0159MediumArrayDynamic ProgrammingBacktrackingKnapsack Problem0-1 Knapsack

Thruster Burn Plans

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A station keeper has a queue of thruster burns already scheduled for the next pass. Burn i will move the station along its track by burns[i] units of drift, and before the pass starts the crew has to point each burn one way or the other: pointed ahead it adds that many units of drift, pointed astern it takes the same amount away. Every burn in the queue fires, in the order given, and none can be cancelled.

A plan is one choice of direction per burn. Two plans count as different as soon as a single burn points the other way, even when both plans finish on the same drift, and even when the burn measures 0 units and so pushes the station nowhere.

Return how many plans finish the pass with a drift of exactly net_shift units.

Examples

Example 1

Input: burns = [2, 2, 2, 2, 2], net_shift = 6
Output: 5

Four burns ahead and one astern land on 8 - 2 = 6, and any one of the five burns can be the one pointed astern, so there are five plans.

Example 2

Input: burns = [0, 5], net_shift = 5
Output: 2

The 5 must point ahead, while the 0 still has two directions to be recorded in, so both plans count.

Example 3

Input: burns = [3, 5], net_shift = -2
Output: 1

Only pointing the 3 ahead and the 5 astern reaches -2, so a single plan works.

Constraints

1 <= burns.length <= 20
0 <= burns[i] <= 1000
0 <= sum(burns) <= 1000
-1000 <= net_shift <= 1000

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def burn_plan_count(burns: list[int], net_shift: int) -> int:

Java

public int burnPlanCount(int[] burns, int netShift)

The intended approach

Time O(n * S) where S is the number of reachable drift values, at most 2 * sum(burns) + 1Space O(S) for the drift table, bounded by the total of all burns

Once a prefix of the queue has been pointed, the only thing that matters about it is the drift it has piled up so far, so plans that agree on that drift can be pooled. Carry a table from reachable drift to the number of prefixes that reach it, and for each new burn spread every entry into two, once ahead and once astern. A burn of size 0 spreads into the same drift twice, which is exactly why it doubles the plan count. The table for the whole queue answers the question by lookup.

Pythonentry point: burn_plan_count

def burn_plan_count(burns, net_shift):
    # drift so far -> how many prefixes of the queue reach it
    ways = {0: 1}
    for b in burns:
        nxt = {}
        for drift, count in ways.items():
            ahead = drift + b
            astern = drift - b
            nxt[ahead] = nxt.get(ahead, 0) + count
            nxt[astern] = nxt.get(astern, 0) + count
        ways = nxt
    # O(n * S) time overall, O(S) space for the table
    return ways.get(net_shift, 0)

The obvious version

Points every burn both ways with plain recursion and tallies the plans that land on the wanted drift. Explores all 2^n plans with no pooling at all.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_burn_plan_count(burns, net_shift):
    n = len(burns)

    def walk(i, drift):
        if i == n:
            if drift == net_shift:
                return 1
            return 0
        return walk(i + 1, drift + burns[i]) + walk(i + 1, drift - burns[i])

    return walk(0, 0)

How this problem was checked

Reference solution executed against 14 test cases — 14 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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