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3220. Odd and Even Transactions

Description

Table: transactions

+------------------+------+
| Column Name      | Type | 
+------------------+------+
| transaction_id   | int  |
| amount           | int  |
| transaction_date | date |
+------------------+------+
The transactions_id column uniquely identifies each row in this table.
Each row of this table contains the transaction id, amount and transaction date.

Write a solution to find the sum of amounts for odd and even transactions for each day. If there are no odd or even transactions for a specific date, display as 0.

Return the result table ordered by transaction_date in ascending order.

The result format is in the following example.

 

Example:

Input:

transactions table:

+----------------+--------+------------------+
| transaction_id | amount | transaction_date |
+----------------+--------+------------------+
| 1              | 150    | 2024-07-01       |
| 2              | 200    | 2024-07-01       |
| 3              | 75     | 2024-07-01       |
| 4              | 300    | 2024-07-02       |
| 5              | 50     | 2024-07-02       |
| 6              | 120    | 2024-07-03       |
+----------------+--------+------------------+

Output:

+------------------+---------+----------+
| transaction_date | odd_sum | even_sum |
+------------------+---------+----------+
| 2024-07-01       | 75      | 350      |
| 2024-07-02       | 0       | 350      |
| 2024-07-03       | 0       | 120      |
+------------------+---------+----------+

Explanation:

  • For transaction dates:
    • 2024-07-01:
      • Sum of amounts for odd transactions: 75
      • Sum of amounts for even transactions: 150 + 200 = 350
    • 2024-07-02:
      • Sum of amounts for odd transactions: 0
      • Sum of amounts for even transactions: 300 + 50 = 350
    • 2024-07-03:
      • Sum of amounts for odd transactions: 0
      • Sum of amounts for even transactions: 120

Note: The output table is ordered by transaction_date in ascending order.

Solutions

Solution 1: Grouping and Summing

We can group the data by transaction_date, and then calculate the sum of transaction amounts for odd and even dates separately. Finally, sort by transaction_date in ascending order.

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# Write your MySQL query statement below
SELECT
    transaction_date,
    SUM(IF(amount % 2 = 1, amount, 0)) AS odd_sum,
    SUM(IF(amount % 2 = 0, amount, 0)) AS even_sum
FROM transactions
GROUP BY 1
ORDER BY 1;

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import pandas as pd


def sum_daily_odd_even(transactions: pd.DataFrame) -> pd.DataFrame:
    transactions["odd_sum"] = transactions["amount"].where(
        transactions["amount"] % 2 == 1, 0
    )
    transactions["even_sum"] = transactions["amount"].where(
        transactions["amount"] % 2 == 0, 0
    )

    result = (
        transactions.groupby("transaction_date")
        .agg(odd_sum=("odd_sum", "sum"), even_sum=("even_sum", "sum"))
        .reset_index()
    )

    result = result.sort_values("transaction_date")

    return result

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