Problem: 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.


Python solution:
import math
import itertools
import bisect
import collections
import string
import heapq
import functools
import sortedcontainers
from typing import List, Dict, Tuple, Iterator

class Solution:
  def boxDelivering(self, boxes: List[List[int]], portsCount: int, maxBoxes: int, maxWeight: int) -> int:
    n = len(boxes)
    dp = [0] * (n + 1)
    trips = 2
    weight = 0

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

      if r > 0 and boxes[r][0] != boxes[r - 1][0]:
        trips += 1

      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

      dp[r + 1] = dp[l] + trips

    return dp[n]


Java solution:
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];
  }
}


C++ solution:
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];
  }
};