Confluent 最新 OA 复盘｜两道题中等偏上强度，全程监控 + 边界坑点详解

最近刚刷完 Confluent 的 OA，整体感觉比预期略难一点，属于那种看上去简单、但细节很多、边界容易出错的类型。整场测试共两道编程题，总时长 65 分钟，题目在 HackerRank 平台上完成。从题型来看，Confluent 这次的 OA 明显更偏逻辑类和数据结构应用类，不是算法陷阱题，而是看你代码思路是否清晰、能否正确处理边界条件。

Problem 1: Generate Two Binary Strings

Problem Description

Given an integer array nums, generate two binary strings string1 and string2 of the same length:

• string1[i] = '1' if nums[i] has appeared before (in nums[0..i-1]), otherwise '0'.
• string2[i] = '1' if nums[i] will appear again later (in nums[i+1..n-1]), otherwise '0'.

Return [string1, string2].

Example

Input: nums = [1, 2, 1, 3]
Output: ["0010", "1000"]
Explanation:
string1 = "0010" because 1 appears again at index 2.
string2 = "1000" because only nums[0] (1) appears later again.

Approach

The most straightforward solution is to use sets to record which elements have already been seen:

• Traverse the array from left to right to build string1:
  • Maintain a seen set.
  • If nums[i] is already in seen, append '1', otherwise '0' and add it to the set.
• Traverse from right to left to build string2:
  • Maintain another set seen_from_right.
  • If nums[i] appears again later (exists in the set), mark '1', else '0' and add it.

This approach takes O(n) time and O(n) space, very clean and efficient.

Key Notes

读题的时候要格外注意，“has appeared before” 和 “will appear again later” 是两个方向完全相反的判断。第一次读英文题容易混淆前后顺序，建议对照 example 理解一下，不然容易反过来写。

Problem 2: Duplicate Message Filter

Problem Description

A message delivery system needs to prevent duplicate messages from being sent within a short period of time.

If the same message has been successfully delivered within the last k seconds, then the new message should be considered a duplicate (False). Otherwise, it should be delivered (True) and update its last delivered timestamp.

You are given:

• timestamps[i]: the arrival time of the i-th message (not necessarily sorted)
• messages[i]: the message text
• k: the time window in seconds

Return a boolean array indicating whether each message should be delivered (True) or dropped (False).

Example

Input:
timestamps = [1, 2, 4, 8, 10]
messages = ["a", "a", "a", "a", "a"]
k = 3

Output:
[True, False, True, True, False]

Explanation:

• At t=1, “a” is new → deliver.
• At t=2, “a” appeared 1s ago (<3) → drop.
• At t=4, difference = 3 → deliver (equal to k is allowed).
• At t=8, difference = 4 → deliver.
• At t=10, difference = 2 (<3) → drop.

Approach

Use a dictionary to record the last delivered timestamp of each message:

last_time = {}
res = []

for t, msg in zip(timestamps, messages):
    if msg not in last_time or t - last_time[msg] >= k:
        res.append(True)
        last_time[msg] = t
    else:
        res.append(False)

This works for both sorted and unsorted timestamps as long as you process them in input order.

Key Pitfall

边界条件是这题最大的陷阱。
题中 “within the last k seconds” 意味着只要时间差小于 k 都算重复，等于 k 是允许发送的。
如果判断写成 > 而不是 >=，会直接导致半数测试不通过。

另外要注意输入可能不是升序的，如果乱序输入，就必须严格按输入顺序判断，而不能先排序，否则逻辑错误。

整体感受与策略

这份 OA 的整体难度属于 Leetcode 中等，第一题偏向逻辑构造，第二题偏工程实现。
两题都没有复杂算法，但都要求写出结构清晰、边界严谨的代码。

建议时间分配：

• 第一题 20 分钟左右足够，代码逻辑非常直接；
• 第二题 40 分钟较稳，尤其要留时间验证边界；
• 最后 5 分钟跑完所有测试、检查格式。

如果英文读题慢或者逻辑容易混淆，建议提前熟悉 HackerRank 的测试界面，例子读一遍后思考输入输出的含义。

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