Problem Description

Alice has n candies where each candy has a type represented by candyType[i]. Since Alice is concerned about her weight, she consulted a doctor who advised her to eat only n/2 of her candies (where n is always even).

Alice wants to maximize the variety of her candy consumption - she wants to eat as many different types of candies as possible while following the doctor's restriction of eating only half of her total candies.

Given an integer array candyType of length n, you need to find the maximum number of different candy types Alice can eat if she consumes exactly n/2 candies.

For example, if Alice has 6 candies of types [1, 1, 2, 2, 3, 3], she can eat 3 candies (which is 6/2). The best strategy would be to eat one candy of each type [1, 2, 3], giving her 3 different types. Even though she has 3 unique types available and is allowed to eat 3 candies, she achieves the maximum variety of 3 different types.

The solution uses a set to count unique candy types. The answer is the minimum between:

• n/2 (the maximum number of candies she can eat)
• The total number of unique candy types available

This ensures Alice eats the maximum variety possible while respecting the doctor's limit.

Intuition

To maximize the variety of candies Alice eats, she should eat as many different types as possible. Since she can only eat n/2 candies total, the optimal strategy is to eat one candy of each type until she either:

1. Runs out of different types to try, or
2. Reaches her limit of n/2 candies

Let's think through the two scenarios:

Scenario 1: Few unique types
If there are fewer unique candy types than n/2, Alice can eat one of each type. For instance, if she has 10 candies but only 3 unique types, and she can eat 5 candies, she'll eat all 3 different types (one of each) and still have 2 slots left. She'll have to repeat some types, but the maximum different types she can eat is still just 3.

Scenario 2: Many unique types
If there are more unique candy types than n/2, Alice is limited by the doctor's restriction. For example, if she has 10 candies with 8 unique types, but can only eat 5 candies, the best she can do is pick 5 different types (one candy each).

This leads us to realize the answer is simply the minimum of:

• The number of unique candy types (found using len(set(candyType)))
• The maximum candies she can eat (n/2 or len(candyType) >> 1)

The min() function naturally handles both scenarios - it gives us the unique types count when that's the limiting factor, or n/2 when the doctor's restriction is the limiting factor.

Solution Approach

The solution uses a hash table (implemented as a set in Python) to efficiently count the unique candy types. Here's how the implementation works:

Step 1: Count unique candy types
We convert the candyType list into a set: set(candyType). This automatically removes all duplicates and gives us only the unique candy types. The time complexity for this operation is O(n) where n is the total number of candies.

Step 2: Calculate the doctor's limit
We need to find n/2, which represents the maximum number of candies Alice can eat. The code uses bit shifting len(candyType) >> 1 which is equivalent to integer division by 2. Right shifting by 1 bit effectively divides the number by 2.

Step 3: Find the minimum
We take the minimum of:

• len(candyType) >> 1 - the maximum candies Alice is allowed to eat
• len(set(candyType)) - the total number of unique candy types available

The min() function returns whichever value is smaller, which represents the maximum variety Alice can achieve.

Why this works:

• If unique types < n/2: Alice can eat one of each type, so the answer is the number of unique types
• If unique types ≥ n/2: Alice is limited by the doctor's restriction, so the answer is n/2

Time Complexity: O(n) for creating the set
Space Complexity: O(n) in worst case when all candies are of different types

Example Walkthrough

Let's walk through a concrete example to illustrate the solution approach.

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

Step 1: Understand the constraints

• Alice has 6 candies total
• She can eat n/2 = 6/2 = 3 candies
• The candy types are: type 1 (appears 2 times), type 2 (appears 1 time), type 3 (appears 3 times)

Step 2: Count unique types

• Convert to set: set([1, 1, 2, 3, 3, 3]) = {1, 2, 3}
• Number of unique types = 3

Step 3: Apply the solution logic

• Maximum candies Alice can eat: 6 >> 1 = 3
• Number of unique candy types: len({1, 2, 3}) = 3
• Answer: min(3, 3) = 3

Why this makes sense: Alice has exactly 3 unique candy types and is allowed to eat 3 candies. The optimal strategy is to eat one candy of each type: one type-1 candy, one type-2 candy, and one type-3 candy. This gives her the maximum variety of 3 different types.

Another example to contrast: candyType = [1, 1, 1, 1, 2, 3]

• Alice can eat 6/2 = 3 candies
• Unique types: {1, 2, 3} = 3 types
• Answer: min(3, 3) = 3

Even though type 1 appears 4 times, Alice still eats one of each type for maximum variety.

Edge case example: candyType = [1, 1, 1, 1, 1, 1]

• Alice can eat 6/2 = 3 candies
• Unique types: {1} = 1 type
• Answer: min(3, 1) = 1

Here, even though Alice can eat 3 candies, she only has 1 type available, so the maximum variety is just 1.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the candyType list. This is because creating a set from the list requires iterating through all n elements once to build the set, which takes O(n) time. The len() operations and the min() comparison are both O(1) operations, and the bit shift operation (>> 1) is also O(1).

The space complexity is O(n) in the worst case. This occurs when all candy types are unique, resulting in a set that contains all n elements from the original list. The set creation requires additional space proportional to the number of unique elements, which can be up to n in the worst case.

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

Common Pitfalls

1. Integer Division Confusion

A common mistake is using regular division / instead of integer division // or bit shifting when calculating n/2. This can lead to float values which cause errors when comparing with integer counts.

Incorrect approach:

max_candies_allowed = len(candyType) / 2  # Returns float (e.g., 3.0)
return min(max_candies_allowed, unique_candy_types)  # Comparing float with int

Correct approach:

max_candies_allowed = len(candyType) // 2  # Integer division
# OR
max_candies_allowed = len(candyType) >> 1  # Bit shift (as in the solution)

2. Attempting to Track Which Specific Candies to Eat

Some developers overcomplicate the problem by trying to determine exactly which candies Alice should eat, creating unnecessary data structures to track selections.

Overcomplicated approach:

def distributeCandies(self, candyType: List[int]) -> int:
    candy_count = {}
    for candy in candyType:
        candy_count[candy] = candy_count.get(candy, 0) + 1
  
    selected = []
    for candy_type in candy_count:
        if len(selected) < len(candyType) // 2:
            selected.append(candy_type)
  
    return len(selected)

Simplified approach (as in the solution):

return min(len(candyType) >> 1, len(set(candyType)))

The problem only asks for the COUNT of different types, not which specific candies to eat.

3. Not Handling Edge Cases

While the problem states n is always even, developers might forget to consider minimal inputs like an empty array or arrays with only one type of candy.

Missing edge case handling:

def distributeCandies(self, candyType: List[int]) -> int:
    unique_types = set(candyType)
    return len(unique_types)  # Forgets the n/2 limit

This would incorrectly return 4 for input [1,2,3,4,4,4,4,4] when the answer should be 4 (since 8/2 = 4).

4. Inefficient Unique Counting

Using nested loops or manual counting instead of leveraging set's built-in duplicate removal.

Inefficient approach:

def distributeCandies(self, candyType: List[int]) -> int:
    unique_types = []
    for candy in candyType:
        if candy not in unique_types:  # O(m) operation for each candy
            unique_types.append(candy)
    return min(len(candyType) // 2, len(unique_types))

This has O(n²) time complexity in worst case, while using set() maintains O(n) complexity.