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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 == n
  • 1 <= n <= 5 * 104
  • 1 <= chargeTimes[i], runningCosts[i] <= 105
  • 1 <= budget <= 1015

Solutions

  • class Solution {
    public int maximumRobots(int[] chargeTimes, int[] runningCosts, long budget) {
        Deque<Integer> q = new ArrayDeque<>();
        int n = chargeTimes.length;
        long s = 0;
        int j = 0;
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            int a = chargeTimes[i], b = runningCosts[i];
            while (!q.isEmpty() && chargeTimes[q.getLast()] <= a) {
                q.pollLast();
            }
            q.offer(i);
            s += b;
            while (!q.isEmpty() && chargeTimes[q.getFirst()] + (i - j + 1) * s > budget) {
                if (q.getFirst() == j) {
                    q.pollFirst();
                }
                s -= runningCosts[j++];
            }
            ans = Math.max(ans, i - j + 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, j = 0, n = chargeTimes.size();
        for (int i = 0; i < n; ++i) {
            int a = chargeTimes[i], b = runningCosts[i];
            while (!q.empty() && chargeTimes[q.back()] <= a) q.pop_back();
            q.push_back(i);
            s += b;
            while (!q.empty() && chargeTimes[q.front()] + (i - j + 1) * s > budget) {
                if (q.front() == j) {
                    q.pop_front();
                }
                s -= runningCosts[j++];
            }
            ans = max(ans, i - j + 1);
        }
        return ans;
    }
};

  • class Solution:
    def maximumRobots(
        self, chargeTimes: List[int], runningCosts: List[int], budget: int
    ) -> int:
        q = deque()
        ans = j = s = 0
        for i, (a, b) in enumerate(zip(chargeTimes, runningCosts)):
            while q and chargeTimes[q[-1]] <= a:
                q.pop()
            q.append(i)
            s += b
            while q and chargeTimes[q[0]] + (i - j + 1) * s > budget:
                if q[0] == j:
                    q.popleft()
                s -= runningCosts[j]
                j += 1
            ans = max(ans, i - j + 1)
        return ans


  • func maximumRobots(chargeTimes []int, runningCosts []int, budget int64) int {
	s := int64(0)
	ans, j := 0, 0
	q := []int{}
	for i, a := range chargeTimes {
		for len(q) > 0 && chargeTimes[q[len(q)-1]] <= a {
			q = q[:len(q)-1]
		}
		q = append(q, i)
		s += int64(runningCosts[i])
		for len(q) > 0 && int64(chargeTimes[q[0]])+int64(i-j+1)*s > budget {
			if q[0] == j {
				q = q[1:]
			}
			s -= int64(runningCosts[j])
			j++
		}
		ans = max(ans, i-j+1)
	}
	return ans
}

  • 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;
    }
}

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