above problem when I tried to create few test cases. But it worked fine for the partial graph. To add further, we can extend the Node structure with second level pointer that […]

@1337c0d3r,

I am not clear about graph representation. If the graph is having more than one SCC, how can it be represented using Node structure? For example, a directed graph with four nodes, A, B, C and D. […]

The precision issue in the second approach can be avoided using equation (don’t use any double expect for finding upper limit of loop),

M + 2*X*Y = (X + Y)^2

Any values of X and Y that satisfy above can be […]

Some how the previous code was not fully printed, here is the code


#include
using namespace std;

struct Node
{
Node(int data) : key(data), lChild(NULL), rChild(NULL)
{}

int key;
[…]

By “boundary” I would expect those nodes that can be seen if any one stands on left side, down side and right side of the tree. In technical terms boundary includes left view, bottom view and right view of a […]

The question is about just the level order traversal, why do we need two queues?

Your explanation is impressive. Thanks for your effort.

Dear 1337c0d3r,
Here is the code without flag. Comments please…

1. Go to left extreme, push nodes on the way
2. Start pop-off and process each node till stack exhausts
3. If pop-ed off node has a right […]