How do you solve Odd and Even Transactions on LeetCode?

Odd and Even Transactions (LeetCode #3220) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using conditional aggregation can simplify the query and improve performance.

What is the time complexity of Odd and Even Transactions?

The optimal solution for Odd and Even Transactions runs in O(n) time and uses O(n) space. This complexity is linear because we only pass through the transactions table once to compute the sums.

What is the brute force solution for Odd and Even Transactions?

The brute force approach for Odd and Even Transactions is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will iterate through the transactions for each date and separately sum the amounts for odd and even transaction IDs. This is straightforward but inefficient because it requires multiple passes over the data. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Odd and Even Transactions?

Odd and Even Transactions uses the following concepts: Database. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Odd and Even Transactions?

Always think about edge cases, like dates with no transactions. Practice writing SQL queries to get comfortable with syntax and functions. Explain your thought process clearly while writing the query.

What are common mistakes when solving Odd and Even Transactions?

Not using GROUP BY correctly, leading to incorrect aggregations. Forgetting to handle cases with no odd/even transactions.

#3220

Odd and Even Transactions

Medium

DatabaseHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses aggregation functions in SQL to compute the sums directly in one pass over the data. This is efficient because it reduces the number of iterations needed to compute the sums for each date.

⚙️

Algorithm

4 steps

1Step 1: Use a single SQL query that groups transactions by transaction_date.
2Step 2: Use conditional aggregation to calculate the odd and even sums in the same query.
3Step 3: Ensure to handle cases where there are no odd or even transactions by using SUM with CASE statements.
4Step 4: Order the results by transaction_date.

solution.py7 lines

1# Full working Python code
2SELECT transaction_date,
3       COALESCE(SUM(CASE WHEN MOD(transaction_id, 2) = 1 THEN amount END), 0) AS odd_sum,
4       COALESCE(SUM(CASE WHEN MOD(transaction_id, 2) = 0 THEN amount END), 0) AS even_sum
5FROM transactions
6GROUP BY transaction_date
7ORDER BY transaction_date;

ℹ

Complexity note: This complexity is linear because we only pass through the transactions table once to compute the sums.

1Using conditional aggregation can simplify the query and improve performance.
2Handling cases with no transactions is crucial to avoid null results.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.