Leetcode 2398 Maximum Number of Robots Within Budget Solution in Java | Hindi Coding Community

 

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

Java Code :

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();

d.addLast(j);

if (times[d.getFirst()] + (j - i + 1) * sum > budget) {

if (d.getFirst() == i)

d.pollFirst();

sum -= costs[i++];

}

}

return n - i;

}