今年秋招 JPMorgan Chase 的 OA 已经出了，题型延续了一贯的风格：思维逻辑+编程实现，不算特别“花哨”，但如果没刷到类似题目，很容易在边界条件上卡壳。下面给大家分享两道真题，都是我们学员第一时间反馈的。
JPMorgan Chase 招聘 / 投递流程（官方 & 经验汇总）
官方流程与政策（JPMorgan Careers “How We Hire”）
- 在 JPM Chase 的官网 “How We Hire” 页面可以看到其招聘流程框架：提交申请 → 面试 → 录用。
- 在 FAQ 部分，有关于申请状态、撤回申请、草稿保存、是否可以更改已提交申请的信息。
- 如果某个岗位表示“需要评估”的（Assessments / 在线测评），你会在申请流程中被通知是否需要做测评。
Signal Filtering
Signal filtering based on frequency is essential to minimize noise outside the desired frequency range. Filters can be combined such that only frequencies within the permissible range of all filters can pass through.
For instance, three filters with frequency ranges (10, 17), (13, 15), and (13, 17) will only allow frequencies between 13 and 15 to pass through, as this is the only overlapping range.
Given n signal frequencies and a sequence of m filters with specified frequency ranges from x to y (inclusive), determine how many signals will pass through all the filters. There will be a single common range where all filters overlap.
Example:
frequencies = [8, 15, 14, 16]
filtersRanges = [[10, 17], [13, …]]
The range that all of the filters overlap is from … to 15, inclusive. The 2 frequencies that will pass through
思路：
典型的区间交集问题。所有 filter 取一个最大下界 L = max(x) 和最小上界 R = min(y)，最终只保留 [L, R] 区间内的频率。遍历一遍 frequencies 统计就行。时间复杂度 O(n+m)。
Two Consignments
A shop has n types of items, where the quantity of the i-th item is denoted by quantity[i]. These items are to be shipped in two consignments:
- The first consignment contains items 1, 2, …, j
- The second consignment contains the remaining items j+1, …, n
The value of j can be chosen such that 1 ≤ j < n (1-based indexing).
The shopkeeper wants to make the total quantities in both consignments equal. To achieve this, they can increase or decrease the quantity of any item type by 1, any number of times. However, the quantity of each item must remain positive.Find the minimum number of operations required to make the quantities of both consignments equal if the items are split optimally.
Example:
n = 3
quantity = [1, 4, 4]
The optimal approach is to increase quantity = [1, 4, 5].
Partition at j = 2 to get consignments [1, 4] and [5].
The answer is 1.
思路：
这道题看似复杂，其实就是在所有分割点 j 中找一个最优的。对于每个分割点，左边和右边的总和分别是 A 和 B。如果两边其中一侧的总和足够大（≥ 类别数），只要调整差值就能搞定，操作数就是 |A-B|。
但如果两边的和都太小（小于该侧的类别数），就必须整体抬升到最低要求，这时公式会变成 2*Lmin - (A+B)。枚举所有 j，取最小值即可。时间复杂度 O(n)。
总结
这两道题一个是区间交集、一个是数组分割平衡，看似独立，其实都考察了边界条件的处理能力：
- Filter 那题如果不注意 inclusive（包含端点），很容易算错。
- Consignment 那题如果忽略了每类数量必须 ≥1，就会漏掉特殊情况。
很多同学平时刷题思路没问题，但一旦在 OA 里卡到这种边界，就容易超时或写崩。
不只是刷题，真正的底气来自这里
我们 programhelp 团队在这类大厂 OA 上有非常多的实战经验：
- 远程无痕辅助：帮你写出 100% 过测的代码，不留痕迹；
- 实时语音助攻：你在做题卡壳时，我们能及时提醒你该注意的边界点；
- VO/面试全流程辅导：不仅限于 OA，还能帮你稳住后续技术面。
已经有不少学员靠着我们的助攻顺利通过 JPMorgan 的 OA，进入后续面试环节。
如果你也在准备今年的秋招，不想在 OA 上栽跟头，可以随时联系我们 。
