Leetcode #1564: Put Boxes Into the Warehouse I
In this guide, we solve Leetcode #1564 Put Boxes Into the Warehouse I 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 labelled 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 = [4,3,4,1], warehouse = [5,3,3,4,1]
Output: 3
Explanation:
We can first put the box of height 1 in room 4. Then we can put the box of height 3 in either of the 3 rooms 1, 2, or 3. Lastly, we can put one box of height 4 in room 0.
There is no way we can fit all 4 boxes in the warehouse.
Python Solution
class Solution:
def maxBoxesInWarehouse(self, boxes: List[int], warehouse: List[int]) -> int:
n = len(warehouse)
left = [warehouse[0]] * n
for i in range(1, n):
left[i] = min(left[i - 1], warehouse[i])
boxes.sort()
i, j = 0, n - 1
while i < len(boxes):
while j >= 0 and left[j] < boxes[i]:
j -= 1
if j < 0:
break
i, j = i + 1, j - 1
return i
Complexity
The time complexity is O(n log n). The space complexity is O(1) to O(n).
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.