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Number of Unique Categories

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ArraysStrings

You are given a 2D string array items of size n x 2 representing items. The format of the items is [[category_i, name_i]], where category_i represents the category of the i^th item and name_i represents the name of the i^th item.

Return the number of unique categories.

Example 1:

Input: items = [["leetcode","phone"],["leetcode","computer"],["leetcode","tablet"],["amazon","phone"],["amazon","computer"]]
Output: 2
Explanation: There are 2 unique categories which are "leetcode" and "amazon".

Example 2:

Input: items = [["leetcode","phone"],["amazon","phone"],["facebook","phone"],["google","phone"]]
Output: 4
Explanation: There are 4 unique categories which are "leetcode", "amazon", "facebook", and "google".

Constraints:

1 <= items.length <= 10⁵
1 <= category_i.length, name_i.length <= 10
category_i and name_i consist of lowercase English letters.

Try coding it in LeetCode

Number of Unique Categories

Medium

Asked by:

31 views

Topics:

ArraysStrings

Return the number of unique categories.

Example 1:

Input: items = [["leetcode","phone"],["leetcode","computer"],["leetcode","tablet"],["amazon","phone"],["amazon","computer"]]
Output: 2
Explanation: There are 2 unique categories which are "leetcode" and "amazon".

Example 2:

Input: items = [["leetcode","phone"],["amazon","phone"],["facebook","phone"],["google","phone"]]
Output: 4
Explanation: There are 4 unique categories which are "leetcode", "amazon", "facebook", and "google".

Constraints:

1 <= items.length <= 10⁵
1 <= category_i.length, name_i.length <= 10
category_i and name_i consist of lowercase English letters.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the data type of the categories? Can they be strings, integers, or something else?
Can the input list be empty or null? What should I return in those cases?
Are the categories case-sensitive if they are strings?
Does the order of categories in the input list matter? Should the output be sorted in any way?
Are there any constraints on the number of unique categories, or the length of category strings?

Brute Force Solution

Approach

The brute force method for finding unique categories involves checking all possible ways to assign items to categories. We will systematically explore every combination to identify the number of distinct category sets that can be formed.

Here's how the algorithm would work step-by-step:

Consider each item one at a time.
For each item, think about assigning it to one of the existing categories, or creating a new category just for it.
Repeat this process for every item, making sure to consider all possible category assignments at each step.
After assigning all items, check if this particular arrangement of items into categories is already counted.
If it's a new arrangement, add it to our collection of unique category sets.
Repeat the entire process, exploring every possible category assignment combination for all the items.
Once we've exhausted all possibilities, the final count of distinct category sets is our answer.

Code Implementation

def number_of_unique_categories_brute_force(items):
    unique_category_sets = set()

    def generate_category_sets(index, current_categories):
        if index == len(items):
            # Base case: all items assigned
            frozen_category_set = tuple(frozenset(category) for category in current_categories)
            unique_category_sets.add(frozen_category_set)
            return

        # Option 1: Add the current item to an existing category.
        for category_index in range(len(current_categories)):
            new_categories = [category[:] for category in current_categories]
            new_categories[category_index].append(items[index])
            generate_category_sets(index + 1, new_categories)

        # Option 2: Create a new category for the current item.
        # Must occur after existing category assignments
        new_categories = [category[:] for category in current_categories]
        new_categories.append([items[index]])
        generate_category_sets(index + 1, new_categories)

    # Start the recursion with no categories.
    generate_category_sets(0, [])

    # Return the total number of unique category sets found.
    return len(unique_category_sets)

Big(O) Analysis

Time Complexity

O(B(n)) – The provided brute force approach explores all possible category assignments for n items. Each item can be assigned to one of the existing categories or to a new category. This results in a Bell number B(n) of possible partitions of the n items into categories. Therefore, the time complexity is directly proportional to the nth Bell number, which grows faster than exponentially.

