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1580. Put Boxes Into the Warehouse II
Description
You are given two arrays of positive integers, boxes
and warehouse
, representing the heights of some boxes of unit width and the heights of n
rooms in a warehouse respectively. The warehouse's rooms are labeled from 0
to n - 1
from left to right where warehouse[i]
(0-indexed) is the height of the ith
room.
Boxes are put into the warehouse by the following rules:
- Boxes cannot be stacked.
- You can rearrange the insertion order of the boxes.
- Boxes can be pushed into the warehouse from either side (left or right)
- If the height of some room in the warehouse is less than the height of a box, then that box and all other boxes behind it will be stopped before that room.
Return the maximum number of boxes you can put into the warehouse.
Example 1:
Input: boxes = [1,2,2,3,4], warehouse = [3,4,1,2] Output: 4 Explanation: We can store the boxes in the following order: 1- Put the yellow box in room 2 from either the left or right side. 2- Put the orange box in room 3 from the right side. 3- Put the green box in room 1 from the left side. 4- Put the red box in room 0 from the left side. Notice that there are other valid ways to put 4 boxes such as swapping the red and green boxes or the red and orange boxes.
Example 2:
Input: boxes = [3,5,5,2], warehouse = [2,1,3,4,5] Output: 3 Explanation: It is not possible to put the two boxes of height 5 in the warehouse since there's only 1 room of height >= 5. Other valid solutions are to put the green box in room 2 or to put the orange box first in room 2 before putting the green and red boxes.
Constraints:
n == warehouse.length
1 <= boxes.length, warehouse.length <= 105
1 <= boxes[i], warehouse[i] <= 109
Solutions
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class Solution { public int maxBoxesInWarehouse(int[] boxes, int[] warehouse) { int n = warehouse.length; int[] left = new int[n]; int[] right = new int[n]; final int inf = 1 << 30; left[0] = inf; right[n - 1] = inf; for (int i = 1; i < n; ++i) { left[i] = Math.min(left[i - 1], warehouse[i - 1]); } for (int i = n - 2; i >= 0; --i) { right[i] = Math.min(right[i + 1], warehouse[i + 1]); } for (int i = 0; i < n; ++i) { warehouse[i] = Math.min(warehouse[i], Math.max(left[i], right[i])); } Arrays.sort(boxes); Arrays.sort(warehouse); int ans = 0, i = 0; for (int x : boxes) { while (i < n && warehouse[i] < x) { ++i; } if (i == n) { break; } ++ans; ++i; } return ans; } }
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class Solution { public: int maxBoxesInWarehouse(vector<int>& boxes, vector<int>& warehouse) { int n = warehouse.size(); const int inf = 1 << 30; vector<int> left(n, inf); vector<int> right(n, inf); for (int i = 1; i < n; ++i) { left[i] = min(left[i - 1], warehouse[i - 1]); } for (int i = n - 2; ~i; --i) { right[i] = min(right[i + 1], warehouse[i + 1]); } for (int i = 0; i < n; ++i) { warehouse[i] = min(warehouse[i], max(left[i], right[i])); } sort(boxes.begin(), boxes.end()); sort(warehouse.begin(), warehouse.end()); int ans = 0; int i = 0; for (int x : boxes) { while (i < n && warehouse[i] < x) { ++i; } if (i == n) { break; } ++ans; ++i; } return ans; } };
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class Solution: def maxBoxesInWarehouse(self, boxes: List[int], warehouse: List[int]) -> int: n = len(warehouse) left = [0] * n right = [0] * n left[0] = right[-1] = inf for i in range(1, n): left[i] = min(left[i - 1], warehouse[i - 1]) for i in range(n - 2, -1, -1): right[i] = min(right[i + 1], warehouse[i + 1]) for i in range(n): warehouse[i] = min(warehouse[i], max(left[i], right[i])) boxes.sort() warehouse.sort() ans = i = 0 for x in boxes: while i < n and warehouse[i] < x: i += 1 if i == n: break ans, i = ans + 1, i + 1 return ans
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func maxBoxesInWarehouse(boxes []int, warehouse []int) (ans int) { n := len(warehouse) left := make([]int, n) right := make([]int, n) const inf = 1 << 30 left[0] = inf right[n-1] = inf for i := 1; i < n; i++ { left[i] = min(left[i-1], warehouse[i-1]) } for i := n - 2; i >= 0; i-- { right[i] = min(right[i+1], warehouse[i+1]) } for i := 0; i < n; i++ { warehouse[i] = min(warehouse[i], max(left[i], right[i])) } sort.Ints(boxes) sort.Ints(warehouse) i := 0 for _, x := range boxes { for i < n && warehouse[i] < x { i++ } if i == n { break } ans++ i++ } return }