How do you solve Duplicate Emails on LeetCode?

Duplicate Emails (LeetCode #182) 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 a hash map can significantly reduce time complexity.

What is the brute force solution for Duplicate Emails?

The brute force approach for Duplicate Emails is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each email against every other email to find duplicates. This is straightforward but inefficient, especially as the number of emails increases. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Duplicate Emails?

Duplicate Emails uses the following concepts: Database. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Duplicate Emails?

Always clarify the problem requirements before jumping into coding. Explain your thought process as you write the code. Consider edge cases, such as all unique emails or all duplicates.

What are common mistakes when solving Duplicate Emails?

Not using GROUP BY correctly in SQL. Failing to account for case sensitivity or whitespace in emails.

#182

Duplicate Emails

Easy

DatabaseHash 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)

Using a hash map allows us to count occurrences of each email efficiently. This way, we can find duplicates in a single pass through the data.

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Algorithm

3 steps

1Step 1: Initialize a hash map to count occurrences of each email.
2Step 2: Loop through each email and update the count in the hash map.
3Step 3: After counting, filter the hash map to find emails with a count greater than 1.

solution.py2 lines

1# Full working Python code
2SELECT email FROM (SELECT email, COUNT(email) as count FROM Person GROUP BY email) as counts WHERE count > 1;

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Complexity note: The time complexity is O(n) because we only pass through the list of emails once to count them. The space complexity is O(n) due to the hash map storing counts for each unique email.

1Using a hash map can significantly reduce time complexity.
2Grouping and counting can be efficiently handled with SQL aggregate functions.

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