Problem Description

You need to deliver boxes from a storage warehouse to various ports using a single ship. The ship has constraints on both the number of boxes it can carry and the total weight capacity.

You're given an array boxes where each element boxes[i] = [ports_i, weight_i] contains:

• ports_i: the destination port for the i-th box
• weight_i: the weight of the i-th box

You're also given:

• portsCount: the total number of ports
• maxBoxes: maximum number of boxes the ship can carry at once
• maxWeight: maximum total weight the ship can carry at once

Important constraints:

• Boxes must be delivered in the exact order they appear in the array
• The ship starts at the storage warehouse

How the delivery process works:

1. The ship loads some consecutive boxes from the front of the queue (respecting both maxBoxes and maxWeight limits)
2. The ship delivers each loaded box in order to its destination port:
   • If the ship is already at the correct port, no trip is needed
   • If the ship needs to go to a different port, it makes one trip
3. After delivering all loaded boxes, the ship makes a return trip to the storage warehouse
4. The process repeats until all boxes are delivered

Counting trips:

• Going from storage to the first port counts as 1 trip
• Moving between different ports counts as 1 trip per move
• Returning from the last port to storage counts as 1 trip
• Staying at the same port for consecutive deliveries counts as 0 trips

Your task is to find the minimum number of trips needed to deliver all boxes to their respective ports.

Example scenario: If the ship takes boxes going to ports [4, 4, 5]:

• Trip 1: Storage → Port 4
• Trip 0: Stay at Port 4 (second box also goes here)
• Trip 1: Port 4 → Port 5
• Trip 1: Port 5 → Storage
• Total: 3 trips for this batch

The challenge is to optimally group consecutive boxes for each loading while respecting the ship's capacity constraints to minimize the total number of trips.

Intuition

Let's think about this problem step by step. Since boxes must be delivered in order, we're essentially deciding where to "cut" the sequence - which groups of consecutive boxes should be loaded together in each batch.

For any batch of consecutive boxes from index j+1 to i, we can calculate the number of trips needed:

• 1 trip from storage to the first port
• Additional trips between different consecutive ports (if port changes)
• 1 trip back to storage
• Total = 2 + (number of port changes within the batch)

This naturally leads to a dynamic programming approach. Let f[i] represent the minimum trips needed to deliver the first i boxes. To compute f[i], we need to consider all valid previous positions j where we could have ended the previous batch, then taken boxes j+1 through i as the current batch.

The key insight is that for any batch [j+1, ..., i]:

• It must satisfy weight constraint: sum(weights[j+1]...weights[i]) ≤ maxWeight
• It must satisfy box count constraint: i - j ≤ maxBoxes
• The trip cost is: 2 + (number of port changes in range)

To efficiently calculate port changes, we can use prefix sums. If we create an array c where c[k] = 1 if port[k] != port[k+1], then the number of port changes from position j+1 to i is simply cs[i-1] - cs[j] where cs is the prefix sum of c.

The state transition becomes: f[i] = min(f[j] + cs[i-1] - cs[j] + 2) for all valid j

However, with data scale reaching 10^5, the O(n²) approach times out. We need optimization.

Looking at the transition equation more carefully: f[i] = min(f[j] - cs[j]) + cs[i-1] + 2

The term cs[i-1] + 2 is constant for a fixed i. We're essentially finding the minimum value of f[j] - cs[j] among all valid j values in the sliding window [i-maxBoxes, i-1] that also satisfy the weight constraint.

This is a classic sliding window minimum problem! We can use a monotonic deque to maintain potential candidates:

• Keep indices in increasing order of f[j] - cs[j] values
• Remove indices that fall outside the box count window
• Remove indices that violate the weight constraint
• The front of the deque gives us the optimal j for current i

This optimization reduces the complexity from O(n²) to O(n), making the solution efficient enough for the given constraints.

Learn more about Segment Tree, Queue, Dynamic Programming, Prefix Sum, Monotonic Queue and Heap (Priority Queue) patterns.

Solution Approach

Let's implement the optimized dynamic programming solution with monotonic queue step by step.

Step 1: Preprocessing

First, we need to prepare auxiliary arrays for efficient calculation:

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))

• ws[i]: prefix sum of weights, where ws[i] = sum of weights from box 0 to box i-1
• c[i]: 1 if port changes between box i and i+1, 0 otherwise
• cs[i]: prefix sum of port changes

Step 2: Dynamic Programming with Monotonic Queue

Initialize the DP array and deque:

f = [0] * (n + 1)  # f[i] = minimum trips to deliver first i boxes
q = deque([0])     # monotonic [queue](/problems/queue_intro) storing indices

Step 3: Main DP Loop

For each position i from 1 to n:

for i in range(1, n + 1):
    # Remove invalid indices from front of deque
    while q and (i - q[0] > maxBoxes or ws[i] - ws[q[0]] > maxWeight):
        q.popleft()

