Number of Prefix Connected Groups - Solution & Explanation

Problem Overview: You receive an array of strings. Two strings belong to the same group if one string is a prefix of another. The task is to count how many distinct prefix-connected groups exist after considering all strings.

Approach 1: Pairwise Prefix Comparison (Brute Force) (Time: O(n^2 * L), Space: O(1))

The direct strategy compares every pair of strings and checks whether one is a prefix of the other. For each pair, iterate character by character until the prefix relationship either matches or fails. If a connection exists, mark them as belonging to the same group using simple visited/group markers. This works for small inputs but becomes slow because every pair of strings is examined. The repeated prefix checks make the runtime quadratic with respect to the number of strings.

Approach 2: Hash Table with Prefix Tracking (Optimal) (Time: O(n * L), Space: O(n))

Use a hash table to store strings that have already been processed. For each word, generate its prefixes from shortest to longest and check if any prefix already exists in the hash set. If a prefix is found, the current string belongs to that existing group. If none of its prefixes exist, this string starts a new group. Insert the full string into the hash table so later strings can connect through it. Since prefix generation costs O(L) per string and hash lookups are O(1) on average, the total complexity becomes O(n * L).

This method leverages the fact that prefix relationships can be detected incrementally while scanning the array once. Instead of comparing every pair, you only examine the prefixes of the current string and perform constant‑time membership checks in the hash structure. The idea is closely related to prefix detection used in string problems and frequency tracking from counting techniques.

Recommended for interviews: Interviewers expect the hash table approach. The brute force solution demonstrates that you understand the prefix relationship definition, but the optimized method shows you can eliminate redundant comparisons using hashing. Recognizing that previously seen prefixes can represent entire groups is the key insight.