#1777
Easy Database Products price for each store
Database
82.1% acceptance
Mar 31, 2026
142
13
No description available.
Solution
Pandas
Time O(1)
Space O(1)
# Table: Products
#
# +-------------+---------+
# | Column Name | Type |
# +-------------+---------+
# | product_id | int |
# | store | enum |
# | price | int |
# +-------------+---------+
# In SQL, (product_id, store) is the primary key for this table.
# store is a category of type ('store1', 'store2', 'store3') where each represents the store this product is available at.
# price is the price of the product at this store.
#
#
#
# Find the price of each product in each store.
#
# Return the result table in any order.
#
# The result format is in the following example.
#
# Example 1:
# Input:
# Products table:
# +-------------+--------+-------+
# | product_id | store | price |
# +-------------+--------+-------+
# | 0 | store1 | 95 |
# | 0 | store3 | 105 |
# | 0 | store2 | 100 |
# | 1 | store1 | 70 |
# | 1 | store3 | 80 |
# +-------------+--------+-------+
# Output:
# +-------------+--------+--------+--------+
# | product_id | store1 | store2 | store3 |
# +-------------+--------+--------+--------+
# | 0 | 95 | 100 | 105 |
# | 1 | 70 | null | 80 |
# +-------------+--------+--------+--------+
# Explanation:
# Product 0 price's are 95 for store1, 100 for store2 and, 105 for store3.
# Product 1 price's are 70 for store1, 80 for store3 and, it's not sold in store2.
import pandas as pd
def products_price(products: pd.DataFrame) -> pd.DataFrame:
return products.pivot(index='product_id', columns='store', values='price').reset_index().rename_axis(None, axis=1)