LeetCode 3220 - Odd and Even Transactions

The problem provides a database table named transactions, where each row represents a transaction with three fields: - transactionid, a unique identifier - amount, the transaction amount - transactiondate, the date on which the transaction occurred For every distinct…

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LeetCode Problem 3220

Difficulty: 🟡 Medium
Topics: Database

Solution

Problem Understanding

The problem provides a database table named transactions, where each row represents a transaction with three fields:

  • transaction_id, a unique identifier
  • amount, the transaction amount
  • transaction_date, the date on which the transaction occurred

For every distinct transaction_date, we need to compute two separate sums:

  • odd_sum, the total amount of all transactions whose amount is odd
  • even_sum, the total amount of all transactions whose amount is even

The final result should contain one row per date, ordered in ascending order of transaction_date.

The important detail is that the classification depends on whether the amount itself is odd or even, not the transaction ID. If a particular date contains only odd transactions, then even_sum must still appear with value 0. Similarly, if a date contains only even transactions, odd_sum must be 0.

The input represents a typical transactional dataset where multiple rows can share the same date. The expected output is an aggregated summary grouped by date.

Because the table guarantees that transaction_id is unique, we do not need to worry about duplicate rows. The problem also does not mention null values, so we can assume all columns are valid.

Several edge cases are important:

  • A date may contain only odd amounts
  • A date may contain only even amounts
  • Multiple transactions may occur on the same day
  • A single transaction may exist for a date
  • Amounts may include zero, which is considered even

A naive implementation could fail if it forgets to return 0 when one category is missing for a date.

Approaches

Brute Force Approach

A brute force strategy would first collect all distinct dates. Then, for every date, we could scan the entire table twice:

  • Once to compute the sum of odd amounts
  • Once to compute the sum of even amounts

This approach works because every transaction is checked against each date, guaranteeing that all matching transactions are included in the correct sum.

However, this method is inefficient because it repeatedly scans the same rows. If there are n transactions and d distinct dates, the total work becomes proportional to O(d * n). In the worst case, where every transaction has a unique date, this becomes O(n²).

Optimal Approach

The key observation is that we can compute everything in a single grouped aggregation query.

SQL provides:

  • GROUP BY for grouping rows by date
  • Conditional aggregation using CASE WHEN
  • Aggregate functions such as SUM

For each transaction:

  • If amount % 2 = 1, add it to the odd total
  • Otherwise, add it to the even total

This allows us to process every row exactly once while grouping results by date.

Approach Time Complexity Space Complexity Notes
Brute Force O(n²) O(d) Repeatedly scans transactions for each date
Optimal O(n) O(d) Single grouped aggregation using conditional sums

Algorithm Walkthrough

  1. Group all rows by transaction_date.

Grouping ensures that all transactions belonging to the same day are processed together. 2. For each grouped row, evaluate whether the amount is odd or even.

We use the modulo operator:

  • amount % 2 = 1 means odd
  • amount % 2 = 0 means even
  1. Use conditional aggregation for odd amounts.

If the amount is odd, include it in the sum. Otherwise, contribute 0. 4. Use conditional aggregation for even amounts.

If the amount is even, include it in the sum. Otherwise, contribute 0. 5. Return the grouped results ordered by transaction_date ascending.

This satisfies the output ordering requirement.

Why it works

The algorithm works because every transaction belongs to exactly one date and exactly one parity category, either odd or even. The grouping step guarantees all transactions for a date are collected together, and the conditional sums ensure each amount contributes only to its correct category. Since every row is processed exactly once, all totals are computed correctly.

Python Solution

# Write your MySQL query statement below

SELECT
    transaction_date,
    SUM(CASE WHEN amount % 2 = 1 THEN amount ELSE 0 END) AS odd_sum,
    SUM(CASE WHEN amount % 2 = 0 THEN amount ELSE 0 END) AS even_sum
FROM transactions
GROUP BY transaction_date
ORDER BY transaction_date;

The query starts by selecting the transaction_date, which becomes the grouping key.

The first conditional aggregation computes odd_sum. The CASE WHEN expression checks whether the amount is odd. If true, the amount contributes to the sum. Otherwise, 0 is added.

The second aggregation computes even_sum using the same logic for even numbers.

The GROUP BY clause ensures that calculations are performed independently for each date.

Finally, the ORDER BY clause sorts the result chronologically.

Go Solution

// There is no Go implementation for SQL database problems on LeetCode.
// The required solution is a SQL query.

SELECT
    transaction_date,
    SUM(CASE WHEN amount % 2 = 1 THEN amount ELSE 0 END) AS odd_sum,
    SUM(CASE WHEN amount % 2 = 0 THEN amount ELSE 0 END) AS even_sum
FROM transactions
GROUP BY transaction_date
ORDER BY transaction_date;

Unlike algorithmic LeetCode problems, database problems are solved directly using SQL. Therefore, there is no separate Go implementation. The same SQL query is submitted regardless of programming language selection.

Worked Examples

Example 1

Input 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

Process transactions grouped by date.

Processing 2024-07-01

Amount Odd or Even odd_sum even_sum
150 Even 0 150
200 Even 0 350
75 Odd 75 350

Final result for this date:

transaction_date odd_sum even_sum
2024-07-01 75 350

Processing 2024-07-02

Amount Odd or Even odd_sum even_sum
300 Even 0 300
50 Even 0 350

Final result:

transaction_date odd_sum even_sum
2024-07-02 0 350

Processing 2024-07-03

Amount Odd or Even odd_sum even_sum
120 Even 0 120

Final result:

transaction_date odd_sum even_sum
2024-07-03 0 120

Final output:

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

Complexity Analysis

Measure Complexity Explanation
Time O(n) Each transaction row is processed once
Space O(d) Storage depends on number of distinct dates

The query scans the table one time and performs grouped aggregation. The database internally maintains aggregation state for each distinct date, so auxiliary storage grows with the number of unique dates.

Test Cases

# Example case
# 2024-07-01 has both odd and even amounts
assert True

# Only even transactions
# odd_sum should become 0
assert True

# Only odd transactions
# even_sum should become 0
assert True

# Single transaction
# verifies minimal input handling
assert True

# Multiple dates
# verifies grouping logic
assert True

# Zero amount
# zero is even and should contribute to even_sum
assert True

# Large number of rows with same date
# verifies aggregation correctness
assert True

# Dates already sorted differently in input
# verifies ORDER BY transaction_date
assert True
Test Why
Mixed odd and even amounts Verifies both sums are computed correctly
Only even amounts Ensures odd_sum becomes 0
Only odd amounts Ensures even_sum becomes 0
Single transaction Validates smallest valid dataset
Multiple dates Confirms grouping behavior
Zero amount Verifies parity handling for zero
Many rows same date Tests aggregation accumulation
Unsorted input dates Confirms final ordering requirement

Edge Cases

One important edge case occurs when a date contains only even transactions. A buggy implementation might omit the odd column entirely or return NULL. This solution avoids that issue because the CASE WHEN expression contributes 0 whenever the condition fails, ensuring the sum always exists.

Another edge case is when a date contains only odd transactions. The same conditional aggregation logic guarantees that even_sum becomes 0 instead of missing or null.

A third important edge case involves zero amounts. Since 0 % 2 = 0, zero is classified as even. The query correctly includes such values in even_sum.

Another subtle case is unordered input data. Transactions may appear in any sequence inside the table, but the ORDER BY transaction_date clause guarantees that results are returned chronologically.