CareerCup

Please clarify the question: which is a longer chain?

"(a,b), (c,d), (e,f)" or "(a,g), (h,z)"

x
  \
   y
  /
z

In order returns XZY yet Y is in the path X->Z

this sln is O(log(n)) not O(log(k)) like the OP wanted.

for(i=l+1;i<=h;i++)
		if(pre[i]>pre[l])
			break;

Hmm this lead to O(n) per level on an unbalanced tree. Which means that this algorithm will take O(n^2). Is there an O(n) solution? Nvm probably not.

there's something wrong with the way you're finding the diameter of the graph... there's no known algorithm that can find the 2 furthest vertices in linear time.

Are there more splits after splitting mod 4? If not this fails for N=10

After the 0 mod 4 and 2 mod 4 split, we have 2,6,10 all in one bucket since they are all 2 mod 4. 6 is the average of 2+10 so the solution is incorrect.

Edit: nvm

"do a binary search to find the largest start time (last conflicting app) which is less than current apps end time."

Your apps are sorted already so why do you need to binarysearch for the largest?

"All the apps between current app and last conflicting app is conflicting"

You have to "touch" the in between apps at some point (perhaps simply to return them) so your algorithm is nlogn+n^2 = O(n^2)

isn't this the majority voting algorithm?

Or what if your target = 7, and array looks like: 1,2,3

Neither side is >=7 but if you use all of the numbers you get 7

If S<K, how can the left side contain your answer? The left side < right side and right side < target

store the nodes you walk in normal BFS in a list. Reverse the list. Print.