LeetCode in Kotlin

2398. Maximum Number of Robots Within Budget

Hard

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

Solution

class Solution {
    // use sliding window to track the largest in a way that the sliding window only grows.
    //   then the maximum size is the size of the sliding window at the end.
    // if condition is met, we just grow the sliding window.
    // if condition is not met, we shift the sliding window with the same size to the next position.
    // e.g., if [0,3] is valid, next time we will try [0,4].
    //       if [0,3] is invalid, next time we will try [1,4],
    //         by adjusting the window to [1,3] first in the current round.
    fun maximumRobots(chargeTimes: IntArray, runningCosts: IntArray, budget: Long): Int {
        val n = chargeTimes.size
        // [front, end).
        val deque = IntArray(n)
        var front = 0
        var end = 0
        var sum: Long = 0
        var left = 0
        var right = 0
        while (right < n) {
            // add right into the sliding window, so the window becomes [left, right].
            // update sliding window max and window sum.
            while (end - front > 0 && chargeTimes[deque[end - 1]] <= chargeTimes[right]) {
                --end
            }
            deque[end++] = right
            sum += runningCosts[right].toLong()
            // if the condition is met in the window, do nothing,
            // so the next window size will become one larger.
            // if the condition is not met in the window, shrink one from the front,
            // so the next window size will stay the same.
            if (chargeTimes[deque[front]] + (right - left + 1) * sum > budget) {
                while (end - front > 0 && deque[front] <= left) {
                    ++front
                }
                sum -= runningCosts[left].toLong()
                ++left
            }
            ++right
        }
        return right - left
    }
}

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