This step removes indices that:

• Violate box count constraint: i - j > maxBoxes
• Violate weight constraint: ws[i] - ws[j] > maxWeight

Step 4: Calculate Minimum Trips

if q:
    f[i] = cs[i - 1] + f[q[0]] - cs[q[0]] + 2

The formula breaks down as:

• f[q[0]]: minimum trips to deliver first q[0] boxes
• cs[i-1] - cs[q[0]]: port changes between boxes q[0]+1 to i
• +2: one trip from storage to first port, one trip back to storage

Step 5: Maintain Monotonic Property

if i < n:
    while q and f[q[-1]] - cs[q[-1]] >= f[i] - cs[i]:
        q.pop()
    q.append(i)

We maintain the deque in increasing order of f[j] - cs[j] values:

• Remove indices from the back if their value is not better than current i
• Add current index i to the deque (only if not the last position)

Why This Works:

The monotonic queue ensures that:

1. The front element always gives the minimum value of f[j] - cs[j] in the valid window
2. Outdated or suboptimal indices are efficiently removed
3. Each index is added and removed at most once, giving O(n) complexity

Time Complexity: O(n) - each element is processed once through the deque Space Complexity: O(n) - for the auxiliary arrays and DP array

The final answer is f[n], representing the minimum trips needed to deliver all n boxes.

Example Walkthrough

Let's walk through a concrete example to understand the solution approach.

Input:

• boxes = [[1,2], [3,3], [3,1], [3,1], [4,4]]
• portsCount = 4, maxBoxes = 3, maxWeight = 6

Step 1: Preprocessing

First, we build our auxiliary arrays:

• Weights: [2, 3, 1, 1, 4]
• Weight prefix sums ws: [0, 2, 5, 6, 7, 11]
• Port changes c: [1, 0, 0, 1] (comparing ports: 1→3, 3→3, 3→3, 3→4)
• Port change prefix sums cs: [0, 1, 1, 1, 2]

Step 2: Initialize DP

• f = [0, 0, 0, 0, 0, 0] (will fill positions 1-5)
• q = deque([0]) (monotonic queue starts with index 0)

Step 3: Process Each Position

i = 1 (Delivering first box [1,2]):

• Check deque front: Can we take boxes from position 0 to 1?
  • Box count: 1 - 0 = 1 ≤ 3 ✓
  • Weight: ws[1] - ws[0] = 2 - 0 = 2 ≤ 6 ✓
• Calculate: f[1] = cs[0] + f[0] - cs[0] + 2 = 0 + 0 - 0 + 2 = 2
  • This means: storage→port1→storage (2 trips)
• Update deque: Add index 1
• State: f = [0, 2, 0, 0, 0, 0], q = [0, 1]

i = 2 (Delivering first two boxes):

• Check deque front (index 0):
  • Box count: 2 - 0 = 2 ≤ 3 ✓
  • Weight: ws[2] - ws[0] = 5 - 0 = 5 ≤ 6 ✓
• Calculate: f[2] = cs[1] + f[0] - cs[0] + 2 = 1 + 0 - 0 + 2 = 3
  • Taking both boxes: storage→port1→port3→storage (3 trips)
• Alternative from index 1: f[2] = cs[1] + f[1] - cs[1] + 2 = 1 + 2 - 1 + 2 = 4
  • Taking second box alone after first: more trips
• Deque maintenance: index 0 gives better value (0 - 0 = 0) than index 1 (2 - 1 = 1)
• State: f = [0, 2, 3, 0, 0, 0], q = [0, 2]

i = 3 (Delivering first three boxes):

• Check deque front (index 0):
  • Box count: 3 - 0 = 3 ≤ 3 ✓
  • Weight: ws[3] - ws[0] = 6 - 0 = 6 ≤ 6 ✓
• Calculate: f[3] = cs[2] + f[0] - cs[0] + 2 = 1 + 0 - 0 + 2 = 3
  • All three boxes: storage→port1→port3(stay)→port3(stay)→storage (3 trips)
• State: f = [0, 2, 3, 3, 0, 0], q = [0, 3]

i = 4 (Delivering first four boxes):

• Check deque front (index 0):
  • Box count: 4 - 0 = 4 > 3 ✗ (exceeds maxBoxes)
  • Remove index 0 from deque
• Next in deque (index 3):
  • Box count: 4 - 3 = 1 ≤ 3 ✓
  • Weight: ws[4] - ws[3] = 7 - 6 = 1 ≤ 6 ✓
• Calculate: f[4] = cs[3] + f[3] - cs[3] + 2 = 1 + 3 - 1 + 2 = 5
  • Take 4th box alone after first three
• State: f = [0, 2, 3, 3, 5, 0], q = [3, 4]

i = 5 (Delivering all five boxes):

• Check deque front (index 3):
  • Box count: 5 - 3 = 2 ≤ 3 ✓
  • Weight: ws[5] - ws[3] = 11 - 6 = 5 ≤ 6 ✓
• Calculate: f[5] = cs[4] + f[3] - cs[3] + 2 = 2 + 3 - 1 + 2 = 6
  • Take last two boxes together after first three
• Final answer: f[5] = 6

Verification:

• Batch 1: Boxes [1,2], [3,3], [3,1] → storage→port1→port3→storage (3 trips)
• Batch 2: Boxes [3,1], [4,4] → storage→port3→port4→storage (3 trips)
• Total: 6 trips ✓

Solution Implementation

Time and Space Complexity

Time Complexity: O(n), where n is the number of boxes.

