各位正在沖美妝行業的朋友們，給大家帶來一手經驗！ 我最近剛剛斬獲了雅诗兰黛（Estée Lauder Companies – ELC）數據分析師（Data Analyst， DA）崗位的 Offer，趁著記憶還熱乎，想和大家聊聊我的技術面經歷，也順便分享一些小技巧。 希望能幫到正在準備的你~
大家都知道，ELC 作為全球頂級的奢侈美妝集團，DA 崗位的競爭非常激烈。 面試環節也比較全面：除了常規的行為面、HR 面，技術面才是真正的考核重點。 它不僅要求我們有紮實的硬技能，還會通過業務場景題去考察你解決問題的思路和邏輯。
技術面整體感受
我這次的技術面基本就是圍繞 SQL + Python（尤其是 Pandas） 展開的，期間還穿插了一些 數據建模 和 可視化 相關的概念。 面試官真的超級 nice，會一步步引導你，但別以為就能輕鬆過關哦~題目的 廣度和深度 都挺考驗功底的。
給大家的建議是：準備的時候千萬別只停留在“會寫語法/會用函數”這個層面，更要理解 背後的邏輯和原理，比如為什麼要這麼設計、這樣寫的最佳實踐是什麼。 這樣在追問的時候，你才能接得住，不會被問懵。
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 notmatter. Assume product names and query words are separated by spaces. This task is crucial for helping customers find the right products quickly, whetherit’s a specific lipstick shade or a new serum.
Example: products = [“Advanced Night Repair Serum”, “Double Wear Foundation”, “Pure ColorEnvy 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-basedcomparison.
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 notmatter. Assume product names and query words are separated by spaces. This task is crucial for helping customers find the right products quickly, whetherit’s a specific lipstick shade or a new serum.
Example: products = [“Advanced Night Repair Serum”, “Double Wear Foundation”, “Pure ColorEnvy 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-basedcomparison.
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 berepresented 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.)
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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 valuesare 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)}")面試真的就像闖關遊戲，準備充分才能一路順利通關
我的這次經歷也算是一個小小的心得分享~如果你現在也在為 OA 或者技術面發愁，其實不用一個人死撐，Programhelp的遠端助攻完全能幫你輕鬆很多，節省精力，把狀態留給關鍵問題。
我們能做什麼？
OA 無痕連線代寫：實時遠端寫題，保證穩定輸出
語音即時助攻：面試過程中小聲提醒，不卡頓不突兀
Mock Interview：提前陪練，幫你熟悉大廠面試套路
面經資料庫：覆蓋 Amazon、Google、Meta 等大廠的最新真題和解析
我們的目標很簡單：
讓你在真正的面試現場不再慌亂，思路清晰，答題有底氣。
