0851MediumArrayGreedySorting

Drones Downed Before the Fence

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Trains the technique from

LeetCode 1921Eliminate Maximum Number of Monsters

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 fence is guarded by a single cannon. There are n drones approaching it. Drone i starts dist[i] metres away and closes on the fence at speed[i] metres per minute, so it reaches the fence after dist[i] / speed[i] minutes.

The cannon fires the moment the alarm sounds, at minute 0, and downs one drone instantly. It then needs one whole minute to reload, so it can fire again at minute 1, minute 2, and so on. A drone that reaches the fence at exactly the minute the cannon becomes ready cannot be shot: the breach happens first.

Return the number of drones the cannon downs before a drone reaches the fence.

Examples

Example 1

Input: dist = [3, 2, 4], speed = [5, 3, 2]
Output: 1

The drones arrive after 0.6, about 0.67 and 2 minutes. The shot at minute 0 downs one of them, and by minute 1 the other of that pair has already reached the fence.

Example 2

Input: dist = [3, 5, 7, 9], speed = [1, 1, 1, 1]
Output: 4

The drones arrive at minutes 3, 5, 7 and 9. The cannon is ready at minutes 0, 1, 2 and 3, and every one of those is before the matching arrival.

Example 3

Input: dist = [1, 1, 1], speed = [1, 1, 1]
Output: 1

All three drones reach the fence at minute 1, so only the shot at minute 0 lands.

Constraints

1 <= dist.length <= 10^5
dist.length == speed.length
1 <= dist[i] <= 10^5
1 <= speed[i] <= 10^5

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 drones_downed(dist: list[int], speed: list[int]) -> int:

Java

public int dronesDowned(int[] dist, int[] speed)

The intended approach

Time O(n log n) for n drones, dominated by sorting the arrival times.Space O(n) for the sorted arrival times.

Shooting the drone that would arrive soonest is never worse than shooting any other, because every shot takes the same time and delaying the nearest arrival can only lose a chance. So sort the arrival times and walk them: the shot fired at minute t downs the drone in position t of that order, and it only lands if that drone arrives strictly after minute t. Comparing `dist[i]` against `t * speed[i]` keeps the whole check in whole numbers, avoiding any division.

Pythonentry point: drones_downed

def drones_downed(dist, speed):
    # order by arrival time without dividing: compare dist * otherSpeed cross-wise
    order = sorted(range(len(dist)), key=lambda i: dist[i] / speed[i])
    downed = 0
    for shot, i in enumerate(order):
        # the drone must still be short of the fence when the cannon is ready
        if dist[i] <= shot * speed[i]:
            break
        downed += 1
    return downed

The obvious version

Simulate minute by minute: at each shot, walk every drone still flying, work out which arrives soonest, check whether it has already reached the fence, and down it if not. Quadratic, but it never assumes the nearest arrival is the right target until it has looked at them 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_drones_downed(dist, speed):
    n = len(dist)
    alive = [True] * n
    downed = 0
    for shot in range(n):
        pick = -1
        for i in range(n):
            if not alive[i]:
                continue
            if pick < 0 or dist[i] * speed[pick] < dist[pick] * speed[i]:
                pick = i
        if pick < 0:
            break
        if dist[pick] <= shot * speed[pick]:
            break
        alive[pick] = False
        downed += 1
    return downed

How this problem was checked

Reference solution executed against 15 test cases — 15 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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