最近真是 OA 密集轟炸模式，一天排好幾場，幾乎沒停過。 昨天剛剛結束一輪 Goldman Sachs online assessment（HackerRank 平臺），整體難度不算誇張，20 分鐘就穩穩做完了，正好來和大家詳細復盤一下，給還在準備的小夥伴一些參考。
這次的題目主要是兩道 Coding，雖然不是那種超難 DP/圖論，但也有一些細節容易掉坑。 分享下思路和注意點。
Problem 1: Minimize Machine Usage
Problem Statement:
You are given a set of machines, each with a certain number of usages. The goal is to bring all usage counts down to zero through a series of operations.
- In one operation, you can select two machines and decrease each of their usage counts by one.
- You need to determine the minimum number of operations required to make all machines reach zero.
Key Insights:
This problem can be efficiently solved with a greedy strategy:
- In every step, pair the machine with the highest usage count with the machine that has the lowest usage count, and reduce both.
- Repeating this ensures that the “heaviest load” is being balanced by the “lightest load,”preventing unnecessary leftover usages.
Special Case Handling:
- If a machine’s usage count becomes very small, it can quickly be reduced to zero without wasting additional operations.
Implementation Details:
- The data size is not very large, so simulation with priority queues is feasible.
- A max heap can be used to always extract the machine with the largest usage count, and a min heap for the smallest one.
- Each operation decreases both selected values, and the total count of operations is accumulated until all values reach zero.
Answer:
The final result is simply the sum of all operations performed.
This greedy pairing guarantees the minimum number of steps.
Problem 2: Pods and Costs Placement
Problem Statement:
You are given a collection of elements, where each element has two attributes:
podcost
The task is to arrange these elements in an array according to the following rules:
- Elements with a higher cost must be placed with higher priority.
- After placing high-cost elements, the lower-cost elements should be assigned to the remaining available positions.
Key Insights:
This is a greedy + data structure problem.
- First, treat each element as a pair
(pod, cost). - Sort all elements in descending order of cost.
- Traverse the sorted list:
- If the position of a high-cost element is available, it should be fixed there.
- For lower-cost elements, if the intended position is already occupied, they must move to the next available free slot.
Optimization:
Naively scanning linearly for the next free position is inefficient. Instead:
- Use a map or disjoint set union (DSU/union-find) to keep track of the “next available position.”
- This allows jumping directly to the nearest empty slot without checking one by one.
Implementation Strategy:
- Sort by cost (descending).
- Insert elements in order, fixing high-cost ones first.
- Maintain the next available slot pointer using DSU or a map for efficient placement of remaining elements.
Takeaway:
The problem combines greedy prioritization (cost ordering) with efficient data structure usage (map/DSU) to handle placement in logarithmic or near-constant time.
Goldman Sachs online assessment 整體難度與感受
說實話，這次 Goldman Sachs OA 算是比較友好的，不像有些公司的 OA 愛考花哨的概率題或者 tricky DP。 這裡的兩道題邏輯都比較直觀，難點主要在實現細節。
我用時大概 20 分鐘，主要時間花在 debug 邊界條件。 比如：
Problem 1 里，機器次數為 0 的處理是否及時跳出迴圈。
Problem 2 里，空位分配如果寫得不夠嚴謹，可能會覆蓋到已經放好的高 cost 元素。
这类细节一旦没踩稳，很容易浪费十几分钟 debug。
給準備高盛 OA 的小建議
- 刷題方向：貪心、排序、map/heap 這種數據結構類的題要重點複習。
- 節奏把握：Goldman Sachs 的 OA 時間並不算很緊張，但如果在細節上卡住，就會很危險。
- 實戰演練：盡量類比真實環境，定時做題，把“卡點思維”提前練熟。
很多同學其實題目會做，但在實戰裡容易慌張，比如：
不知道什麼時候該果斷放棄 brute force。
debug 時陷入死迴圈，時間就被吃光。
没有规划好写代码的节奏，导致本来 15 分钟能写完的题，拖到 30 分钟还没过样例。
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