Problem Description

You have n elements numbered from 0 to n - 1, where each element belongs to a certain category. Your goal is to determine how many unique categories exist among these elements.

To help you solve this problem, you're provided with a CategoryHandler object that has one method:

• haveSameCategory(a, b): This method takes two integers a and b and returns true if both elements belong to the same category, or false if they belong to different categories. The method also returns false if either a or b is invalid (less than 0 or greater than or equal to n).

The solution uses a Union-Find (Disjoint Set Union) data structure to group elements that belong to the same category. Here's how it works:

1. Initialize parent array: Create an array p where each element initially points to itself as its parent (p[i] = i), meaning each element starts in its own category.

2. Find function with path compression: The find(x) function returns the root representative of element x's category. It uses path compression (p[x] = find(p[x])) to optimize future lookups.

3. Group elements by category: For every pair of elements (a, b) where a < b, check if they belong to the same category using haveSameCategory(a, b). If they do, union their sets by making the root of a point to the root of b.

4. Count unique categories: After processing all pairs, count how many elements are their own root (i == p[i]). Each root represents a unique category.

The time complexity is O(n² × α(n)) where α(n) is the inverse Ackermann function (practically constant), since we check all pairs of elements. The space complexity is O(n) for the parent array.

Intuition

The key insight is recognizing that this is fundamentally a grouping problem. We need to identify which elements belong together in the same category, even though we don't know what the categories actually are - we can only check if two elements share the same category.

Think of it like this: imagine you have a collection of colored balls, but you're blindfolded. You can't see the colors, but you have a helper who can tell you whether any two balls you pick have the same color. How would you figure out how many different colors there are?

The natural approach is to compare elements pairwise and group them together when they match. If element a has the same category as element b, and element b has the same category as element c, then all three must be in the same category. This transitivity property suggests using a Union-Find data structure.

Why Union-Find? Because:

1. We start with each element potentially being in its own category (worst case: n categories if all elements are different)
2. As we discover that two elements share a category, we need to merge their groups
3. We need to efficiently track which elements belong to the same group even after multiple merges
4. At the end, we need to count the number of distinct groups

The Union-Find structure elegantly handles all these requirements. By checking all possible pairs (a, b) and unioning them when they share a category, we progressively build up the complete picture of which elements belong together. The final count of root elements (elements that are their own parent) gives us the number of unique categories, since each group has exactly one root representing that entire category.

Learn more about Union Find patterns.

Solution Approach

The solution implements a Union-Find (Disjoint Set Union) data structure with path compression to efficiently group elements by their categories.

Step 1: Initialize the parent array

p = list(range(n))

Create a parent array where p[i] = i for all elements from 0 to n-1. Initially, each element is its own parent, representing that it's in its own separate category.

Step 2: Define the find function with path compression

def find(x: int) -> int:
    if p[x] != x:
        p[x] = find(p[x])
    return p[x]

This recursive function finds the root representative of element x's category. The path compression optimization (p[x] = find(p[x])) directly connects x to its root, flattening the tree structure for faster future lookups.

Step 3: Check all pairs and union categories

for a in range(n):
    for b in range(a + 1, n):
        if categoryHandler.haveSameCategory(a, b):
            p[find(a)] = find(b)

Iterate through all unique pairs (a, b) where a < b. For each pair:

• Call haveSameCategory(a, b) to check if they belong to the same category
• If they do, perform a union operation: p[find(a)] = find(b)
• This makes the root of a's group point to the root of b's group, effectively merging the two groups

Step 4: Count unique categories

return sum(i == x for i, x in enumerate(p))

After processing all pairs, count how many elements satisfy p[i] == i. These are the root elements, each representing a unique category. The total count gives us the number of distinct categories.

The algorithm ensures that all elements belonging to the same category will share the same root in the Union-Find structure, making the final count accurate and efficient.

Example Walkthrough

Let's walk through a small example with n = 5 elements, where the actual categories are:

• Elements 0, 2, 4 belong to category A
• Elements 1, 3 belong to category B

Initial Setup:

Parent array p = [0, 1, 2, 3, 4]
Each element points to itself

Step 1: Check pairs and union same categories

Checking pair (0, 1): haveSameCategory(0, 1) returns false (different categories)

• No union needed
• p = [0, 1, 2, 3, 4]

Checking pair (0, 2): haveSameCategory(0, 2) returns true (both in category A)

• Union: p[find(0)] = find(2) → p[0] = 2
• p = [2, 1, 2, 3, 4]

