Eliminate Maximum Number of Monsters
Problem
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.
Examples
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.
Solution
Method 1 - Sorting and Greedy
We first compute the time each monster takes to reach the city and store these as their arrival times, then sort this array.
At each minute, we eliminate the monster with the earliest arrival; if a monster arrives before it can be eliminated, we stop. For example, with arrival = [1,3,4], the monster at index 0 must be eliminated first since it arrives soonest.
Code
C++
class Solution {
public:
int eliminateMaximum(vector<int>& dist, vector<int>& speed) {
int n = dist.size();
vector<int> arrival(n);
for (int i = 0; i < n; ++i) {
arrival[i] = (dist[i] + speed[i] - 1) / speed[i]; // ceil division
}
sort(arrival.begin(), arrival.end());
int eliminated = 0;
for (int i = 0; i < n; ++i) {
if (arrival[i] > i) {
++eliminated;
} else {
break;
}
}
return eliminated;
}
};
Go
func eliminateMaximum(dist []int, speed []int) int {
n := len(dist)
arrival := make([]int, n)
for i := 0; i < n; i++ {
arrival[i] = (dist[i] + speed[i] - 1) / speed[i] // ceil division
}
sort.Ints(arrival)
eliminated := 0
for i := 0; i < n; i++ {
if arrival[i] > i {
eliminated++
} else {
break
}
}
return eliminated
}
Java
class Solution {
public int eliminateMaximum(int[] dist, int[] speed) {
int n = dist.length;
int[] arrival = new int[n];
for(int i = 0; i < n; i++){
arrival[i] = (int)Math.ceil(dist[i] * 1.0 / speed[i]);
}
Arrays.sort(arrival);
int eliminated = 0;
// At i = 0, minute = 0 ( therefore, we can use i in place of minute )
for(int i = 0; i < n; i++){
// if current arrival time is more than current minute,
// eliminate the monstor
if(arrival[i] > i){
eliminated++;
}else{
break; // Monster reached the city
}
}
return eliminated;
}
}
Kotlin
class Solution {
fun eliminateMaximum(dist: IntArray, speed: IntArray): Int {
val n = dist.size
val arrival = IntArray(n) { i -> (dist[i] + speed[i] - 1) / speed[i] }
arrival.sort()
var eliminated = 0
for (i in 0 until n) {
if (arrival[i] > i) {
eliminated++
} else {
break
}
}
return eliminated
}
}
Python
class Solution:
def eliminateMaximum(self, dist: List[int], speed: List[int]) -> int:
n = len(dist)
arrival = [(dist[i] + speed[i] - 1) // speed[i] for i in range(n)] # ceil division
arrival.sort()
eliminated = 0
for i in range(n):
if arrival[i] > i:
eliminated += 1
else:
break
return eliminated
Rust
impl Solution {
pub fn eliminate_maximum(dist: Vec<i32>, speed: Vec<i32>) -> i32 {
let n = dist.len();
let mut arrival: Vec<i32> = dist.iter()
.zip(speed.iter())
.map(|(&d, &s)| (d + s - 1) / s)
.collect();
arrival.sort();
for i in 0..n {
if arrival[i] as usize <= i {
return i as i32;
}
}
n as i32
}
}
Complexity
- ⏰ Time complexity:
O(n log n) - 🧺 Space complexity:
O(1)
Method 2 - Using minHeap
Compute when each monster will arrive at the city, then use a min-heap to always eliminate the monster with the earliest arrival each minute, stopping if a monster reaches the city before it can be eliminated.
Code
C++
class Solution {
public:
int eliminateMaximum(vector<int>& dist, vector<int>& speed) {
priority_queue<double, vector<double>, greater<double>> minHeap;
for (int i = 0; i < dist.size(); ++i) {
minHeap.push((double)dist[i] / speed[i]);
}
double min = 0.0;
int count = 0;
while (!minHeap.empty() && minHeap.top() > min) {
min += 1.0;
minHeap.pop();
count++;
}
return count;
}
};
Go
type Float64Heap []float64
func (h Float64Heap) Len() int { return len(h) }
func (h Float64Heap) Less(i, j int) bool { return h[i] < h[j] }
func (h Float64Heap) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
func (h *Float64Heap) Push(x interface{}) { *h = append(*h, x.(float64)) }
func (h *Float64Heap) Pop() interface{} {
old := *h
n := len(old)
x := old[n-1]
*h = old[:n-1]
return x
}
func eliminateMaximum(dist []int, speed []int) int {
h := &Float64Heap{}
heap.Init(h)
for i := 0; i < len(dist); i++ {
heap.Push(h, float64(dist[i])/float64(speed[i]))
}
min := 0.0
count := 0
for h.Len() > 0 && (*h)[0] > min {
heap.Pop(h)
min += 1.0
count++
}
return count
}
Java
class Solution {
public int eliminateMaximum(int[] dist, int[] speed) {
PriorityQueue<Double> minHeap = new PriorityQueue<>();
for (int i = 0; i < dist.length; i++) {
minHeap.add(dist[i] * 1.0 / speed[i]);
}
double min = 0.0;
int count = 0;
while (!minHeap.isEmpty() && minHeap.poll() > min) {
min += 1.0;
count++;
}
return count;
}
}
Kotlin
class Solution {
fun eliminateMaximum(dist: IntArray, speed: IntArray): Int {
val minHeap = PriorityQueue<Double>()
for (i in dist.indices) {
minHeap.add(dist[i].toDouble() / speed[i])
}
var min = 0.0
var count = 0
while (minHeap.isNotEmpty() && minHeap.peek() > min) {
min += 1.0
minHeap.poll()
count++
}
return count
}
}
Python
class Solution:
def eliminateMaximum(self, dist: List[int], speed: List[int]) -> int:
min_heap = [d / s for d, s in zip(dist, speed)]
heapq.heapify(min_heap)
min_time = 0.0
count = 0
while min_heap and min_heap[0] > min_time:
heapq.heappop(min_heap)
min_time += 1.0
count += 1
return count
Rust
impl Solution {
pub fn eliminate_maximum(dist: Vec<i32>, speed: Vec<i32>) -> i32 {
let mut heap: Vec<f64> = dist.iter().zip(speed.iter())
.map(|(&d, &s)| d as f64 / s as f64)
.collect();
heap.sort_by(|a, b| a.partial_cmp(b).unwrap());
let mut min = 0.0;
let mut count = 0;
for &time in &heap {
if time > min {
min += 1.0;
count += 1;
} else {
break;
}
}
count
}
}
Complexity
- ⏰ Time complexity:
O(n log n) - 🧺 Space complexity:
O(n)