How do you solve Find Loyal Customers on LeetCode?

Find Loyal Customers (LeetCode #3657) 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 refund rates is crucial for determining loyalty.

What is the time complexity of Find Loyal Customers?

The optimal solution for Find Loyal Customers runs in O(n) time and uses O(n) space. Single pass through transactions allows for linear time complexity.

What is the brute force solution for Find Loyal Customers?

The brute force approach for Find Loyal Customers is Brute Force, with O(n²) time complexity and O(1) space complexity. Check each customer individually, counting purchases and refunds. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Find Loyal Customers?

Clarify the problem requirements before coding. Discuss your thought process and approach with the interviewer. Test edge cases, such as customers with no transactions.

What are common mistakes when solving Find Loyal Customers?

Not accounting for both purchases and refunds in calculations. Overlooking the date difference for activity duration.

#3657

Find Loyal Customers

Medium

Hash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

Use a single pass to aggregate data for each customer, reducing the need for multiple scans.

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Algorithm

3 steps

1Step 1: Create a summary table for each customer with counts of purchases, refunds, and transaction dates.
2Step 2: Calculate the refund rate and check the activity duration in one pass.
3Step 3: Filter customers based on loyalty criteria and return the result.

solution.py1 lines

1SELECT customer_id FROM (SELECT customer_id, COUNT(CASE WHEN transaction_type = 'purchase' THEN 1 END) AS purchase_count, COUNT(CASE WHEN transaction_type = 'refund' THEN 1 END) AS refund_count, MAX(transaction_date) AS last_date, MIN(transaction_date) AS first_date FROM customer_transactions GROUP BY customer_id) AS stats WHERE purchase_count >= 3 AND refund_count / (purchase_count + refund_count) < 0.2 AND DATEDIFF(last_date, first_date) >= 30;

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Complexity note: Single pass through transactions allows for linear time complexity.

1Understanding refund rates is crucial for determining loyalty.
2Tracking transaction dates helps assess customer activity duration.

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