Maximum Number of Robots Within Budget

Maximum Number of Robots Within Budget

You have $n$ robots. You are given two 0-indexed integer arrays, chargeTimes and runningCosts , both of length $n$. The i^{th} 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:

$chargeTimes.length == runningCosts.length == n$
$1 \leq n \leq 5 \times {10}^{4}$
$1 \leq chargeTimes[i], runningCosts[i] \leq {10}^{5}$
$1 \leq budget \leq {10}^{15}$

 

解题思路

　　可以发现答案具有二段性，因此可以二分答案。假设最优解为长度为$ans$的区间，把这个区间右端点的元素去掉，可以发现区间最大值不会变大，而区间和变小（每个元素都是正整数），因此总代价减小。即如果最优解为$ans$，那么长度小于$ans$的区间也一定满足条件。因此具有二段性。

　　然后在二分确定了区间长度后，每个区间的大小都是固定的，check函数就需要遍历所有长度相同的区间，并求每个区间的最大值和区间和，区间最值可以用单调队列来维护，区间和可以用一个变量来维护。

　　AC代码如下，时间复杂度为$O(nlogn)$：

class Solution {
public:
    int n;
    long long m;

    bool check(int len, vector<int>& chargeTimes, vector<int>& runningCosts) {
        deque<int> q;
        long long s = 0;
        for (int i = 0; i < n; i++) {
            // 维护区间最值
            if (!q.empty() && i - q.front() + 1 > len) q.pop_front();
            while (!q.empty() && chargeTimes[q.back()] <= chargeTimes[i]) {
                q.pop_back();
            }
            q.push_back(i);

            // 维护区间和
            s += runningCosts[i];
            if (i - len >= 0) s -= runningCosts[i - len];
            
            if (i - len + 1 >= 0 && chargeTimes[q.front()] + len * s <= m) return true;
        }

        return false;
    }

    int maximumRobots(vector<int>& chargeTimes, vector<int>& runningCosts, long long budget) {
        n = chargeTimes.size();
        m = budget;
        
        int l = 0, r = n;
        while (l < r) {
            int mid = l + r + 1 >> 1;
            if (check(mid, chargeTimes, runningCosts)) l = mid;
            else r = mid - 1;
        }

        return l;
    }
};

　　还可以用双指针。定义指针$j$为当指针$i$固定后，满足条件的最左边的位置。当$i$往右边移动，$j$也应该往右边移动。假设$i$往右移动为$i'$，对应的最左边的位置为$j'$，且$j'$在$j$的左边，由于$j' \sim i'$这个区间满足要求，而$j' \sim i$也满足要求（前面二段性的证明），因此就与$j$为指针$i$最左边的位置矛盾了。

　　AC代码如下，时间复杂度为$O(n)$：

class Solution {
public:
    int maximumRobots(vector<int>& chargeTimes, vector<int>& runningCosts, long long budget) {
        int n = chargeTimes.size();
        long long m = budget;
        
        deque<int> q;
        long long s = 0;
        int ret = 0;
        for (int i = 0, j = 0; i < n; i++) {
            s += runningCosts[i];
            while (!q.empty() && chargeTimes[q.back()] <= chargeTimes[i]) {
                q.pop_back();
            }
            q.push_back(i);

            while (!q.empty() && chargeTimes[q.front()] + (i - j + 1) * s > m) {
                s -= runningCosts[j];
                if (q.front() == j) q.pop_front();
                j++;
            }

            ret = max(ret, i - j + 1);
        }

        return ret;
    }
};

　　最后扩展一下，题目要求的子序列是连续的，如果子序列可以不连续那应该怎么做呢？

　　也是二分答案。先对所有机器人按照取最大值的那项 chargeTimes 按照从小到大的顺序排序，根据二分的值选出对应个数的机器人，假设是$len$个，枚举最大值选哪个，假设当前枚举到第$i$个，然后剩下的$len-1$个机器人一定是从排序后的第$i$个位置之前的机器人中选（如果从第$i$个位置后面选的话，那么最大值就不是第$i$个机器人了）。为了满足条件，应该从前面的机器人中选择$len-1$个最小的 runningCosts （总和应该尽可能小），这个可以用堆来维护。每次往堆插入一个元素，如果发现堆中的元素个数大于$k$，那么就把堆中最大的元素弹出。

　　代码如下，时间复杂度为$O(n \cdot {log}^{2} n)$：

class Solution {
public:
    int n;
    long long m;
    vector<int> idx;

    bool check(int len, vector<int>& chargeTimes, vector<int>& runningCosts) {
        priority_queue<int> pq;
        long long s = 0;
        for (int i = 0; i < n; i++) {
            if (pq.size() == len) {
                s -= pq.top();
                pq.pop();
            }
            pq.push(runningCosts[idx[i]]);
            s += runningCosts[idx[i]];
            if (pq.size() == len && chargeTimes[idx[i]] + len * s <= m) return true;
        }

        return false;
    }

    int maximumRobots(vector<int>& chargeTimes, vector<int>& runningCosts, long long budget) {
        n = chargeTimes.size();
        m = budget;

        for (int i = 0; i < n; i++) {
            idx.push_back(i);
        }
        sort(idx.begin(), idx.end(), [&](int a, int b) {
            return chargeTimes[a] < chargeTimes[b];
        });

        int l = 0, r = n;
        while (l < r) {
            int mid = l + r + 1 >> 1;
            if (check(mid, chargeTimes, runningCosts)) l = mid;
            else r = mid - 1;
        }

        return l;
    }
};

 

参考资料

　　力扣第86场双周赛：https://www.bilibili.com/video/BV11G41157DN