2279. Maximum Bags With Full Capacity of Rocks
Explanation:
To maximize the number of bags that could have full capacity, we should prioritize filling the bags with the least amount of rocks currently in them. This way, we can distribute the additional rocks more evenly and achieve the highest number of bags at full capacity.
- Sort the bags based on the difference between capacity and current rocks.
- Iterate through the sorted bags and add additional rocks to each bag until it reaches full capacity.
- Count the bags that have reached full capacity and return the count as the result.
Time Complexity: O(n log n) where n is the number of bags
Space Complexity: O(n) for sorting
:
import java.util.Arrays;
class Solution {
public int maxBagsWithFullCapacity(int[] capacity, int[] rocks, int additionalRocks) {
int n = capacity.length;
int[][] bags = new int[n][2];
for (int i = 0; i < n; i++) {
bags[i][0] = capacity[i] - rocks[i];
bags[i][1] = i;
}
Arrays.sort(bags, (a, b) -> a[0] - b[0]);
int fullCapacityBags = 0;
for (int i = 0; i < n && additionalRocks > 0; i++) {
int idx = bags[i][1];
int remainingCapacity = capacity[idx] - rocks[idx];
int addRocks = Math.min(remainingCapacity, additionalRocks);
rocks[idx] += addRocks;
additionalRocks -= addRocks;
if (rocks[idx] == capacity[idx]) {
fullCapacityBags++;
}
}
return fullCapacityBags;
}
}Code Editor (Testing phase)
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