How do you solve Subdomain Visit Count on LeetCode?

Subdomain Visit Count (LeetCode #811) 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 subdomains are structured helps in generating them efficiently.

What is the time complexity of Subdomain Visit Count?

The optimal solution for Subdomain Visit Count runs in O(n) time and uses O(n) space. This complexity is due to the single pass through the input and the efficient counting using a hash map.

What is the brute force solution for Subdomain Visit Count?

The brute force approach for Subdomain Visit Count is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves iterating through each count-paired domain and generating all possible subdomains. This is straightforward but inefficient as it checks every subdomain for every input. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Subdomain Visit Count?

Subdomain Visit Count uses the following concepts: Array, Hash Table, String, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Subdomain Visit Count?

Always clarify the problem requirements and constraints before jumping into coding. Think about edge cases, such as domains with no subdomains or multiple entries for the same domain. Discuss your thought process with the interviewer to demonstrate your understanding.

What are common mistakes when solving Subdomain Visit Count?

Not considering all subdomains when generating counts. Forgetting to convert counts back to string format when preparing the output.

#811

Subdomain Visit Count

Medium

ArrayHash TableStringCountingHash 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 approach uses a hash map to efficiently count visits for each subdomain while iterating through the input only once. This reduces the time complexity significantly.

⚙️

Algorithm

4 steps

1Step 1: Initialize a hash map to store the visit counts for each subdomain.
2Step 2: For each count-paired domain, split it into count and domain.
3Step 3: Generate subdomains and update their counts in the hash map.
4Step 4: Convert the hash map into the required output format.

solution.py13 lines

1# Full working Python code
2from collections import defaultdict
3
4def subdomainVisits(cpdomains):
5    count_map = defaultdict(int)
6    for cp in cpdomains:
7        count, domain = cp.split()
8        count = int(count)
9        parts = domain.split('.');
10        for i in range(len(parts)):
11            subdomain = '.'.join(parts[i:])
12            count_map[subdomain] += count
13    return [f'{count} {subdomain}' for subdomain, count in count_map.items()]

ℹ

Complexity note: This complexity is due to the single pass through the input and the efficient counting using a hash map.

1Understanding how subdomains are structured helps in generating them efficiently.
2Using a hash map allows for quick updates and retrievals of counts.

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