Approach 1: Stack


Clearly, we need to focus on how to make each query faster than a linear scan. In a typical case, we get a new element like 7, and there are some previous elements like 11, 3, 9, 5, 6, 4. Let's try to create some relationship between this query and the next query.

If (after getting 7) we get an element like 2, then the answer is 1. So in general, whenever we get a smaller element, the answer is 1.

If we get an element like 8, the answer is 1 plus the previous answer (for 7), as the 8 "stops" on the same value that 7 does (namely, 9).

If we get an element like 10, the answer is 1 plus the previous answer, plus the answer for 9.

Notice throughout this evaluation, we only care about elements that occur in increasing order - we "shortcut" to them. That is, from adding an element like 10, we cut to 7 [with "weight" 4], then to 9 [with weight 2], then cut to 11 [with weight 1].

A stack is the ideal data structure to maintain what we care about efficiently.


Let's maintain a weighted stack of decreasing elements. The size of the weight will be the total number of elements skipped. For example, 11, 3, 9, 5, 6, 4, 7 will be (11, weight=1), (9, weight=2), (7, weight=4).

When we get a new element like 10, this helps us count the previous values faster by popping weighted elements off the stack. The new stack at the end will look like (11, weight=1), (10, weight=7).

Complexity Analysis

  • Time Complexity: , where is the number of calls to In total, there are pushes to the stack, and at most pops.

  • Space Complexity: .

Analysis written by: @awice.