How do you solve Monthly Transactions I on LeetCode?

Monthly Transactions I (LeetCode #1193) 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: Understanding how to use SQL aggregation functions can significantly reduce the complexity of queries.

What is the time complexity of Monthly Transactions I?

The optimal solution for Monthly Transactions I runs in O(n) time and uses O(n) space. The complexity is O(n) because we only scan the Transactions table once to compute the aggregates, which is efficient.

What is the brute force solution for Monthly Transactions I?

The brute force approach for Monthly Transactions I is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we can iterate through the transactions for each month and country, counting the total transactions and their amounts. This is straightforward but inefficient as it requires multiple passes through the data. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Monthly Transactions I?

Monthly Transactions I uses the following concepts: Database. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Monthly Transactions I?

Always clarify the requirements and expected output format before writing your query. Practice writing SQL queries that involve aggregation and grouping to build confidence. Be prepared to explain your thought process and the reasoning behind your SQL choices.

What are common mistakes when solving Monthly Transactions I?

Not using the correct aggregation functions, leading to incorrect results. Failing to group by all necessary columns, which can result in incorrect data aggregation.

#1193

Monthly Transactions I

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 SQL's aggregation functions and grouping capabilities to efficiently compute the required values in a single scan of the Transactions table. This avoids the need for nested loops and reduces the overall complexity.

⚙️

Algorithm

3 steps

1Step 1: Use the DATE_FORMAT function to extract the month from trans_date.
2Step 2: Group the results by the formatted month and country.
3Step 3: Use COUNT and SUM functions to calculate the total transactions, approved transactions, and their respective amounts.

solution.py8 lines

1SELECT DATE_FORMAT(trans_date, '%Y-%m') AS month,
2       country,
3       COUNT(*) AS trans_count,
4       SUM(state = 'approved') AS approved_count,
5       SUM(amount) AS trans_total_amount,
6       SUM(CASE WHEN state = 'approved' THEN amount ELSE 0 END) AS approved_total_amount
7FROM Transactions
8GROUP BY month, country;

ℹ

Complexity note: The complexity is O(n) because we only scan the Transactions table once to compute the aggregates, which is efficient.

1Understanding how to use SQL aggregation functions can significantly reduce the complexity of queries.
2Grouping data effectively allows for efficient computation of totals and counts.

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