How do you solve Number of Prefix Connected Groups on LeetCode?

Number of Prefix Connected Groups (LeetCode #3839) 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: Prefix grouping reduces complexity.

What is the time complexity of Number of Prefix Connected Groups?

The optimal solution for Number of Prefix Connected Groups runs in O(n) time and uses O(n) space. We traverse the list once and use a hash map for counting, leading to linear time complexity.

What is the brute force solution for Number of Prefix Connected Groups?

The brute force approach for Number of Prefix Connected Groups is Brute Force, with O(n²) time complexity and O(1) space complexity. This approach checks every pair of words to see if they share the same prefix of length k. It's straightforward but inefficient for large inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Prefix Connected Groups?

Number of Prefix Connected Groups uses the following concepts: Array, Hash Table, String, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Number of Prefix Connected Groups?

Explain your thought process clearly. Discuss edge cases upfront. Practice coding without an IDE.

What are common mistakes when solving Number of Prefix Connected Groups?

Ignoring words shorter than k. Not counting duplicates correctly.

#3839

Number of Prefix Connected Groups

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)

Utilizing a hash map to group words by their prefixes allows us to quickly count connected groups without redundant comparisons.

⚙️

Algorithm

3 steps

1Step 1: Filter out words shorter than k.
2Step 2: Use a hash map to count occurrences of each prefix of length k.
3Step 3: Count the number of prefixes with two or more words.

solution.py7 lines

1def countConnectedGroups(words, k):
2    prefix_count = {}
3    for word in words:
4        if len(word) >= k:
5            prefix = word[:k]
6            prefix_count[prefix] = prefix_count.get(prefix, 0) + 1
7    return sum(1 for count in prefix_count.values() if count > 1)

ℹ

Complexity note: We traverse the list once and use a hash map for counting, leading to linear time complexity.

1Prefix grouping reduces complexity.
2Filtering short words optimizes processing.

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