LeetCode 575: Distribute Candies Solution in Python – A Step-by-Step Guide

What Is LeetCode 575: Distribute Candies?

In LeetCode 575: Distribute Candies, you’re given an integer array candyType of even length n, representing different types of candies (each integer is a type ID), and your task is to return the maximum number of unique candy types one sibling can have when the candies are split evenly between two siblings (each gets n/2 candies). For example, with candyType = [1,1,2,2,3,3], one sibling can have 3 unique types (1, 2, 3), limited by the n/2 = 3 candies they receive. This problem builds on LeetCode 349: Intersection of Two Arrays for set operations but focuses on maximizing unique elements under a quantity constraint.

Problem Statement

• Input: candyType (List[int])—array of candy type IDs, length n (even).
• Output: Integer—maximum number of unique candy types one sibling can have.
• Rules: Split n candies evenly (n/2 each); maximize unique types for one sibling; each integer is a type ID.

Constraints

• 2 <= candyType.length <= 10⁴
• candyType.length is even
• -10⁵ <= candyType[i] <= 10⁵

Examples

• Input: candyType = [1,1,2,2,3,3]
• Output: 3
• Total = 6, each gets 3; unique types = 3 (1, 2, 3).
• Input: candyType = [1,1,2,3]
• Output: 2
• Total = 4, each gets 2; unique types = 3, limited to 2.
• Input: candyType = [1,1,1,1]
• Output: 1
• Total = 4, each gets 2; unique types = 1.

Understanding the Problem: Sharing Candies

To solve LeetCode 575: Distribute Candies in Python, we need to maximize the number of unique candy types one sibling can have when splitting an even-length array candyType into two equal parts (n/2 candies each). With lengths up to 10⁴, a brute-force approach counting frequencies and distributing candies could work but might overcomplicate things. Instead, a key insight—max unique types is limited by either the total unique types or n/2—leads to an efficient solution using a hash set. We’ll explore:

• Best Solution (Hash Set with Limit Checking): O(n) time, O(n) space—fast and optimal (n = array length).
• Alternative Solution (Brute-Force Frequency Counting): O(n) time, O(n) space—thorough but less direct.

Let’s dive into the hash set solution with a friendly breakdown!

Best Solution: Hash Set with Limit Checking

Why Hash Set Wins

The hash set with limit checking solution is the best for LeetCode 575 because it computes the maximum unique candy types in O(n) time and O(n) space by using a set to count unique types in a single pass, then taking the minimum of the set size and n/2, elegantly handling the constraint without needing complex distribution logic. It’s like tossing candies into a unique-type basket and checking how many fit within half the pile—all in one quick, clever step!

How It Works

Think of this as a candy-type collector:

• Step 1: Calculate Limit:
• Total candies n, each sibling gets n/2.
• Step 2: Count Unique Types:
• Use a set to store unique candy types from candyType.
• Step 3: Determine Maximum:
• Max unique types = min(number of unique types, n/2).
• Step 4: Return Result:
• Integer representing max unique types.
• Why It Works:
• Unique types can’t exceed n/2 (candy limit per sibling).
• Set ensures O(1) insertion and counts uniques efficiently.

It’s like a candy-sharing optimizer!

Step-by-Step Example

Example: candyType = [1,1,2,2,3,3]

• Step 1: n = 6, limit = n/2 = 3.
• Step 2: Unique types: set([1,1,2,2,3,3]) = {1, 2, 3}, size = 3.
• Step 3: Max = min(3, 3) = 3.
• Result: 3.

Example: candyType = [1,1,2,3]

• Step 1: n = 4, limit = 2.
• Step 2: Unique types: {1, 2, 3}, size = 3.
• Step 3: Max = min(3, 2) = 2.
• Result: 2.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

Copy codeclass Solution:
    def distributeCandies(self, candyType: List[int]) -> int:
        # Step 1: Calculate total candies and limit
        n = len(candyType)
        limit = n // 2

        # Step 2: Count unique candy types
        unique_types = len(set(candyType))

        # Step 3: Return maximum unique types possible
        return min(unique_types, limit)

• Lines 4-5: Get total length n, compute limit n/2.
• Line 8: Convert array to set, get size (unique types).
• Line 11: Return min of unique types and limit.
• Time Complexity: O(n)—set construction is linear.
• Space Complexity: O(n)—set storage for unique types.

It’s like a candy-type maximizer!

Alternative Solution: Brute-Force Frequency Counting

Why an Alternative Approach?

The brute-force frequency counting approach uses a dictionary to count occurrences of each candy type, then simulates distributing candies to maximize unique types, running in O(n) time and O(n) space. It’s thorough but less concise, making it a good baseline for understanding or when explicit distribution is desired.

How It Works

Picture this as a candy-frequency sorter:

• Step 1: Count frequencies of each candy type.
• Step 2: Distribute up to n/2 candies, prioritizing unique types.
• Step 3: Count unique types distributed.
• Step 4: Return count.

It’s like a candy-distribution planner!

Step-by-Step Example

Example: candyType = [1,1,2,2,3,3]

• Step 1: Frequencies: {1: 2, 2: 2, 3: 2}, n = 6, limit = 3.
• Step 2: Distribute:
• Take 1 (count = 1).
• Take 2 (count = 2).
• Take 3 (count = 3).
• Step 3: Unique types = 3.
• Result: 3.

Code for Brute-Force Approach

Copy codeclass Solution:
    def distributeCandies(self, candyType: List[int]) -> int:
        # Step 1: Count frequencies
        freq = {}
        for candy in candyType:
            freq[candy] = freq.get(candy, 0) + 1

        # Step 2: Distribute up to n/2 candies
        n = len(candyType)
        limit = n // 2
        unique_count = 0
        candies_used = 0

        for candy, count in freq.items():
            if candies_used < limit:
                unique_count += 1
                candies_used += min(count, limit - candies_used)

        # Step 3: Return unique types
        return unique_count

• Lines 4-6: Build frequency dictionary.
• Lines 9-10: Set limit to n/2.
• Lines 13-16: Distribute candies, increment unique count.
• Line 19: Return unique types.
• Time Complexity: O(n)—single pass for counting.
• Space Complexity: O(n)—dictionary storage.

It’s a frequency-based candy sharer!

Comparing the Two Solutions

• Hash Set (Best):
• Pros: O(n), O(n), concise.
• Cons: Abstract limit logic.
• Brute-Force (Alternative):
• Pros: O(n), O(n), explicit distribution.
• Cons: More code, less elegant.

Hash set wins for simplicity!

Additional Examples and Edge Cases

• [1,1]: 1.
• [1,2,3,4]: 2.
• All same: Limited to 1 type.

Hash set handles them all!

Complexity Recap

• Hash Set: Time O(n), Space O(n).
• Brute-Force: Time O(n), Space O(n).

Hash set’s the elegance champ!

Key Takeaways

• Hash Set: Type counting—learn at Python Basics!
• Brute-Force: Frequency detail.
• Candies: Sharing is fun.
• Python: Set or freq, your pick!

Final Thoughts: Share Those Candies!

LeetCode 575: Distribute Candies in Python is a delightful array challenge. Hash set with limit checking is your fast track, while brute-force offers a thorough dive. Want more? Try LeetCode 349: Intersection of Two Arrays or LeetCode 217: Contains Duplicate. Ready to divvy? Head to Solve LeetCode 575 on LeetCode and maximize those unique candies today!