You’re given an online stream of numbers. At any point of time if K numbers have already appeared, you need to find out floor(K/3)^{th} largest number.

QUICK EXPLANATION:

Maintain two heaps - one min heap for top floor(K/3) numbers and other max heap for all remaining numbers.

DETAILED EXPLANATION:

We can maintain two different heaps - one min heap for top 1/3^{th} of votes and
one max heap for all other votes. Once we have this, we can simulate actual
voting itself. The reason is after every vote, at maximum 1 vote moves between
the heaps.

To understand this, let’s say that at some point of time we’ve x votes in top heap
and N - x votes in other heap. If a new vote comes push it in one of the two halves
by comparing its value to the lowest value in top heap. Now assume it went in top
heap. Number of votes in top heap might be more than floor( (N+1) / 3) now in which case we’d need to transfer some numbers to the other heap. But difference
is only of 1 vote as number of votes in top heap <= 1 + floor(N/3) and hence only 1 vote needs
to goto bottom heap. That one vote has to be the minimum value of this heap.
By similar argument, had the vote gone to bottom heap, again only its topmost value
need to be transfered to top heap, if at all.

At any query, all we have to do is find out the smallest value from top heap and print it.

Complexity of our solution is O(N log N) as we take O(log N) time per query.

I tried the same solution with TreeSet and myInteger class to take care of the duplicates. Because it gave me TLE, I tested only insertion in one treeset, and it gave me TLE again. Why did this happen? They’re suppose to be equivalent in their complexities for insertion with the priority queue and to be balanced to get the logn… (I used fast io)

i am not happy with codechef. When i tried with cin and cout i got TLE and when i tried with printf() and scanf() i got AC. I think Codechef should know that its a algorithmic contest not a Humpty-Dumpty Language contest…

An easy way to solve the problem is to store the first position of box(n/3), say var and see the shifts of this element. Only two cases are held. One case is if n divisible by 3 and var greater than new value inserted, we shift one behind. Its symmetric case is if n not divisible by 3 and var less than or equal to new value inserted, we shift one ahead.
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