Leetcode #2701: Consecutive Transactions with Increasing Amounts
In this guide, we solve Leetcode #2701 Consecutive Transactions with Increasing Amounts in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Table: Transactions +------------------+------+ | Column Name | Type | +------------------+------+ | transaction_id | int | | customer_id | int | | transaction_date | date | | amount | int | +------------------+------+ transaction_id is the primary key of this table. Each row contains information about transactions that includes unique (customer_id, transaction_date) along with the corresponding customer_id and amount.
Quick Facts
- Difficulty: Hard
- Premium: Yes
- Tags: Database
Intuition
The task is relational in nature, which maps cleanly to DataFrame operations in Python.
By treating tables as DataFrames, joins and group-bys become concise and readable.
Approach
Load the inputs as DataFrames and apply the appropriate merge, filter, or group-by.
Select or rename the columns to match the required output.
Steps:
- Load inputs as DataFrames.
- Apply merge/groupby/filter operations.
- Select the output columns.
Example
+------------------+------+
| Column Name | Type |
+------------------+------+
| transaction_id | int |
| customer_id | int |
| transaction_date | date |
| amount | int |
+------------------+------+
transaction_id is the primary key of this table.
Each row contains information about transactions that includes unique (customer_id, transaction_date) along with the corresponding customer_id and amount.
Python Solution
import duckdb
import pandas as pd
def solution(transactions: pd.DataFrame) -> pd.DataFrame:
con = duckdb.connect()
con.register("Transactions", transactions)
return con.execute("""WITH
T AS (
SELECT
t1.*,
SUM(
CASE
WHEN t2.customer_id IS NULL THEN 1
ELSE 0
END
) OVER (ORDER BY customer_id, transaction_date) AS s
FROM
Transactions AS t1
LEFT JOIN Transactions AS t2
ON t1.customer_id = t2.customer_id
AND t1.amount > t2.amount
AND DATEDIFF(t1.transaction_date, t2.transaction_date) = 1
)
SELECT
customer_id,
MIN(transaction_date) AS consecutive_start,
MAX(transaction_date) AS consecutive_end
FROM T
GROUP BY customer_id, s
HAVING COUNT(1) >= 3
ORDER BY customer_id;""").df()
Complexity
The time complexity is O(n log n) (typical). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.