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#
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| 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
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| 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;
}
}
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| 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;
}
};
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| 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
}
|
