1687. Delivering Boxes from Storage to Ports LeetCode Solution

Last updated: February 24, 2025

In this guide, you will get 1687. Delivering Boxes from Storage to Ports LeetCode Solution with the best time and space complexity. The solution to Delivering Boxes from Storage to Ports problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

  1. Problem Statement
  2. Complexity Analysis
  3. Delivering Boxes from Storage to Ports solution in C++
  4. Delivering Boxes from Storage to Ports solution in Java
  5. Delivering Boxes from Storage to Ports solution in Python
  6. Additional Resources
1687. Delivering Boxes from Storage to Ports LeetCode Solution image

Problem Statement of Delivering Boxes from Storage to Ports

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.
You are given an array boxes, where boxes[i] = [ports​​i​, weighti], and three integers portsCount, maxBoxes, and maxWeight.

ports​​i is the port where you need to deliver the ith box and weightsi is the weight of the ith box.
portsCount is the number of ports.
maxBoxes and maxWeight are the respective box and weight limits of the ship.

The boxes need to be delivered in the order they are given. The ship will follow these steps:

The ship will take some number of boxes from the boxes queue, not violating the maxBoxes and maxWeight constraints.
For each loaded box in order, the ship will make a trip to the port the box needs to be delivered to and deliver it. If the ship is already at the correct port, no trip is needed, and the box can immediately be delivered.
The ship then makes a return trip to storage to take more boxes from the queue.

The ship must end at storage after all the boxes have been delivered.
Return the minimum number of trips the ship needs to make to deliver all boxes to their respective ports.

Example 1:

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).

Example 2:

Input: boxes = [[1,2],[3,3],[3,1],[3,1],[2,4]], portsCount = 3, maxBoxes = 3, maxWeight = 6
Output: 6
Explanation: The optimal strategy is as follows:
– The ship takes the first box, goes to port 1, then returns to storage. 2 trips.
– The ship takes the second, third and fourth boxes, goes to port 3, then returns to storage. 2 trips.
– The ship takes the fifth box, goes to port 2, then returns to storage. 2 trips.
So the total number of trips is 2 + 2 + 2 = 6.

Example 3:

Input: boxes = [[1,4],[1,2],[2,1],[2,1],[3,2],[3,4]], portsCount = 3, maxBoxes = 6, maxWeight = 7
Output: 6
Explanation: The optimal strategy is as follows:
– The ship takes the first and second boxes, goes to port 1, then returns to storage. 2 trips.
– The ship takes the third and fourth boxes, goes to port 2, then returns to storage. 2 trips.
– The ship takes the fifth and sixth boxes, goes to port 3, then returns to storage. 2 trips.
So the total number of trips is 2 + 2 + 2 = 6.

Constraints:

1 <= boxes.length <= 105
1 <= portsCount, maxBoxes, maxWeight <= 105
1 <= ports​​i <= portsCount
1 <= weightsi <= maxWeight

Complexity Analysis

  • Time Complexity: O(n)
  • Space Complexity: O(n)

1687. Delivering Boxes from Storage to Ports LeetCode Solution in C++

class Solution {
 public:
  int boxDelivering(vector<vector<int>>& boxes, int portsCount, int maxBoxes,
                    int maxWeight) {
    const int n = boxes.size();
    // dp[i] := the minimum trips to deliver boxes[0..i) and return to the
    // storage
    vector<int> dp(n + 1);
    int trips = 2;
    int weight = 0;

    for (int l = 0, r = 0; r < n; ++r) {
      weight += boxes[r][1];

      // The current box is different from the previous one, need to make one
      // more trip.
      if (r > 0 && boxes[r][0] != boxes[r - 1][0])
        ++trips;

      while (r - l + 1 > maxBoxes || weight > maxWeight ||
             // Loading boxes[l] in the previous turn is always no bad than
             // loading it in this turn.
             (l < r && dp[l + 1] == dp[l])) {
        weight -= boxes[l][1];
        if (boxes[l][0] != boxes[l + 1][0])
          --trips;
        ++l;
      }

      //   min trips to deliver boxes[0..r]
      // = min trips to deliver boxes[0..l) + trips to deliver boxes[l..r]
      dp[r + 1] = dp[l] + trips;
    }

    return dp[n];
  }
};
/* code provided by PROGIEZ */

1687. Delivering Boxes from Storage to Ports LeetCode Solution in Java

class Solution {
  public int boxDelivering(int[][] boxes, int portsCount, int maxBoxes, int maxWeight) {
    final int n = boxes.length;
    // dp[i] := the minimum trips to deliver boxes[0..i) and return to the
    // storage
    int[] dp = new int[n + 1];
    int trips = 2;
    int weight = 0;

    for (int l = 0, r = 0; r < n; ++r) {
      weight += boxes[r][1];

      // The current box is different from the previous one, need to make one
      // more trip.
      if (r > 0 && boxes[r][0] != boxes[r - 1][0])
        ++trips;

      while (r - l + 1 > maxBoxes || weight > maxWeight ||
             // Loading boxes[l] in the previous turn is always no bad than
             // loading it in this turn.
             (l < r && dp[l + 1] == dp[l])) {
        weight -= boxes[l][1];
        if (boxes[l][0] != boxes[l + 1][0])
          --trips;
        ++l;
      }

      //   the minimum trips to deliver boxes[0..r]
      // = the minimum trips to deliver boxes[0..l) +
      //               trips to deliver boxes[l..r]
      dp[r + 1] = dp[l] + trips;
    }

    return dp[n];
  }
}
// code provided by PROGIEZ

1687. Delivering Boxes from Storage to Ports LeetCode Solution in Python

class Solution:
  def boxDelivering(
      self,
      boxes: list[list[int]],
      portsCount: int,
      maxBoxes: int,
      maxWeight: int,
  ) -> int:
    n = len(boxes)
    # dp[i] := the minimum trips to deliver boxes[0..i) and return to the
    # storage
    dp = [0] * (n + 1)
    trips = 2
    weight = 0

    l = 0
    for r in range(n):
      weight += boxes[r][1]

      # The current box is different from the previous one, need to make one
      # more trip.
      if r > 0 and boxes[r][0] != boxes[r - 1][0]:
        trips += 1

      # Loading boxes[l] in the previous turn is always no bad than loading it
      # in this turn
      while r - l + 1 > maxBoxes or weight > maxWeight or (
              l < r and dp[l + 1] == dp[l]):
        weight -= boxes[l][1]
        if boxes[l][0] != boxes[l + 1][0]:
          trips -= 1
        l += 1

      #   min trips to deliver boxes[0..r]
      # = min trips to deliver boxes[0..l) + trips to deliver boxes[l..r]
      dp[r + 1] = dp[l] + trips

    return dp[n]
 # code by PROGIEZ

Additional Resources

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