You are assigned to put some amount of boxes onto one truck. You are given a 2D array boxTypes
, where boxTypes[i] = [numberOfBoxesi, numberOfUnitsPerBoxi]
:
numberOfBoxesi
is the number of boxes of typei
.numberOfUnitsPerBoxi
is the number of units in each box of the typei
.
You are also given an integer truckSize
, which is the maximum number of boxes that can be put on the truck. You can choose any boxes to put on the truck as long as the number of boxes does not exceed truckSize
.
Return the maximum total number of units that can be put on the truck.
Example 1:
Input: boxTypes = [[1,3],[2,2],[3,1]], truckSize = 4 Output: 8 Explanation: There are: - 1 box of the first type that contains 3 units. - 2 boxes of the second type that contain 2 units each. - 3 boxes of the third type that contain 1 unit each. You can take all the boxes of the first and second types, and one box of the third type. The total number of units will be = (1 * 3) + (2 * 2) + (1 * 1) = 8.
Example 2:
Input: boxTypes = [[5,10],[2,5],[4,7],[3,9]], truckSize = 10 Output: 91
Constraints:
1 <= boxTypes.length <= 1000
1 <= numberOfBoxesi, numberOfUnitsPerBoxi <= 1000
1 <= truckSize <= 106
Solution: Greedy
Sort by unit in descending order.
Time complexity: O(nlogn)
Space complexity: O(1)
C++
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 |
// Author: Huahua class Solution { public: int maximumUnits(vector<vector<int>>& boxTypes, int truckSize) { sort(begin(boxTypes), end(boxTypes), [](const auto& a, const auto& b){ return a[1] > b[1]; // Sort by unit DESC }); int ans = 0; for (const auto& b : boxTypes) { ans += b[1] * min(b[0], truckSize); if ((truckSize -= b[0]) <= 0) break; } return ans; } }; |
请尊重作者的劳动成果,转载请注明出处!花花保留对文章/视频的所有权利。
如果您喜欢这篇文章/视频,欢迎您捐赠花花。
If you like my articles / videos, donations are welcome.
Be First to Comment