2398. Maximum Number of Robots Within Budget
Description
You have n robots. You are given two 0-indexed integer arrays, chargeTimes and runningCosts, both of length n. The ith robot costs chargeTimes[i] units to charge and costs runningCosts[i] units to run. You are also given an integer budget.
The total cost of running k chosen robots is equal to max(chargeTimes) + k * sum(runningCosts), where max(chargeTimes) is the largest charge cost among the k robots and sum(runningCosts) is the sum of running costs among the k robots.
Return the maximum number of consecutive robots you can run such that the total cost does not exceed budget.
Example 1:
Input: chargeTimes = [3,6,1,3,4], runningCosts = [2,1,3,4,5], budget = 25 Output: 3 Explanation: It is possible to run all individual and consecutive pairs of robots within budget. To obtain answer 3, consider the first 3 robots. The total cost will be max(3,6,1) + 3 * sum(2,1,3) = 6 + 3 * 6 = 24 which is less than 25. It can be shown that it is not possible to run more than 3 consecutive robots within budget, so we return 3.
Example 2:
Input: chargeTimes = [11,12,19], runningCosts = [10,8,7], budget = 19 Output: 0 Explanation: No robot can be run that does not exceed the budget, so we return 0.
Constraints:
chargeTimes.length == runningCosts.length == n1 <= n <= 5 * 1041 <= chargeTimes[i], runningCosts[i] <= 1051 <= budget <= 1015
Solutions
Solution 1: Two Pointers + Monotonic Queue
The problem is essentially finding the maximum value within a sliding window, which can be solved using a monotonic queue.
We only need to use binary search to enumerate the size of the window $k$, and find the largest $k$ that satisfies the problem requirements.
In the implementation process, we don't actually need to perform binary search enumeration. We just need to change the fixed window to a non-fixed window with double pointers.
The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the number of robots in the problem.
Python3
class Solution:
def maximumRobots(
self, chargeTimes: List[int], runningCosts: List[int], budget: int
) -> int:
q = deque()
ans = s = l = 0
for r, (t, c) in enumerate(zip(chargeTimes, runningCosts)):
s += c
while q and chargeTimes[q[-1]] <= t:
q.pop()
q.append(r)
while q and (r - l + 1) * s + chargeTimes[q[0]] > budget:
if q[0] == l:
q.popleft()
s -= runningCosts[l]
l += 1
ans = max(ans, r - l + 1)
return ans
Java
class Solution {
public int maximumRobots(int[] chargeTimes, int[] runningCosts, long budget) {
Deque<Integer> q = new ArrayDeque<>();
int n = chargeTimes.length;
int ans = 0;
long s = 0;
for (int l = 0, r = 0; r < n; ++r) {
s += runningCosts[r];
while (!q.isEmpty() && chargeTimes[q.peekLast()] <= chargeTimes[r]) {
q.pollLast();
}
q.offerLast(r);
while (!q.isEmpty() && (r - l + 1) * s + chargeTimes[q.peekFirst()] > budget) {
if (q.peekFirst() == l) {
q.pollFirst();
}
s -= runningCosts[l++];
}
ans = Math.max(ans, r - l + 1);
}
return ans;
}
}
C++
class Solution {
public:
int maximumRobots(vector<int>& chargeTimes, vector<int>& runningCosts, long long budget) {
deque<int> q;
long long s = 0;
int ans = 0;
int n = chargeTimes.size();
for (int l = 0, r = 0; r < n; ++r) {
s += runningCosts[r];
while (q.size() && chargeTimes[q.back()] <= chargeTimes[r]) {
q.pop_back();
}
q.push_back(r);
while (q.size() && (r - l + 1) * s + chargeTimes[q.front()] > budget) {
if (q.front() == l) {
