IBM Coding Question – Solved

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To approach this problem, begin by fully understanding what it means for two strings to be anagrams. Anagrams are strings composed of the same characters with the same frequency but potentially in a different order. For example, "duel" and "dule" are anagrams because they both contain the characters 'd', 'u', 'e', and 'l' exactly once. Identifying anagrams involves comparing character compositions, not just string equality.

The main task is to efficiently find all products that are anagrams of each query string. Since there can be many products and queries, a brute force approach—comparing every query against every product by checking if they are anagrams—would be inefficient, potentially O(n*q*m log m), where m is the length of the strings. This would become computationally expensive when the product list and query list are large.

Instead, focus on preprocessing the products to allow fast anagram lookups. A reliable method to detect anagrams is to transform each string into a canonical form, which acts as an identifier for all strings that are anagrams of each other. The most common canonical form is the sorted version of the string’s characters. For example, the canonical form of "duel" and "dule" would be "delu" after sorting the characters. All strings that have the same canonical form belong to the same anagram group.

Step 1: Preprocessing products
- Iterate over each product string.
- For each product, sort its characters to create the canonical form.
- Use a data structure such as a hash map or dictionary where the key is this canonical form and the value is a list of all products that match this canonical form.
- Append the product to the corresponding list in the dictionary.
- After processing all products, sort each list of products alphabetically. This ensures the output for any query is already sorted without needing to sort repeatedly.

This preprocessing step converts the problem into a simple lookup problem. Once you have grouped all products by their sorted character strings, for any query, you just need to compute its canonical form and retrieve the corresponding list of products from your map.

Step 2: Processing queries
- For each query string, similarly compute its canonical form by sorting the characters.
- Use this canonical form as a key to look up the dictionary created during preprocessing.
- If the canonical form exists in the dictionary, return the corresponding sorted list of products.
- If it doesn’t exist, return an empty list since there are no anagram matches.

Step 3: Return the results as a list of lists, where each inner list corresponds to the matched products for the respective query.

This approach is efficient because the expensive operation of sorting strings and grouping anagrams is done once upfront during preprocessing. Each query then benefits from O(m log m) time (for sorting the query string) plus O(1) average time for dictionary lookup, which is much faster than comparing each query against all products.

Key points to consider:
- Sorting characters of strings is critical for grouping anagrams effectively.
- Storing the results in a dictionary keyed by the sorted character string creates instant retrieval of all anagrams.
- Sorting each product group beforehand avoids sorting multiple times when answering queries.
- If the products or queries contain uppercase or mixed case letters, decide whether to normalize case before sorting to treat them uniformly as anagrams or treat case-sensitive anagrams as distinct groups.
- Handle edge cases like empty strings or queries with no matching products gracefully by returning empty lists.

In summary, think of the problem as grouping products by their sorted characters, then quickly mapping queries to these groups using the same transformation. This transforms a potentially costly pairwise comparison problem into a much more scalable dictionary lookup task with upfront preprocessing.
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