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Problem solution · Java

Delivering Boxes from Storage to Ports

Delivering Boxes from Storage to Ports: a Java solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Dynamic programming
Source
walkccc LeetCode Solutions
Length
37 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Dynamic programming

For Delivering Boxes from Storage to Ports, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 37 lines of Java from the credited upstream file 1687.java.
  • The implementation visibly relies on sequence storage, cached states.
  • 2 loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeDelivering Boxes from Storage to Ports · JavaJava
Use this to learn the idea, then write your own version.
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];  }}

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