各位正在冲美妆行业的朋友们，给大家带来一手经验！我最近刚刚斩获了雅诗兰黛（Estée Lauder Companies – ELC）数据分析师（Data Analyst, DA）岗位的Offer，趁着记忆还热乎，想和大家聊聊我的技术面经历，也顺便分享一些小技巧。希望能帮到正在准备的你～
大家都知道，ELC作为全球顶级的奢侈美妆集团，DA岗位的竞争非常激烈。面试环节也比较全面：除了常规的行为面、HR面，技术面才是真正的考核重点。它不仅要求我们有扎实的硬技能，还会通过业务场景题去考察你解决问题的思路和逻辑。
技术面整体感受
我这次的技术面基本就是围绕 SQL + Python（尤其是Pandas） 展开的，期间还穿插了一些 数据建模 和 可视化 相关的概念。面试官真的超级 nice，会一步步引导你，但别以为就能轻松过关哦～题目的 广度和深度 都挺考验功底的。
给大家的建议是：准备的时候千万别只停留在“会写语法/会用函数”这个层面，更要理解 背后的逻辑和原理，比如为什么要这么设计、这样写的最佳实践是什么。这样在追问的时候，你才能接得住，不会被问懵。
Estee Lauder interview questions
Estée Lauder Technical Interview Questions
Question 1: Inventory Search
Prompt: Given a list of product names and a search query, implement a function searchProducts(products, query) that returns all product names that contain all words from the query. The search should be case-insensitive, and the order of words in the query does not matter. Assume product names and query words are separated by spaces. This task is crucial for helping customers find the right products quickly, whether it’s a specific lipstick shade or a new serum.
Example: products = [“Advanced Night Repair Serum”, “Double Wear Foundation”, “Pure Color Envy Lipstick”, “Revitalizing Supreme+ Youth Power Creme”] query = “repair serum” Output: [“Advanced Night Repair Serum”]
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Solution: To solve this, you need to first normalize both the product names and the query.
- Normalize the query: Convert the query to lowercase and split it into a list of words. This gives you a set of target words to look for.
- Iterate through products: Go through each product name in the
productslist. - Normalize each product name: For each product name, convert it to lowercase and split it into a list of words.
- Check for matches: For a product name to be a match, every single word from the normalized query must be present in the normalized list of words for that product. A simple way to do this is to use a nested loop or a set-based comparison.
def searchProducts(products, query):
"""
Finds products containing all words from a query.
Args:
products (list): A list of product name strings.
query (str): The search query string.
Returns:
list: A list of matching product names.
"""
query_words = set(query.lower().split())
matching_products = []
for product in products:
product_words = set(product.lower().split())
# Check if all query words are in the product words
if query_words.issubset(product_words):
matching_products.append(product)
return matching_products
# Example usage:
products = ["Advanced Night Repair Serum", "Double Wear Foundation", "Pure Color Envy Lipstick", "Revitalizing Supreme+ Youth Power Creme"]
query = "repair serum"
print(f"Products: {products}")
print(f"Query: \"{query}\"")
print(f"Result: {searchProducts(products, query)}")Estée Lauder Technical Interview Questions
Question 1: Inventory Search
Prompt: Given a list of product names and a search query, implement a function searchProducts(products, query) that returns all product names that contain all words from the query. The search should be case-insensitive, and the order of words in the query does not matter. Assume product names and query words are separated by spaces. This task is crucial for helping customers find the right products quickly, whether it’s a specific lipstick shade or a new serum.
Example: products = [“Advanced Night Repair Serum”, “Double Wear Foundation”, “Pure Color Envy Lipstick”, “Revitalizing Supreme+ Youth Power Creme”] query = “repair serum” Output: [“Advanced Night Repair Serum”]
<br>
Solution: To solve this, you need to first normalize both the product names and the query.
Normalize the query: Convert the query to lowercase and split it into a list of words. This gives you a set of target words to look for.
Iterate through products: Go through each product name in the products list.
Normalize each product name: For each product name, convert it to lowercase and split it into a list of words.
Check for matches: For a product name to be a match, every single word from the normalized query must be present in the normalized list of words for that product. A simple way to do this is to use a nested loop or a set-based comparison.
Python
def searchProducts(products, query):
"""
Finds products containing all words from a query.