q.pop_front();
}
s -= runningCosts[l++];
}
ans = max(ans, r - l + 1);
}
return ans;
}
};
Go
func maximumRobots(chargeTimes []int, runningCosts []int, budget int64) (ans int) {
q := Deque{}
s := int64(0)
l := 0
for r, t := range chargeTimes {
s += int64(runningCosts[r])
for !q.Empty() && chargeTimes[q.Back()] <= t {
q.PopBack()
}
q.PushBack(r)
for !q.Empty() && int64(r-l+1)*s+int64(chargeTimes[q.Front()]) > budget {
if q.Front() == l {
q.PopFront()
}
s -= int64(runningCosts[l])
l++
}
ans = max(ans, r-l+1)
}
return
}
// template
type Deque struct{ l, r []int }
func (q Deque) Empty() bool {
return len(q.l) == 0 && len(q.r) == 0
}
func (q Deque) Size() int {
return len(q.l) + len(q.r)
}
func (q *Deque) PushFront(v int) {
q.l = append(q.l, v)
}
func (q *Deque) PushBack(v int) {
q.r = append(q.r, v)
}
func (q *Deque) PopFront() (v int) {
if len(q.l) > 0 {
q.l, v = q.l[:len(q.l)-1], q.l[len(q.l)-1]
} else {
v, q.r = q.r[0], q.r[1:]
}
return
}
func (q *Deque) PopBack() (v int) {
if len(q.r) > 0 {
q.r, v = q.r[:len(q.r)-1], q.r[len(q.r)-1]
} else {
v, q.l = q.l[0], q.l[1:]
}
return
}
func (q Deque) Front() int {
if len(q.l) > 0 {
return q.l[len(q.l)-1]
}
return q.r[0]
}
func (q Deque) Back() int {
if len(q.r) > 0 {
return q.r[len(q.r)-1]
}
return q.l[0]
}
func (q Deque) Get(i int) int {
if i < len(q.l) {
return q.l[len(q.l)-1-i]
}
return q.r[i-len(q.l)]
}
TypeScript
function maximumRobots(chargeTimes: number[], runningCosts: number[], budget: number): number {
const q = new Deque<number>();
const n = chargeTimes.length;
let [ans, s] = [0, 0];
for (let l = 0, r = 0; r < n; ++r) {
s += runningCosts[r];
while (!q.isEmpty() && chargeTimes[q.backValue()!] <= chargeTimes[r]) {
q.popBack();
}
q.pushBack(r);
while (!q.isEmpty() && (r - l + 1) * s + chargeTimes[q.frontValue()!] > budget) {
if (q.frontValue() === l) {
q.popFront();
}
s -= runningCosts[l++];
}
ans = Math.max(ans, r - l + 1);
}
return ans;
}
class Node<T> {
value: T;
next: Node<T> | null;
prev: Node<T> | null;
constructor(value: T) {
this.value = value;
this.next = null;
this.prev = null;
}
}
class Deque<T> {
private front: Node<T> | null;
private back: Node<T> | null;
private size: number;
constructor() {
this.front = null;
this.back = null;
this.size = 0;
}
pushFront(val: T): void {
const newNode = new Node(val);
if (this.isEmpty()) {
this.front = newNode;
this.back = newNode;
} else {
newNode.next = this.front;
this.front!.prev = newNode;
this.front = newNode;
}
this.size++;
}
pushBack(val: T): void {
const newNode = new Node(val);
if (this.isEmpty()) {
this.front = newNode;
this.back = newNode;
} else {
newNode.prev = this.back;
this.back!.next = newNode;
this.back = newNode;
}
this.size++;
}
popFront(): T | undefined {
if (this.isEmpty()) {
return undefined;
}
const value = this.front!.value;
this.front = this.front!.next;
if (this.front !== null) {
this.front.prev = null;
} else {
this.back = null;
}
this.size--;
return value;
}
popBack(): T | undefined {
if (this.isEmpty()) {
return undefined;
}
const value = this.back!.value;
this.back = this.back!.prev;
if (this.back !== null) {
this.back.next = null;
} else {
this.front = null;
}
this.size--;
return value;
}
frontValue(): T | undefined {
return this.front?.value;
}
backValue(): T | undefined {
return this.back?.value;
}
getSize(): number {
return this.size;
}
isEmpty(): boolean {
return this.size === 0;
}
}