Roblox Online Assessment 面经分享 | 两道真题+思路+代码详解

最近很多同学在准备 Roblox Online Assessment 测试，这里分享一位同学最新的 Roblox OA 复盘，涵盖两道高频题：矩阵缺失值排序题和股票收益最大化题。本文将完整呈现题目思路、代码实现和潜在考点，帮助大家从逻辑推理到代码优化全面掌握。

Roblox OA 概览

Roblox 的 OA 通常包含 2 道编程题，平台为 CoderPad 或 HackerRank，总时长约 75 分钟。题目风格偏向中等难度，强调数据结构理解与数组处理能力。
这次的题目非常典型，考察了矩阵操作和滑动窗口优化两个方向。

Problem 1: Missing Value in 4×4 Squares and Rearrangement

Problem Description

You are given a large matrix composed of several 4×4 submatrices. Each 4×4 block contains unique numbers from 1 to 16, but one number is missing in each block. Your task is to find the missing value in each 4×4 submatrix, then sort all submatrices in ascending order based on their missing value (keeping the original order for ties), and finally reassemble them into one complete matrix.

Approach

Python 代码实现

def solution(mat):
    n = len(mat[0]) // 4
    sub_matrices = []
    for i in range(n):
        sub = [row[4*i:4*(i+1)] for row in mat]
        total = 136
        current_sum = sum(sum(row) for row in sub)
        missing = total - current_sum
        sub_matrices.append((missing, sub))
    sub_matrices.sort(key=lambda x: x[0])
    result = []
    for row in range(4):
        new_row = []
        for _, sub in sub_matrices:
            new_row.extend(sub[row])
        result.append(new_row)
    return result

考点总结

• 二维数组切片与重组
• 排序稳定性（tiebreak by original order）
• Python 列表推导式与矩阵操作熟练度

Problem 2: Maximize Stock Bot Revenue

Problem Description

You are given two arrays, prices and algo, where prices[i] represents the price of the stock on day i, and algo[i] indicates the bot’s action: 0 means “buy”, and 1 means “sell”.
You can choose a consecutive window of k days and force the bot to sell (set algo[i] = 1 within that window). Return the maximum total revenue achievable after applying this change.

Approach

1. Compute original revenue
   • If algo[i] == 1, add price[i]
   • If algo[i] == 0, subtract price[i]
2. Use sliding window to find best gain
   • Converting a buy (0) to sell (1) increases revenue by 2 * price[i]
   • Compute the maximum sum of such changes over any k-day window.
3. Final revenue
   • final_revenue = original_revenue + max_window_gain

Python 代码实现

def solution(prices, algo, k):
    original_revenue = 0
    for price, action in zip(prices, algo):
        original_revenue += price if action == 1 else -price
    changes = [2 * price if action == 0 else 0 for price, action in zip(prices, algo)]
    max_change = current_change = sum(changes[:k])
    for i in range(k, len(changes)):
        current_change += changes[i] - changes[i - k]
        if current_change > max_change:
            max_change = current_change
    return original_revenue + max_change

考点总结

• 滑动窗口技巧
• 时间复杂度优化 O(n)
• Python 中 zip 与列表推导式应用
• 投资收益场景逻辑建模

总结与经验建议

Roblox OA 的难度主要体现在对数据结构的灵活运用和算法思维的完整性。
这两道题虽然语义不复杂，但实现时容易卡在矩阵分割与窗口移动逻辑。建议刷题时重点掌握：

• 数组切片、矩阵拼接操作
• Prefix Sum 与 Sliding Window 技巧
• 代码稳定性与边界控制

从实战经验来看，大多数通过 Roblox OA 的同学在逻辑清晰度和代码规范性上都表现较好。时间充裕的情况下可以多练类似 Google Kick Start 或 LeetCode Medium 难度的题，题型风格相似。

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