1921. Eliminate Maximum Number of Monsters
Description
You are playing a video game where you are defending your city from a group of n
monsters. You are given a 0-indexed integer array dist
of size n
, where dist[i]
is the initial distance in kilometers of the ith
monster from the city.
The monsters walk toward the city at a constant speed. The speed of each monster is given to you in an integer array speed
of size n
, where speed[i]
is the speed of the ith
monster in kilometers per minute.
You have a weapon that, once fully charged, can eliminate a single monster. However, the weapon takes one minute to charge.The weapon is fully charged at the very start.
You lose when any monster reaches your city. If a monster reaches the city at the exact moment the weapon is fully charged, it counts as a loss, and the game ends before you can use your weapon.
Return the maximum number of monsters that you can eliminate before you lose, or n
if you can eliminate all the monsters before they reach the city.
Example 1:
Input: dist = [1,3,4], speed = [1,1,1] Output: 3 Explanation: In the beginning, the distances of the monsters are [1,3,4]. You eliminate the first monster. After a minute, the distances of the monsters are [X,2,3]. You eliminate the second monster. After a minute, the distances of the monsters are [X,X,2]. You eliminate the thrid monster. All 3 monsters can be eliminated.
Example 2:
Input: dist = [1,1,2,3], speed = [1,1,1,1] Output: 1 Explanation: In the beginning, the distances of the monsters are [1,1,2,3]. You eliminate the first monster. After a minute, the distances of the monsters are [X,0,1,2], so you lose. You can only eliminate 1 monster.
Example 3:
Input: dist = [3,2,4], speed = [5,3,2] Output: 1 Explanation: In the beginning, the distances of the monsters are [3,2,4]. You eliminate the first monster. After a minute, the distances of the monsters are [X,0,2], so you lose. You can only eliminate 1 monster.
Constraints:
n == dist.length == speed.length
1 <= n <= 105
1 <= dist[i], speed[i] <= 105
Solution
class Solution:
def eliminateMaximum(self, dist: List[int], speed: List[int]) -> int:
for i, (d,s) in enumerate(zip(dist, speed)):
tt = d // s + int(d % s != 0)
dist[i] = tt
dist.sort()
res = t = 0
for tt in dist:
if tt > t:
res += 1
else:
break
t += 1
return res