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: Greedy
Sort by arrival time, and see how many we can eliminate.
Time complexity: O(nlogn)
Space complexity: O(n)
C++
1 2 3 4 5 6 7 8 9 10 11 12 13 14 |
// Author: Huahua class Solution { public: int eliminateMaximum(vector<int>& dist, vector<int>& speed) { const int n = dist.size(); vector<int> t(n); for (int i = 0; i < n; ++i) t[i] = (dist[i] + speed[i] - 1) / speed[i]; sort(begin(t), end(t)); for (int i = 0; i < n; ++i) if (t[i] <= i) return i; return n; } }; |
请尊重作者的劳动成果,转载请注明出处!花花保留对文章/视频的所有权利。
如果您喜欢这篇文章/视频,欢迎您捐赠花花。
If you like my articles / videos, donations are welcome.
Be First to Comment