Hi,

I am stuck at 2001D - Longest Max Min Subsequence. My logic is similar to other accepted solutions. Also, my solution works for all the small test cases I can find from other people's accepted solutions. Some test cases are big and truncated on CF, so I could not test my code with those cases.

Can someone please point out the bug/ provide a test case for which the code does not work?

Thanks in advance.

My submission: 280104081

import heapq


def solve(nums):
    uq = set()
    cum = [0] * len(nums)
    for i, v in enumerate(nums[::-1]):
        uq.add(v)
        cum[i] = len(uq)
    cum = cum[::-1]

    min_heap, max_heap = [], []
    for i, v in enumerate(nums):
        heapq.heappush(min_heap, (-cum[i], v, i))
        heapq.heappush(max_heap, (-cum[i], -v, i))

    last_idx = -1
    ans, used = [], set()
    for i in range(len(uq)):
        if i%2: # pick small
            while min_heap and (min_heap[0][2] < last_idx or min_heap[0][1] in used):
                heapq.heappop(min_heap)
            now = heapq.heappop(min_heap)
            ans.append(now[1])
            used.add(now[1])
            last_idx = now[2]
            continue

        while max_heap and (max_heap[0][2] < last_idx or -max_heap[0][1] in used):
            heapq.heappop(max_heap)
        now = heapq.heappop(max_heap)
        ans.append(-now[1])
        used.add(-now[1])
        last_idx = now[2]

    return ans

if __name__ == '__main__':
    for _ in range(int(input())):
        n = int(input())
        num = list(map(int, input().split()))

        ans_ = solve(num)
        print(len(ans_))
        print(*ans_)

This is not a programming competition problem. This is a paraphrase of a problem I am trying to solve at my job.

A graph G consists of N nodes where every node has some positive value. Graph G is a connected, dense graph where edges are bi-directional and the number of edges can go up to O(N^2).

To traverse an edge we have to pay a tax equal to the edge's weight. The weight of an edge depends on its two node's values. If the value of two connected nodes is vA and vB, calc_weight(vA, vB) returns the weight of the edge. Please refer to the clarification section to understand why we can not pre-calculate all the weights.

We start with some initial value (let's call it amount). From a predefined source node, our goal is to reach the predefined destination node, with minimum tax paying. Both the source and destination are present in the graph.

Clarification:

At some arbitrary moment, let's assume we are at nodeA and the value of nodeA is valA. And at that moment our amount is amtA. Let's consider nodeB has a bi-directional edge from nodeA. If we want to go nodeA -> nodeB, we first have to calculate the edge's weight.

The catch is when we are at a node, we have to call weightAB = calc_weight(amtA+valA, valB). See how our presence affects the parameters of the function.

we then pay weightAB from amtA and when we reach nodeB our amount becomes amtB = (amtA-weightAB). If we want to go to nodeB -> nodeC, the weight will be calc_weight(amtB+valB, valC).

I can solve this problem using a bitmaskDP. But N can reach up to 2000.

Thanks in advance for any ideas/suggestions. Also if you have seen any problem that closely resembles this, please point me in that direction.