Asked in: AMAZON
Image of the Question
All Testcases Passed ✔
Solution
def minimizeMaxParcelLoad(initialParcels, additional):
def helper(mid):
diff = 0
// ... rest of solution available after purchaseExplanation
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This problem asks us to distribute a large number of additional parcels among delivery agents who already have some initial load, aiming to minimize the maximum load assigned to any single agent after distribution. The key challenge is to find the minimal possible maximum load across all agents after distributing all additional parcels optimally.
To approach this problem effectively, you need to think in terms of searching for a feasible "maximum load" value and then checking if it is possible to distribute the additional parcels such that no agent exceeds this load. This naturally suggests a binary search on the answer (the maximum load), which is a common technique in optimization problems involving feasibility checks.
Here is a detailed direction on how to think about and solve this problem:
1. **Understanding the problem constraints:**
- You have `n` agents, each with some initial parcel count.
- You need to assign `additionalParcels` parcels among them.
- You want to minimize the maximum number of parcels assigned to any agent after assignment.
- The answer must be at least the maximum initial load because no agent's load can be less than what they already have.
- The answer can be as high as the sum of all parcels (initial plus additional) if you assign everything to one agent.
2. **Binary search on the maximum load:**
- Since the answer is a number that lies between the maximum initial parcels (lower bound) and the sum of all parcels (upper bound), use binary search on this range.
- Define a function that, given a candidate maximum load `mid`, checks whether it's possible to distribute all `additionalParcels` such that no agent's total load exceeds `mid`.
3. **Feasibility check for a candidate maximum load:**
- For each agent, calculate how many additional parcels they can receive without exceeding the candidate load `mid`.
- This is `max(0, mid - initialParcels[i])`.
- Sum these possible additional capacities over all agents.
- If the total capacity to take additional parcels at this load `mid` is at least `additionalParcels`, then it is possible to distribute all additional parcels without exceeding `mid`.
- Otherwise, `mid` is too small to accommodate the additional parcels fairly.
4. **Adjusting binary search bounds:**
- If the feasibility check passes for `mid`, move the upper bound down to try to find a smaller maximum load.
- If it fails, increase the lower bound to allow more capacity.
- Continue until the binary search converges.
5. **Why binary search is efficient:**
- Directly trying all possible maximum loads or simulating assignments would be too expensive.
- Binary searching the maximum load reduces the problem to O(n log M), where M is the difference between the upper and lower bound of maximum load.
- The feasibility check per step is O(n), which is efficient for `n` up to 10^5.
6. **Handling large inputs and data types:**
- The values involved can be large (e.g., additionalParcels up to 10^15), so use appropriate data types (like 64-bit integers) to avoid overflow.
- Carefully handle summations to prevent integer overflow.
7. **Starting boundaries for binary search:**
- The minimum load to consider is the maximum value in the initialParcels array, because no load can be less than the maximum already assigned.
- The maximum load can be the sum of initialParcels plus additionalParcels, meaning all parcels are assigned to one agent (worst case).
8. **Example walkthrough:**
- Suppose initialParcels = [7,5,1,9,1] and additionalParcels = 25.
- Maximum initial load = 9.
- Sum total = 7+5+1+9+1+25 = 48.
- Binary search between 9 and 48.
- Mid could be 28, check if each agent can take additional parcels to not exceed 28:
- Agent 1: can take 28-7=21
- Agent 2: can take 28-5=23
- Agent 3: can take 28-1=27
- Agent 4: can take 28-9=19
- Agent 5: can take 28-1=27
- Total possible additional parcels = 21+23+27+19+27=117 which is more than 25 → feasible.
- Try lower mid, eventually narrow down to the minimum possible max load.
9. **Edge cases to consider:**
- When additionalParcels is zero (no parcels to assign), the answer is just max(initialParcels).
- When all initial parcels are equal and additionalParcels is divisible evenly.
- Very large additionalParcels compared to initial load, requiring balancing across many agents.
- Single agent scenarios.
10. **Final output:**
- After the binary search concludes, you have the minimum maximum load achievable.
- This value guarantees the parcels can be assigned so that no agent has more than this number of parcels in total.
In summary, the problem is a classic application of binary search on the answer combined with a greedy feasibility check. By transforming the question into checking if a candidate load can accommodate the additional parcels, you leverage the ordered numeric search space to efficiently find the minimal maximum load achievable. This approach scales well for large input sizes and large numbers of parcels, making it suitable for the constraints provided.
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