Space Complexity

O(B(N)) – The described brute force approach explores every possible category assignment for N items. This inherently requires tracking all possible arrangements of items into categories which can be represented as Bell numbers, B(N), a rapidly growing sequence. Since we need to keep track of arrangements that are already counted, we'd need to store these arrangements in some data structure like a set. The space used to store these unique arrangements would grow proportionally to the number of possible arrangements, B(N). Therefore, the auxiliary space complexity is O(B(N)), where B(N) represents the Nth Bell number.

Optimal Solution

Approach

The goal is to find how many different types of items are in a list. To do this quickly, we avoid counting the same category multiple times by using a technique that remembers which categories we've already seen.

Here's how the algorithm would work step-by-step:

Start with a blank list to keep track of the unique categories we've encountered.
Go through the list of items one by one.
For each item, check if its category is already in our 'unique categories' list.
If the category is not in the list, add it to the list.
If the category is already there, skip it and move to the next item.
Once we've checked every item, count how many categories are in our 'unique categories' list. This is the number of unique categories.

Code Implementation

def count_unique_categories(items):
    unique_categories_list = []

    for item in items:
        # We assume each item has a 'category' attribute.
        category = item.category

        # Avoid duplicates in counting.
        if category not in unique_categories_list:

            unique_categories_list.append(category)

    # Return the total count of unique categories.
    number_of_unique_categories = len(unique_categories_list)
    return number_of_unique_categories

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through a list of n items. For each item, it checks if its category is already present in a set of unique categories. Checking for membership in a set takes O(1) time on average. Therefore, the dominant operation is iterating through the n items, resulting in O(n) time complexity.

Space Complexity

O(N) – The provided solution uses a list to store unique categories encountered while iterating through the input list. In the worst-case scenario, where all items in the input list belong to different categories, this list of unique categories will grow linearly with the number of items (N) in the input list. Therefore, the auxiliary space required is proportional to the number of unique categories, which can be up to N. Thus, the space complexity is O(N).

Edge Cases

Case	How to Handle
Null or empty input list	Return 0, indicating no categories.
List containing only one category	Return 1, as there is only one unique category.
List with a very large number of categories (memory constraints)	Use a memory-efficient data structure like a hash set and consider streaming if the input is too large.
List containing only duplicate categories	The hash set will ensure only one instance of that category is counted, resulting in 1.
List with extremely long category strings	Ensure the underlying string comparison method is efficient and handle potential memory issues.
Categories are case-sensitive (e.g., 'Food' vs. 'food')	Convert all categories to lowercase or uppercase before adding to the set if case-insensitivity is required.
Input list is read-only or immutable	Copy the input list into a mutable list before processing, or work directly with the input list if it supports iteration and checking existence without modification.
Categories containing special characters or unicode	Ensure that the string comparison and hashing functions correctly handle special characters and unicode without errors.

Null or empty input list

How to Handle:

Return 0, indicating no categories.

List containing only one category

How to Handle:

Return 1, as there is only one unique category.

List with a very large number of categories (memory constraints)

How to Handle:

Use a memory-efficient data structure like a hash set and consider streaming if the input is too large.

List containing only duplicate categories

How to Handle:

The hash set will ensure only one instance of that category is counted, resulting in 1.

List with extremely long category strings

How to Handle:

Ensure the underlying string comparison method is efficient and handle potential memory issues.

Categories are case-sensitive (e.g., 'Food' vs. 'food')

How to Handle:

Convert all categories to lowercase or uppercase before adding to the set if case-insensitivity is required.

Input list is read-only or immutable

How to Handle:

Copy the input list into a mutable list before processing, or work directly with the input list if it supports iteration and checking existence without modification.

