LeetCode 2398: Maximum Number of Robots Within Budget
LeetCode 2398 Solution Explanation
Explanation:
To solve this problem, we can use a sliding window approach. We will maintain a window of consecutive robots and keep track of the total charge cost and total running cost within the window. We will iterate through the robots and continuously expand the window until the total cost exceeds the budget. At each step, we will update the maximum number of consecutive robots that can be run within the budget.
- Sort the robots based on their charge times.
- Initialize variables to track current window's charge cost, running cost, maximum number of robots, and the current number of robots.
- Iterate through the robots, updating the window's cost and number of robots.
- If the total cost exceeds the budget, shrink the window from the left side and update the current window's cost and number of robots.
- Keep track of the maximum number of robots that can be run within the budget.
- Return the maximum number of robots.
Time Complexity: O(n log n) where n is the number of robots (for sorting) Space Complexity: O(1)
:
LeetCode 2398 Solutions in Java, C++, Python
import java.util.Arrays;
class Solution {
public int maxNumberOfRobots(int[] chargeTimes, int[] runningCosts, int budget) {
int n = chargeTimes.length;
int[][] robots = new int[n][2];
for (int i = 0; i < n; i++) {
robots[i] = new int[]{chargeTimes[i], runningCosts[i]};
}
Arrays.sort(robots, (a, b) -> a[0] - b[0]);
int maxRobots = 0;
long totalCharge = 0, totalRunning = 0;
int left = 0;
for (int right = 0; right < n; right++) {
totalCharge += robots[right][0];
totalRunning += robots[right][1];
while (totalCharge + (long)totalRunning * (right - left) > budget) {
totalCharge -= robots[left][0];
totalRunning -= robots[left][1];
left++;
}
maxRobots = Math.max(maxRobots, right - left + 1);
}
return maxRobots;
}
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