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# Maximum Number of Robots Within Budget LeetCode Solution

## Problem – Maximum Number of Robots Within Budget LeetCode Solution

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`

## Maximum Number of Robots Within Budget LeetCode Solution in C++

``````    int maximumRobots(vector<int>& times, vector<int>& costs, long long budget) {
long long i = 0, j, sum = 0, n = times.size();
multiset<int> s;
for (int j = 0; j < n; ++j) {
sum += costs[j];
s.insert(times[j]);
if (*rbegin(s) + sum * (j - i + 1) > budget) {
sum -= costs[i];
s.erase(s.find(times[i++]));
}
}
return n - i;
}
``````

## Maximum Number of Robots Within Budget LeetCode Solution in Python

``````    def maximumRobots(self, times: List[int], costs: List[int], budget: int) -> int:
cur = i = 0
n = len(times)
s = SortedList()
for j in range(n):
cur += costs[j]
if s[-1] + (j - i + 1) * cur > budget:
s.remove(times[i])
cur -= costs[i]
i += 1
return n - i
``````

## Maximum Number of Robots Within Budget LeetCode Solution in Java

``````    public int maximumRobots(int[] times, int[] costs, long budget) {
long sum = 0;
int i = 0, n = times.length;
Deque<Integer> d = new LinkedList<Integer>();
for (int j = 0; j < n; ++j) {
sum += costs[j];
while (!d.isEmpty() && times[d.peekLast()] <= times[j])
d.pollLast();
if (times[d.getFirst()] + (j - i + 1) * sum > budget) {
if (d.getFirst() == i)
d.pollFirst();
sum -= costs[i++];
}
}
return n - i;
}
``````
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