KTHMAX - Editorial

PROBLEM LINK:

Practice

Contest

Author: Rishabh

Tester: Kevin Charles Atienza

Editorialist: Vaibhav Tulsyan

DIFFICULTY:

Medium

PREREQUISITES:

Stacks, Binary Search

PROBLEM:

For an array of N integers, you are asked to answer M queries. In each query, you are asked the first number in the p^{th} sub-array,
after sorting all numbers in each sub-array in descending order and then sorting the sub-arrays in descending order.

QUICK EXPLANATION:

For each i (1 \le i \le N), calculate the number of subarrays for which a_i is the maximum element.
Store indexes j (1 \le j \lt i) and k (i + 1 \le k \le N), such that:

  • a_j > a_i
  • a_k > a_i
  • All integers between i and j are smaller than a_i.
  • All integers between i and k and smaller than a_i.
    Number of subarrays for which a_i is max is equal to (i - j) * (k - i).
    Sort the elements in descending order and maintain a cumulative sum of no. of subarrays uptil the i^{th} index.
    Use binary search to find the value associated with the p^{th} sub-array.

EXPLANATION:

Subtask 1:

  • Iterate over all possible sub-arrays of length 1, 2, 3, ... , N.
  • After fixing the end-points, iterate over the array to find the maximum element.
  • Store the maximum element of each subarray in a list L.
  • Sort L in descending order.
  • The maximum element of the p^{th} sub-array would be the p^{th} element in this list L.

Complexity: O(N^3 + N^2.log_2(N) + M)

This naive solution would time out for Subtasks 2 and 3 and would hence fetch only 20 points.

Subtask 2:

  • Maintain a list L to store the maximum values of all sub-arrays.
  • Iterate over all possible sub-arrays by fixing the end-points of the array, say X and Y.
  • Keep a running maximum element present in all sub-arrays whose left boundary is X.
  • For each sub-array, store the newly computed maximum in the list L.
  • Sort L in descending order.
  • The maximum element of the p^{th} sub-array would be the p^{th} element in this list L.

Complexity: O(N^2 + N^2.log_2(N) + M)

This solution would give a total of 50 points.

Subtask 3:

In order to find the number of subarrays whose max is a_i, we can make a useful observation.
For each i, we can determine a left boundary j and a right boundary k such all integers a_x (j \lt x \lt i) and a_y (i \lt x \lt k)
are less than or equal to a_i.
In other words, let [j+1 .. k-1] represent the longest segment in which no element is greater than a_i.
Then, the no. of sub-arrays in which a_i is maximum is equal to (i - j) * (k - i).

How do we find these boundaries for all i in O(N) or O(N.log(N)) time?

We will use a stack to find the boundaries.
Let’s try to find only the left boundary for each i first.
If the current element being processed, say Z, is less than or equal to the element at the top of the stack, say T,
then the left boundary of Z would be the index of T.
Otherwise, we keep popping elements from the stack until we find a suitable boundary.

Pseudo-code:

stack = [-1]
left_boundary = [null for i in (1…n)]
for i in [1…n]:
while (stack is not empty) and (stack.top <= a*):
stack.pop()
left_boundary* = stack.top
stack.push(i)

Similarly, we can also compute the right boundary for each i.

We have computed the boundaries of all i in O(N) time. This was the slowest step in our previous naive solutions.
Now, we need to store the no. of subarrays for which each a_i is maximum, and sort this list in descending order of a_i.

The next question is: how do we find the maximum element associated with the p^{th} sub-array?

Since N can be 10^5, the number of subarrays can be of the order 10^10. A linear search would result in TLE.

If we maintain a cumulative sum of the no. of subarrays processed in this sorted list, we will get a monotonically increasing sequence.
We can apply binary search on this sequence to find the required answer.

Pseudo-code:

data = [(a[1], count[1]), (a[2], count[2]), … , (a[k], count[k])]
for i in [1…len(data) - 1]:
count[i + 1] += count*
for query in [1…m]:
p = input()
low = 0, high = n
while high - low > 1:
mid = (low + high) >> 1
if count[mid] > p:
low = mid
else:
high = mid
return a[high]

Complexity: O(N + log_2(N)

AUTHOR’S AND TESTER’S SOLUTIONS:

Setter’s solution can be found here.

Tester’s solution can be found here.

Editorialist’s solution can be found here.