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This approach involves computing the exact time each monster will take to reach the city and then eliminating them in the order of their arrival times. The goal is to eliminate as many monsters as possible before any monster reaches the city.
Time Complexity: O(n log n) due to sorting the times.
Space Complexity: O(n) to store the arrival times.
1import java.util.Arrays;
2
3public class MonsterElimination {
4
5 public int eliminateMaximum(int[] dist, int[] speed) {
6 int n = dist.length;
7 int[] time = new int[n];
8 for (int i = 0; i < n; i++) {
9 time[i] = (dist[i] + speed[i] - 1) / speed[i]; // ceiling of (dist[i] / speed[i])
10 }
11 Arrays.sort(time);
12 for (int i = 0; i < n; i++) {
13 if (time[i] <= i) {
14 return i;
15 }
16 }
17 return n;
18 }
19
20 public static void main(String[] args) {
21 MonsterElimination me = new MonsterElimination();
22 int[] dist = {1, 3, 4};
23 int[] speed = {1, 1, 1};
24 System.out.println(me.eliminateMaximum(dist, speed));
25 }
26}The Java version involves computing arrival times and sorting them. Each iteration step checks whether the elapsed time matches or exceeds a monster’s arrival time, at which point it stops.
This approach improves the efficiency of eliminating the maximum number of monsters by using a binary search method. It pre-calculates when a monster will reach the city and uses efficient search techniques to make decisions dynamically during the elimination process.
Time Complexity: O(n log n) due to the sort operation.
Space Complexity: O(n) to store arrival times.
1function eliminateMaximum(dist, speed) {
2
This solution uses a binary search-inspired thought process but leverages JavaScript's sort and conditional checks with the built-in array methods for a straightforward solution.