The algorithm processes each box exactly once through the following operations:

• Building the weight prefix sum array ws takes O(n) time
• Creating the consecutive port difference array c takes O(n) time
• Building the prefix sum array cs takes O(n) time
• The main dynamic programming loop runs n iterations
  • Each iteration performs deque operations (append, pop, popleft)
  • While there are nested while loops, each element is added to and removed from the deque at most once throughout the entire algorithm
  • This amortizes to O(1) per iteration
• Total: O(n) + O(n) + O(n) + O(n) = O(n)

Space Complexity: O(n), where n is the number of boxes.

The algorithm uses the following data structures:

• ws: weight prefix sum array of size n + 1
• c: consecutive port difference array of size n - 1
• cs: prefix sum of differences array of size n
• f: dynamic programming array of size n + 1
• q: deque that can contain at most min(n, maxBoxes) elements, which is O(n) in the worst case
• Total: O(n) + O(n) + O(n) + O(n) + O(n) = O(n)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Off-by-One Errors in Index Management

Pitfall: The most common mistake is confusion between box indices and DP array indices. The DP array dp[i] represents the state after delivering the first i boxes (0-indexed), while boxes[i-1] is the i-th box. This leads to errors when:

• Calculating weight sums: Using weight_prefix_sum[i] - weight_prefix_sum[j] for boxes j to i-1
• Accessing port changes: Using changes_prefix_sum[i-1] for changes up to box i-1

Solution: Always be explicit about what each index represents:

# When dp[i] represents first i boxes delivered:
# - The last box included is boxes[i-1]
# - Weight from box j to i-1: weight_prefix_sum[i] - weight_prefix_sum[j]
# - Port changes up to box i-1: changes_prefix_sum[i-1]

2. Incorrect Monotonic Queue Maintenance

Pitfall: Forgetting to skip adding the last index n to the deque. Since dp[n] is the final answer and won't be used as a starting point for future batches, adding it causes unnecessary comparisons and potential errors.

Solution: Add the guard condition before deque maintenance:

if i < n:  # Critical: Don't add the last position to deque
    while deque_indices and (...):
        deque_indices.pop()
    deque_indices.append(i)

3. Misunderstanding Port Change Calculation

Pitfall: Incorrectly calculating port changes when the ship delivers multiple boxes. Remember that consecutive boxes to the same port require 0 additional trips, not 1 trip each.

Example Error:

# WRONG: Counting each box delivery as a separate trip
trips = number_of_boxes + 1  # Incorrect!

# CORRECT: Only count trips when port changes
trips = port_changes_in_batch + 2  # +2 for initial and return trips

Solution: Use the preprocessed port_changes array which correctly marks transitions:

port_changes = [int(boxes[i][0] != boxes[i+1][0]) for i in range(n-1)]

4. Missing Edge Cases in Constraint Validation

Pitfall: Not handling the case where even a single box exceeds constraints, or when the deque becomes empty after removing invalid indices.

Solution: Always check if the deque is non-empty before accessing:

if deque_indices:  # Essential check
    dp[i] = changes_prefix_sum[i - 1] + dp[deque_indices[0]] - 
            changes_prefix_sum[deque_indices[0]] + 2
else:
    # This shouldn't happen with valid inputs, but good to handle
    dp[i] = float('inf')  # Or handle the error appropriately

5. Inefficient Pairwise Implementation

Pitfall: For Python versions before 3.10, pairwise isn't available in itertools, causing runtime errors.

Solution: Provide a fallback implementation or use manual iteration:

# Option 1: Manual iteration
port_changes = [int(boxes[i][0] != boxes[i+1][0]) for i in range(len(boxes)-1)]

# Option 2: Custom pairwise for older Python versions
def pairwise(iterable):
    a, b = tee(iterable)
    next(b, None)
    return zip(a, b)

6. Integer Overflow in Weight Calculations

Pitfall: When dealing with large weight values and many boxes, prefix sums might overflow in languages with fixed integer sizes (though Python handles this automatically).

Solution: In Python this isn't an issue, but in other languages, use appropriate data types:

# For other languages, ensure using long/int64 for weight calculations
weight_prefix_sum = [0]
for box in boxes:
    weight_prefix_sum.append(weight_prefix_sum[-1] + box[1])