Ship Packages Within D Days

1. Step 1: Initialize Solution.shipWithinDays

   Replace the empty starter with the first real state owned by Solution.shipWithinDays. A small, named state is easier to verify than a complete algorithm. Establish it before adding the branch or loop that changes it.

2. Step 2: Pass the Three Days case

   Complete the readable core algorithm for one representative Interview case. Binary-search capacity using a greedy ordered packing simulation as a monotone predicate.

3. Step 3: Harden the Five Days boundary

   Repair the reviewed boundary and pass the complete submission contract. Greedy packing uses the fewest possible days for a fixed capacity because delaying a package cannot reduce later load. Feasible capacities form a suffix, so lower-bound search returns the minimum feasible one.

• The lower bound is max(weights), because a package cannot be split across days.
• arrays strings two pointers sliding window

• LeetCode 1011 — Capacity To Ship Packages Within D Days

• Minimize a Feasible Value(opens in a new tab)

  Minimize a Feasible Value isolates left is the first unresolved candidate and right is always a feasible candidate, so the smallest feasible value remains inside [left, right]. That focused state discipline is required when implementing ship packages within days as a complete Interview Problem.

Binary-search capacity between the heaviest package and total weight. Greedily start a new day whenever the next package would exceed a proposed capacity.

Why it works: Greedy packing uses the fewest possible days for a fixed capacity because delaying a package cannot reduce later load. Feasible capacities form a suffix, so lower-bound search returns the minimum feasible one.

class Solution:
    def shipWithinDays(self, weights, days):
        """
        Checkpoint 1: initialize the state owned by this Interview contract.
        Checkpoint 2: assemble the primary transition without hiding the boundary.
        """
        left, right = max(weights), sum(weights)
        while left < right:
            capacity = (left + right) // 2
            used_days = 1
            load = 0
            for weight in weights:
                if load + weight > capacity:
                    used_days += 1
                    load = 0
                load += weight
            if used_days <= days:
                right = capacity
            else:
                left = capacity + 1
        return left

• Koko Eating Bananas(opens in a new tab)