Args:
products (list): A list of product name strings.
query (str): The search query string.
Returns:
list: A list of matching product names.
"""
query_words = set(query.lower().split())
matching_products = []
for product in products:
product_words = set(product.lower().split())
# Check if all query words are in the product words
if query_words.issubset(product_words):
matching_products.append(product)
return matching_products
# Example usage:
products = ["Advanced Night Repair Serum", "Double Wear Foundation", "Pure Color Envy Lipstick", "Revitalizing Supreme+ Youth Power Creme"]
query = "repair serum"
print(f"Products: {products}")
print(f"Query: \"{query}\"")
print(f"Result: {searchProducts(products, query)}")
Question 2: Supply Chain Network Analysis
Prompt: At Estée Lauder, we have a complex supply chain network connecting manufacturing sites, distribution centers, and retail stores. This network can be represented as a graph where nodes are locations and edges are transportation routes. Given a list of connections connections (where connections[i] = [locationA, locationB] means locationA and locationB are directly connected), and a starting location startLocation, implement a function find_next_tier_locations(connections, startLocation) that returns all unique locations that are ‘next-tier locations’ (i.e., locations directly connected to the startLocation‘s direct connections, but not the startLocation itself or its direct connections). Assume all connections are bidirectional.
Example: connections = [[‘plant_A’, ‘dc_1’], [‘plant_A’, ‘dc_2’], [‘dc_1’, ‘store_A’], [‘dc_2’, ‘store_B’], [‘store_A’, ‘store_C’]] startLocation = ‘plant_A’ Output: [‘store_A’, ‘store_B’] (Locations ‘dc_1’ and ‘dc_2’ are direct connections to ‘plant_A’. ‘store_A’ is connected to ‘dc_1’ and ‘store_B’ is connected to ‘dc_2’. ‘store_C’ is a connection of a connection of a connection.)
<br>
Solution: This problem is a classic graph traversal question.
Build the graph: First, convert the list of connections into a more usable graph representation, like an adjacency list (a dictionary where keys are locations and values are lists of connected locations). Since connections are bidirectional, you must add both A -> B and B -> A to your graph.
Find direct connections: Start at the startLocation and find all of its direct neighbors.
Find next-tier connections: Iterate through each of the direct connections you just found. For each of these, find all of their neighbors.
Filter the results: The final result set should include all the unique locations found in the previous step. You must exclude the startLocation itself and all its direct connections. A set is the ideal data structure for this to ensure uniqueness and efficient filtering.
from collections import defaultdict
def find_next_tier_locations(connections, startLocation):
"""
Finds unique locations that are 'next-tier' in a network.
Args:
connections (list): A list of connection pairs.
startLocation (str): The starting location.
Returns:
list: A list of unique next-tier locations.
"""
# 1. Build the adjacency list graph
graph = defaultdict(list)
for loc1, loc2 in connections:
graph[loc1].append(loc2)
graph[loc2].append(loc1)
# 2. Find direct connections
direct_connections = set(graph[startLocation])
# Keep track of all visited nodes to prevent cycles and duplicates
visited = {startLocation} | direct_connections
# 3. Find next-tier connections
next_tier_locations = set()
for direct_loc in direct_connections:
for next_loc in graph[direct_loc]:
if next_loc not in visited:
next_tier_locations.add(next_loc)
return list(next_tier_locations)
# Example usage:
connections = [['plant_A', 'dc_1'], ['plant_A', 'dc_2'], ['dc_1', 'store_A'], ['dc_2', 'store_B'], ['store_A', 'store_C']]
startLocation = 'plant_A'
print(f"Connections: {connections}")
print(f"Start Location: \"{startLocation}\"")
print(f"Result: {find_next_tier_locations(connections, startLocation)}")面试真的就像闯关游戏，准备充分才能一路顺利通关
我的这次经历也算是一个小小的心得分享～如果你现在也在为OA或者技术面发愁，其实不用一个人死撑，Programhelp的远程助攻完全能帮你轻松很多，节省精力，把状态留给关键问题。
我们能做什么？
OA无痕联机代写：实时远程写题，保证稳定输出
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Mock Interview：提前陪练，帮你熟悉大厂面试套路
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我们的目标很简单：
让你在真正的面试现场不再慌乱，思路清晰，答题有底气。