Checking pair (0, 3): haveSameCategory(0, 3) returns false

• No union needed
• p = [2, 1, 2, 3, 4]

Checking pair (0, 4): haveSameCategory(0, 4) returns true (both in category A)

• find(0) returns 2 (follows 0→2)
• Union: p[find(0)] = find(4) → p[2] = 4
• p = [2, 1, 4, 3, 4]

Checking pair (1, 2): haveSameCategory(1, 2) returns false

• No union needed

Checking pair (1, 3): haveSameCategory(1, 3) returns true (both in category B)

• Union: p[find(1)] = find(3) → p[1] = 3
• p = [2, 3, 4, 3, 4]

Checking pair (1, 4): haveSameCategory(1, 4) returns false

• No union needed

Checking pair (2, 3): haveSameCategory(2, 3) returns false

• No union needed

Checking pair (2, 4): haveSameCategory(2, 4) returns true (both in category A)

• find(2) returns 4 (follows 2→4)
• find(4) returns 4
• Union: p[4] = 4 (already connected)

Checking pair (3, 4): haveSameCategory(3, 4) returns false

• No union needed

Step 2: Count unique categories

Final parent array with path compression applied:

• find(0) = 4 (follows 0→2→4)
• find(1) = 3 (follows 1→3)
• find(2) = 4 (follows 2→4)
• find(3) = 3 (itself)
• find(4) = 4 (itself)

After path compression: p = [4, 3, 4, 3, 4]

Count elements where i == p[i]:

• i=0: 0 ≠ 4 ✗
• i=1: 1 ≠ 3 ✗
• i=2: 2 ≠ 4 ✗
• i=3: 3 = 3 ✓
• i=4: 4 = 4 ✓

Result: 2 unique categories (elements 3 and 4 are roots representing categories B and A respectively)

Solution Implementation

Time and Space Complexity

Time Complexity: O(n² × α(n))

The algorithm uses a Union-Find (Disjoint Set Union) data structure with path compression. Breaking down the time complexity:

• The outer loop runs n times
• The inner loop runs approximately n/2 times on average (from a+1 to n)
• Total number of iterations: n × (n-1) / 2 = O(n²)
• For each iteration, we call haveSameCategory() which takes O(1) time
• When categories match, we perform two find() operations and one union operation
• With path compression, find() operations have an amortized time complexity of O(α(n)), where α is the inverse Ackermann function (practically constant, but technically grows very slowly)
• The final counting loop takes O(n) time

Overall: O(n²) iterations × O(α(n)) per iteration = O(n² × α(n))

For practical purposes, this can be considered O(n²) since α(n) is effectively constant for all reasonable values of n.

Space Complexity: O(n)

The space usage consists of:

• The parent array p which stores n integers: O(n)
• The recursive call stack for find() function: worst case O(n) if the tree becomes a chain before path compression occurs
• No additional data structures are used

Total space complexity: O(n)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Not Finding Roots Before Union Operation

A critical mistake is directly connecting nodes without finding their roots first. This creates incorrect connections and fails to properly merge entire groups.

Incorrect Implementation:

# WRONG: Directly connecting nodes instead of roots
if categoryHandler.haveSameCategory(item_a, item_b):
    parent[item_a] = item_b  # This is incorrect!

Why it fails: If item_a is already part of a larger group with root r1, and item_b is part of another group with root r2, this code only connects item_a to item_b, leaving the rest of item_a's group disconnected from item_b's group.

Correct Implementation:

if categoryHandler.haveSameCategory(item_a, item_b):
    root_a = find_root(item_a)
    root_b = find_root(item_b)
    parent[root_a] = root_b  # Connect the roots, not the nodes

2. Incorrect Final Count Due to Not Updating Parent Array

After all union operations, the parent array might not have all nodes directly pointing to their ultimate root due to path compression only happening during find operations.

Problem: Counting parent[i] == i without ensuring all paths are compressed can give incorrect results.

Solution: Call find_root on all nodes before counting to ensure complete path compression:

# Ensure all paths are compressed
for i in range(n):
    find_root(i)

# Now count the roots
num_categories = sum(1 for i in range(n) if parent[i] == i)

3. Redundant Union Operations

Performing union operations when two elements are already in the same category wastes computation.

Optimization:

if categoryHandler.haveSameCategory(item_a, item_b):
    root_a = find_root(item_a)
    root_b = find_root(item_b)
    if root_a != root_b:  # Only union if they're in different sets
        parent[root_a] = root_b

This check prevents unnecessary writes to the parent array when elements are already grouped together, which can happen frequently as the algorithm progresses.