What is the time complexity of Customer Who Visited but Did Not Make Any Transactions?

The optimal solution for Customer Who Visited but Did Not Make Any Transactions runs in O(n + m) time and uses O(n) space. This complexity is linear because we are iterating through both the Visits and Transactions tables once, leading to a combined time complexity of O(n + m).

What is the brute force solution for Customer Who Visited but Did Not Make Any Transactions?

The brute force approach for Customer Who Visited but Did Not Make Any Transactions is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will check each visit to see if there are any corresponding transactions. If there are none, we will count that visit for the customer. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n + m).

What algorithm or data structure is used to solve Customer Who Visited but Did Not Make Any Transactions?

Customer Who Visited but Did Not Make Any Transactions uses the following concepts: Database. The recommended patterns to study are: Hash Map, Aggregation, JOIN operations.

What are the interview tips for Customer Who Visited but Did Not Make Any Transactions?

Always explain your thought process clearly while writing queries. Practice writing SQL queries without an IDE to improve your fluency. Be prepared to discuss the trade-offs of different approaches.

What are common mistakes when solving Customer Who Visited but Did Not Make Any Transactions?

Not considering customers with no visits at all. Overlooking NULL values when counting transactions.

#1581

Customer Who Visited but Did Not Make Any Transactions

Easy

DatabaseHash MapAggregationJOIN operations

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n + m)Space O(n)

By using a HashMap to track transactions, we can efficiently determine which customers have made no transactions in a single pass through both tables.

⚙️

Algorithm

4 steps

1Step 1: Create a HashMap to count the number of visits for each customer.
2Step 2: Iterate through the Visits table and populate the HashMap with customer IDs and their visit counts.
3Step 3: Iterate through the Transactions table and reduce the count for each customer who has made a transaction.
4Step 4: Filter the HashMap to include only those customers with a count greater than zero, indicating no transactions.

solution.py16 lines

1# Full working Python code
2WITH VisitCounts AS (
3    SELECT customer_id, COUNT(*) AS visit_count
4    FROM Visits
5    GROUP BY customer_id
6),
7TransactionCounts AS (
8    SELECT v.customer_id, COUNT(*) AS trans_count
9    FROM Transactions t
10    JOIN Visits v ON t.visit_id = v.visit_id
11    GROUP BY v.customer_id
12)
13SELECT vc.customer_id, vc.visit_count AS count_no_trans
14FROM VisitCounts vc
15LEFT JOIN TransactionCounts tc ON vc.customer_id = tc.customer_id
16WHERE tc.trans_count IS NULL;

ℹ

Complexity note: This complexity is linear because we are iterating through both the Visits and Transactions tables once, leading to a combined time complexity of O(n + m).

1Understanding how to efficiently join and aggregate data is crucial in SQL.
2Using EXISTS or JOIN can significantly improve performance over nested loops.

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