LeetCode in Kotlin
1921. Eliminate Maximum Number of Monsters
Medium
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.length1 <= n <= 1051 <= dist[i], speed[i] <= 105
Solution
class Solution {
fun eliminateMaximum(dist: IntArray, speed: IntArray): Int {
for (i in dist.indices) {
dist[i] = (dist[i] - 1) / speed[i] + 1
}
dist.sort()
var ans = 1
var time = 1
for (i in 1 until dist.size) {
if (dist[i] > time) {
ans++
time++
} else {
return ans
}
}
return ans
}
}