Leetcode #1687: Delivering Boxes from Storage to Ports
In this guide, we solve Leetcode #1687 Delivering Boxes from Storage to Ports 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 have the task of delivering some boxes from storage to their ports using only one ship. However, this ship has a limit on the number of boxes and the total weight that it can carry.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Segment Tree, Queue, Array, Dynamic Programming, Prefix Sum, Monotonic Queue, Heap (Priority Queue)
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: boxes = [[1,1],[2,1],[1,1]], portsCount = 2, maxBoxes = 3, maxWeight = 3
Output: 4
Explanation: The optimal strategy is as follows:
- The ship takes all the boxes in the queue, goes to port 1, then port 2, then port 1 again, then returns to storage. 4 trips.
So the total number of trips is 4.
Note that the first and third boxes cannot be delivered together because the boxes need to be delivered in order (i.e. the second box needs to be delivered at port 2 before the third box).
Python Solution
# 33/39
class Solution:
def boxDelivering(
self, boxes: List[List[int]], portsCount: int, maxBoxes: int, maxWeight: int
) -> int:
n = len(boxes)
ws = list(accumulate((box[1] for box in boxes), initial=0))
c = [int(a != b) for a, b in pairwise(box[0] for box in boxes)]
cs = list(accumulate(c, initial=0))
f = [inf] * (n + 1)
f[0] = 0
for i in range(1, n + 1):
for j in range(max(0, i - maxBoxes), i):
if ws[i] - ws[j] <= maxWeight:
f[i] = min(f[i], f[j] + cs[i - 1] - cs[j] + 2)
return f[n]
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.