| comments | difficulty | edit_url | rating | source | tags | |||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|
true |
困难 |
1917 |
第 86 场双周赛 Q4 |
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你有 n 个机器人，给你两个下标从 0 开始的整数数组 chargeTimes 和 runningCosts ，两者长度都为 n 。第 i 个机器人充电时间为 chargeTimes[i] 单位时间，花费 runningCosts[i] 单位时间运行。再给你一个整数 budget 。
运行 k 个机器人 总开销 是 max(chargeTimes) + k * sum(runningCosts) ，其中 max(chargeTimes) 是这 k 个机器人中最大充电时间，sum(runningCosts) 是这 k 个机器人的运行时间之和。
请你返回在 不超过 budget 的前提下，你 最多 可以 连续 运行的机器人数目为多少。
示例 1：
输入：chargeTimes = [3,6,1,3,4], runningCosts = [2,1,3,4,5], budget = 25 输出：3 解释： 可以在 budget 以内运行所有单个机器人或者连续运行 2 个机器人。 选择前 3 个机器人，可以得到答案最大值 3 。总开销是 max(3,6,1) + 3 * sum(2,1,3) = 6 + 3 * 6 = 24 ，小于 25 。 可以看出无法在 budget 以内连续运行超过 3 个机器人，所以我们返回 3 。
示例 2：
输入：chargeTimes = [11,12,19], runningCosts = [10,8,7], budget = 19 输出：0 解释：即使运行任何一个单个机器人，还是会超出 budget，所以我们返回 0 。
提示：
chargeTimes.length == runningCosts.length == n1 <= n <= 5 * 1041 <= chargeTimes[i], runningCosts[i] <= 1051 <= budget <= 1015
问题实际上是求滑动窗口内的最大值，可以用单调队列来求解。
我们只需要二分枚举窗口
实现过程中，实际上不需要进行二分枚举，只需要将固定窗口改为双指针非固定窗口即可。
时间复杂度
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 ansclass 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;
}
}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;
}
};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)]
}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;
}
}