Categories containing special characters or unicode

How to Handle:

Ensure that the string comparison and hashing functions correctly handle special characters and unicode without errors.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the data type of the categories? Can they be strings, integers, or something else?
Can the input list be empty or null? What should I return in those cases?
Are the categories case-sensitive if they are strings?
Does the order of categories in the input list matter? Should the output be sorted in any way?
Are there any constraints on the number of unique categories, or the length of category strings?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Consider each item one at a time.
For each item, think about assigning it to one of the existing categories, or creating a new category just for it.
Repeat this process for every item, making sure to consider all possible category assignments at each step.
After assigning all items, check if this particular arrangement of items into categories is already counted.
If it's a new arrangement, add it to our collection of unique category sets.
Repeat the entire process, exploring every possible category assignment combination for all the items.
Once we've exhausted all possibilities, the final count of distinct category sets is our answer.

Code Implementation

def number_of_unique_categories_brute_force(items):
    unique_category_sets = set()

    def generate_category_sets(index, current_categories):
        if index == len(items):
            # Base case: all items assigned
            frozen_category_set = tuple(frozenset(category) for category in current_categories)
            unique_category_sets.add(frozen_category_set)
            return

        # Option 1: Add the current item to an existing category.
        for category_index in range(len(current_categories)):
            new_categories = [category[:] for category in current_categories]
            new_categories[category_index].append(items[index])
            generate_category_sets(index + 1, new_categories)

        # Option 2: Create a new category for the current item.
        # Must occur after existing category assignments
        new_categories = [category[:] for category in current_categories]
        new_categories.append([items[index]])
        generate_category_sets(index + 1, new_categories)

    # Start the recursion with no categories.
    generate_category_sets(0, [])

    # Return the total number of unique category sets found.
    return len(unique_category_sets)

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Start with a blank list to keep track of the unique categories we've encountered.
Go through the list of items one by one.
For each item, check if its category is already in our 'unique categories' list.
If the category is not in the list, add it to the list.
If the category is already there, skip it and move to the next item.
Once we've checked every item, count how many categories are in our 'unique categories' list. This is the number of unique categories.

Code Implementation

def count_unique_categories(items):
    unique_categories_list = []

    for item in items:
        # We assume each item has a 'category' attribute.
        category = item.category

        # Avoid duplicates in counting.
        if category not in unique_categories_list:

            unique_categories_list.append(category)

    # Return the total count of unique categories.
    number_of_unique_categories = len(unique_categories_list)
    return number_of_unique_categories

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input list	Return 0, indicating no categories.
List containing only one category	Return 1, as there is only one unique category.
List with a very large number of categories (memory constraints)	Use a memory-efficient data structure like a hash set and consider streaming if the input is too large.
List containing only duplicate categories	The hash set will ensure only one instance of that category is counted, resulting in 1.
List with extremely long category strings	Ensure the underlying string comparison method is efficient and handle potential memory issues.
Categories are case-sensitive (e.g., 'Food' vs. 'food')	Convert all categories to lowercase or uppercase before adding to the set if case-insensitivity is required.
Input list is read-only or immutable	Copy the input list into a mutable list before processing, or work directly with the input list if it supports iteration and checking existence without modification.
Categories containing special characters or unicode	Ensure that the string comparison and hashing functions correctly handle special characters and unicode without errors.

Null or empty input list

How to Handle:

Return 0, indicating no categories.

List containing only one category

How to Handle:

Return 1, as there is only one unique category.

List with a very large number of categories (memory constraints)

How to Handle:

Use a memory-efficient data structure like a hash set and consider streaming if the input is too large.

List containing only duplicate categories

How to Handle:

The hash set will ensure only one instance of that category is counted, resulting in 1.

List with extremely long category strings

How to Handle:

Ensure the underlying string comparison method is efficient and handle potential memory issues.

Categories are case-sensitive (e.g., 'Food' vs. 'food')

How to Handle:

Convert all categories to lowercase or uppercase before adding to the set if case-insensitivity is required.

Input list is read-only or immutable

How to Handle:

Copy the input list into a mutable list before processing, or work directly with the input list if it supports iteration and checking existence without modification.

Categories containing special characters or unicode

How to Handle:

Ensure that the string comparison and hashing functions correctly handle special characters and unicode without errors.