Leetcode #1580: Put Boxes Into the Warehouse II
In this guide, we solve Leetcode #1580 Put Boxes Into the Warehouse II in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Greedy, Array, Sorting
Intuition
A locally optimal choice leads to a globally optimal result for this structure.
That means we can commit to decisions as we scan without backtracking.
Approach
Sort or preprocess if needed, then repeatedly take the best available local choice.
Maintain the minimal state necessary to validate the greedy decision.
Steps:
- Sort or preprocess as needed.
- Iterate and pick the best local option.
- Track the current solution.
Example
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.
Python Solution
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
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.