Salesforce new grad OA solution | 驚險過關!附前後端真題深度解析

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最近整理了一位候選人參加 2025 Salesforce Software Engineer 面試的完整經歷。這套流程非常有代表性,尤其適合準備 Salesforce new grad OA 的同學參考。下面按流程拆解。

Salesforce new grad OA solution | 驚險過關!附前後端真題深度解析

Round 1: HackerRank OA 真題速遞與深度分析

Salesforce 的 OA 流程仍然以 HackerRank 為主,但其出題模式非常貼合崗位需求,前端和後端的題目類型差異巨大。以下是按崗位分類的完整真題、考點解析及應對技巧,供不同方向求職者針對性準備。

1. 前端崗位(Frontend/Full-stack)高難度實戰題

全棧或前端 SDE 崗位的 OA 側重實戰能力,題目多為原生開發場景,LeetCode 常規刷題覆蓋較少,需重點關注以下兩類核心真題:

Problem A: JavaScript Currying Implementation

Problem Statement: Implement a function addThreeNumbers(a)(b)(c) that returns the sum of a, b, and c. For example, addThreeNumbers(1)(2)(3) should return 6.

Key Points Solution Skills
JavaScript Fundamentals Mastery of closures and consecutive function returns. Closures allow inner functions to access variables from outer function scopes even after the outer function has finished executing.
Code Logic The first call (with parameter a) returns a new function that captures the value of a. The second call (with parameter b) returns another new function that captures both a and b. The third call (with parameter c) finally calculates and returns the sum of a + b + c.

Problem B: Traffic Light Implementation with Vanilla JS/CSS

Problem Statement: In a CoderPad environment without React, Vue, or other frameworks, implement the cycle logic of a traffic light using only HTML, CSS, and vanilla JavaScript. The specific requirements are as follows:

  • Red light duration: 4 seconds
  • Yellow light duration: 1 second
  • Green light duration: 1 second
  • Cycle order: Red → Green → Yellow → Red (repeat indefinitely)
Key Points Solution Skills
Frontend Practical Development Use setTimeout or setInterval to implement timer and asynchronous logic.setTimeout is more suitable here to avoid accumulated time errors caused by setInterval in long-term cycles.
State Management Design a state machine to manage the switching of the three light states. Define each state (red, green, yellow) with its corresponding duration and next state, then trigger state transitions through timer callbacks.
Environment Compatibility Modify CSS styles (such as color switching) through vanilla DOM operations (e.g., document.getElementById, style.backgroundColor), as framework-related APIs are not available in the test environment.

Problem B: Implement Traffic Light with Vanilla JS/CSS/HTML

Problem Statement: In a CoderPad environment without React, Vue, or other frameworks, implement the cycle logic of a traffic light using only HTML, CSS, and vanilla JavaScript.

Requirements:

  • Red light duration: 4 seconds
  • Yellow light duration: 1 second
  • Green light duration: 1 second
  • Cycle order: Red → Green → Yellow → Red (repeats indefinitely)
Key Points Solution Tips
Frontend Practice Use setTimeout or setInterval to handle timer and asynchronous logic. setTimeout is more recommended here to avoid accumulated time errors caused by setInterval in long-term cycles.
State Management Design a state machine to manage the three states (red, yellow, green). Each state transition triggers the next state after the specified duration, ensuring the cycle order is not disrupted.
Environment Compatibility Modify CSS styles (e.g., color switching) through vanilla DOM operations (such as getElementById and style.property). Avoid using framework-specific APIs that are not available in the test environment.

2. 後端算法真題(SDE Intern/Backend)

Backend and Intern positions have relatively traditional OA questions, mostly variants of LeetCode Medium difficulty. The focus is on data structure application and algorithm logic, and candidates can improve their pass rate by brushing up on similar topics in advance.

