For all of you who are rushing the beauty industry, here's a handful of experiences! I just recently choppedEstee Lauder (name)(Estée Lauder Companies - ELC) Data Analyst (DA) position Offer, while the memory is still hot, would like to talk to you about my technical interview experience, and also share some tips along the way. I hope it will help you to prepare for the job.
As we all know, ELC, as the top luxury beauty group in the world, the competition for DA positions is very fierce. The interview session is also relatively comprehensive: in addition to the regular behavioral and HR interviews, the technical interview is the real focus of the assessment. It not only requires us to have solid hard skills, but also examines your problem-solving ideas and logic through business scenario questions.
Overall Technical Feeling
My technical interview was basically about SQL + Python (especially Pandas), with some data modeling and visualization related concepts thrown in. The interviewers are really nice and will guide you step by step, but don't think you can get through it easily - the breadth and depth of the questions are quite a test.
My advice to you is: when preparing, don't just focus on the level of "knowing how to write syntax/functions", but understand the logic and principles behind it, such as why it was designed this way, and what are the best practices for writing it this way. This way, you will be able to catch the questions and not be confused.
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) The search should be case-insensitive, and the order of words in the query does not matter. 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 product 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
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. 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.
query (str): The search query string.
Returns: list: A list of matching product names.
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)
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) The search should be case-insensitive, and the order of words in the query does not matter. 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 product 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. 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.
query (str): The search query string.
Returns: list: A list of matching product names.
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)
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 Assume all connections are bidirectional.) 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 '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 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 list of connections that are 'next-tier' in a network.
startLocation (str): The starting location.
Returns: list: A list of unique next-tier locations.
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_connections.
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)}")Interviews are really like a game of chance, you have to be well prepared in order to get all the way through!
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