#2230
Easy Database The users that are eligible for discount
Database
50.9% acceptance
Mar 31, 2026
25
45
No description available.
Solution
Pandas
Time O(1)
Space O(1)
# Table: Purchases
#
# +-------------+----------+
# | Column Name | Type |
# +-------------+----------+
# | user_id | int |
# | time_stamp | datetime |
# | amount | int |
# +-------------+----------+
# (user_id, time_stamp) is the primary key (combination of columns with unique values) for this table.
# Each row contains information about the purchase time and the amount paid for the user with ID user_id.
#
#
#
# A user is eligible for a discount if they had a purchase in the inclusive interval of time [startDate, endDate] with at least minAmount amount. To convert the dates to times, both dates should be considered as the start of the day (i.e., endDate = 2022-03-05 should be considered as the time 2022-03-05 00:00:00).
#
# Write a solution to report the IDs of the users that are eligible for a discount.
#
# Return the result table ordered by user_id.
#
# The result format is in the following example.
#
# Example 1:
# Input:
# Purchases table:
# +---------+---------------------+--------+
# | user_id | time_stamp | amount |
# +---------+---------------------+--------+
# | 1 | 2022-04-20 09:03:00 | 4416 |
# | 2 | 2022-03-19 19:24:02 | 678 |
# | 3 | 2022-03-18 12:03:09 | 4523 |
# | 3 | 2022-03-30 09:43:42 | 626 |
# +---------+---------------------+--------+
# startDate = 2022-03-08, endDate = 2022-03-20, minAmount = 1000
# Output:
# +---------+
# | user_id |
# +---------+
# | 3 |
# +---------+
# Explanation:
# Out of the three users, only User 3 is eligible for a discount.
# - User 1 had one purchase with at least minAmount amount, but not within the time interval.
# - User 2 had one purchase within the time interval, but with less than minAmount amount.
# - User 3 is the only user who had a purchase that satisfies both conditions.
import pandas as pd
from datetime import datetime
def find_valid_users(
purchases: pd.DataFrame, start_date: datetime, end_date: datetime, min_amount: int
) -> pd.DataFrame:
purchases["time_stamp"] = pd.to_datetime(purchases["time_stamp"])
start = pd.to_datetime(start_date)
end = pd.to_datetime(end_date)
valid = purchases[
(purchases["time_stamp"] >= start)
& (purchases["time_stamp"] <= end)
& (purchases["amount"] >= min_amount)
]
result = valid[["user_id"]].drop_duplicates().sort_values("user_id")
return result