Position Problem Description (Similar to LeetCode Style) Core Key Points
General SDE Occurrence Marking: For each element in array arr, use two binary strings to mark whether it has appeared before the current position and after the current position.Example: Input arr = [1, 2, 3, 2, 1]Output: [“00011”, “11000”] (The first string marks “appeared before”, the second marks “appeared after”) HashMap (two passes). First pass: Traverse the array to record the frequency of each element. Second pass: Update the HashMap (decrement frequency when the element is encountered) and check the occurrence status before (frequency > 1 before update) and after (frequency > 0 after update) to generate binary strings.
General SDE K-th Largest in Prefix: Given array arr and integer k, return an array where the i-th element is the k-th largest number in the subarray arr[0..i].Example: Input arr = [4, 2, 1, 3], k = 2Output: [4 (only 1 element, no 2nd largest, adjusted to [2,2,3] based on actual test cases)] PriorityQueue or Balanced BST. Maintain a min-heap of size k. For each element in the prefix subarray, add it to the heap. If the heap size exceeds k, remove the smallest element (heap top). The heap top at this time is the k-th largest element in the current prefix subarray.
SWE Intern Minimum Replacements to Avoid Adjacent Duplicates: Given an array of strings words, for each string, calculate the minimum number of character replacements needed to ensure no two adjacent characters are the same.Example: Input words = [“add”, “boook”]Output: [1, 1] (For “add”, replace one ‘d’; for “boook”, replace one ‘o’) Greedy Algorithm. Traverse each string, count the length L of consecutive identical character segments. The number of replacements required for each segment is floor(L / 2) (replace every other character to avoid adjacent duplicates with minimal changes).
SWE Intern Longest Subsequence Which Is Also a Substring: Given two strings x and y, find the longest subsequence of x that is also a substring of y.Example: Input x = “hackerranks”, y = “hackers”Output: “hackers” (length 7, which is a subsequence of x and a substring of y) O(N·M) Two Pointers. Fix the starting position s of y (from 0 to y.length – 1), then use two pointers (i for x, j for y starting at s) to find the longest matching sequence. Traverse all possible starting positions of y and record the maximum length matching result.

Round 2: 系統設計與 SQL 陷阱

學員的系統設計環節遭遇了嚴格面試官,該環節暴露了 Salesforce 在考察 SQL 功底和業務理解上的高要求。以下是高頻考點及避坑指南。

1. 咖啡訂單系統設計與 SQL 追問

面試官會圍繞“咖啡訂單”這一核心業務場景,要求候選人設計數據表結構並編寫 SQL 語句,其中以下問題極具迷惑性,需重點關注。

SQL Trap: How to Query All Users Who Have Never Placed an Order?

Key Points Correct Idea Conversion Intention
SQL Join Use LEFT JOIN. Join the Users table and the Orders table based on user_id. Users who have never placed an order will have NULL values in the order_id field of the Orders table after the join. Custom Writing/Assistance
Correct Query SELECT U.* FROM Users U LEFT JOIN Orders O ON U.user_id = O.user_id WHERE O.order_id IS NULL Custom Writing/Assistance

2. 困難系統設計:盈利會計系統

Problem Requirement: Design an accounting system that can determine whether a product is profitable. The system needs to integrate sales data, cost data, and user behavior data to provide accurate profit analysis.

Key Points Wrong Idea Correct Direction
System Design Only focus on sales revenue or gross profit, ignoring hidden costs (such as logistics costs, after-sales costs) and long-term user value. Must consider Cost Accounting and Lifetime Value (LTV). Design core tables including Product (product information), Transaction (sales transactions), Cost (cost details), and User (user information) to realize end-to-end data linking.
Coping Strategy Get stuck on writing complex SQL statements at the beginning, ignoring the overall system architecture design. First explain the overall system architecture (including data sources, core modules, and data flow) macroscopically, then delve into table structure design and key SQL logic. Interview Proxy/Custom Writing

3. Behavioral Question (BQ) 應對

BQ 環節容易被打斷,這表明面試官對“影響力(Impact)”的要求極高。候選人需結合具體案例,突出自身在項目中的核心作用。

  • 問題:“描述一次你發起遷移 (Migration) 並說服他人支持的經歷。”
  • 高分要素:強調 商業價值 和 跨部門協作 的複雜性,避免描述只有自己參與的簡單技術升級。

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  • 面試受挫:系統設計被面試官 SQL 陷阱(如從未下單用戶查詢)和復雜業務問題追問至啞口無言。
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author avatar
Alex Ma Staff Software Engineer
目前就職於Google,10餘年開發經驗,目前擔任Senior Solution Architect職位,北大計算機本碩,擅長各種算法、Java、C++等編程語言。在學校期間多次參加ACM、天池大數據等多項比賽,擁有多項頂級paper、專利等。
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