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Maximum Tastiness of Candy Basket #4 Most Asked

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Topics:
ArraysBinary SearchGreedy Algorithms

You are given an array of positive integers price where price[i] denotes the price of the ith candy and a positive integer k.

The store sells baskets of k distinct candies. The tastiness of a candy basket is the smallest absolute difference of the prices of any two candies in the basket.

Return the maximum tastiness of a candy basket.

Example 1:

Input: price = [13,5,1,8,21,2], k = 3
Output: 8
Explanation: Choose the candies with the prices [13,5,21].
The tastiness of the candy basket is: min(|13 - 5|, |13 - 21|, |5 - 21|) = min(8, 8, 16) = 8.
It can be proven that 8 is the maximum tastiness that can be achieved.

Example 2:

Input: price = [1,3,1], k = 2
Output: 2
Explanation: Choose the candies with the prices [1,3].
The tastiness of the candy basket is: min(|1 - 3|) = min(2) = 2.
It can be proven that 2 is the maximum tastiness that can be achieved.

Example 3:

Input: price = [7,7,7,7], k = 2
Output: 0
Explanation: Choosing any two distinct candies from the candies we have will result in a tastiness of 0.

Constraints:

  • 2 <= k <= price.length <= 105
  • 1 <= price[i] <= 109

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:

  1. What are the constraints on the tastiness values of the candies? Can they be negative, zero, or floating-point numbers?
  2. What is the maximum number of candies allowed in the basket? Should I be concerned about integer overflow?
  3. If it's impossible to achieve a tastiness level of at least k, what should the function return?
  4. Are there any constraints on k, the minimum tastiness level?
  5. Are duplicate tastiness values allowed in the `tastiness` array, and if so, how should I handle them when determining the maximum possible minimum tastiness?

Brute Force Solution

Approach

The goal is to find the best possible candy basket by trying every single combination of candies. We'll consider all possible baskets and choose the one with the highest 'tastiness' score.

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

  1. First, consider a basket with only the first candy. Calculate its tastiness.
  2. Next, consider a basket with only the second candy. Calculate its tastiness.
  3. Do this for every single candy individually.
  4. Now, consider all possible pairs of candies. Calculate the tastiness of each pair.
  5. Then, consider all possible groups of three candies. Calculate the tastiness of each group.
  6. Continue this process, adding one more candy to the group each time, until you've considered every possible group of candies, up to the entire collection of candies in one giant basket.
  7. Finally, compare the tastiness scores of all the candy baskets you considered, and choose the basket with the highest score.

Code Implementation

def maximum_tastiness_brute_force(candies):
    number_of_candies = len(candies)
    maximum_tastiness = 0

    # Iterate through all possible subset sizes
    for subset_size in range(1, number_of_candies + 1):
        # Generate all combinations of candies of the current size
        combinations = get_combinations(candies, subset_size)

        # Calculate the tastiness of each combination and update the maximum
        for combination in combinations:
            current_tastiness = sum(combination)
            maximum_tastiness = max(maximum_tastiness, current_tastiness)

    return maximum_tastiness

def get_combinations(items, subset_size):
    if subset_size == 0:
        return [[]]

    if not items:
        return []

    combinations = []
    first_item = items[0]
    rest_of_items = items[1:]

    # Include the first item in the subset
    combinations_with_first = get_combinations(rest_of_items, subset_size - 1)
    for combination in combinations_with_first:
        combinations.append([first_item] + combination)

    # Exclude the first item from the subset
    combinations_without_first = get_combinations(rest_of_items, subset_size)
    combinations.extend(combinations_without_first)

    return combinations

# Example Usage
# candies = [1, 2, 3, 4, 5]
# result = maximum_tastiness_brute_force(candies)
# print(f"Maximum Tastiness: {result}")

Big(O) Analysis

Time Complexity
O(2^n)The provided approach considers all possible subsets of the candy collection. For a collection of n candies, there are 2^n possible subsets (including the empty set). Calculating the tastiness of each subset takes varying time but is at least O(1). Since we must iterate through all 2^n possible subsets to find the maximum tastiness, the overall time complexity is O(2^n). This arises because we examine every possible combination of candies, from single candies to the full set of candies.
Space Complexity
O(1)The provided solution iterates through all possible combinations of candies but doesn't explicitly store these combinations in auxiliary data structures. It seems to only maintain a variable to track the maximum tastiness encountered so far and potentially some temporary variables for calculating the tastiness of the current combination. Therefore, the auxiliary space used is constant and independent of the number of candies, N.

Optimal Solution

Approach

The problem asks to pick candies such that the smallest tastiness difference between any two chosen candies is as big as possible. The clever approach is to first sort the candies and then use a process to find the maximum possible minimum difference without testing all the combinations.

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

  1. First, arrange the candies in order of their tastiness, from least to most tasty.
  2. Imagine a range of possible minimum tastiness differences. Use a process that efficiently checks if a given minimum difference is achievable, such as binary search.
  3. For a given minimum difference, start picking candies. Always pick the candy with the lowest tastiness that still allows for the minimum difference to be maintained with the candies already chosen.
  4. If you can pick all the required number of candies with the selected minimum difference, then that difference is possible. Try a larger difference.
  5. If you cannot pick all the required number of candies, then that difference is not possible. Try a smaller difference.
  6. Repeat this process of adjusting the difference and checking if it's possible until you find the largest possible minimum tastiness difference.

Code Implementation

def maximum_tastiness(tastiness, k):
    tastiness.sort()
    left_value = 0
    right_value = tastiness[-1] - tastiness[0]
    max_tastiness = 0

    while left_value <= right_value:
        mid_value = (left_value + right_value) // 2

        # Check if it's possible to pick candies with this min tastiness.
        if can_pick(tastiness, k, mid_value):
            # If possible, try a larger tastiness.
            max_tastiness = mid_value
            left_value = mid_value + 1
        else:
            # If not possible, try a smaller tastiness.
            right_value = mid_value - 1

    return max_tastiness

def can_pick(tastiness, k, min_tastiness):
    count = 1
    previous_candy = tastiness[0]

    # Greedily pick candies.
    for i in range(1, len(tastiness)):
        if tastiness[i] - previous_candy >= min_tastiness:

            # Only increment when candy is picked.
            count += 1
            previous_candy = tastiness[i]

        if count == k:
            return True

    return False

Big(O) Analysis

Time Complexity
O(n log n)Sorting the candies initially takes O(n log n) time. The binary search for the maximum tastiness difference iterates approximately log(max_tastiness - min_tastiness) times, which we can consider log n in the worst case, where n represents the range of tastiness values. Inside the binary search, the isPossible function iterates through the sorted candies array once, taking O(n) time. Therefore, the binary search contributes O(n log n) to the overall time complexity. Since O(n log n) + O(n log n) simplifies to O(n log n), the overall time complexity is O(n log n).
Space Complexity
O(1)The algorithm sorts the candies in place, which doesn't use extra memory proportional to the input size. Binary search uses a few variables to track the search range, which take constant space. The 'picking candies' process uses a few more variables to keep track of the last candy picked and the count of candies, and these also take constant space. Thus, the auxiliary space used is independent of the number of candies, N.

Edge Cases

Empty candies array
How to Handle:
Return 0 as no tastiness is possible with no candies.
candies array with only one element
How to Handle:
Return 0 as a basket needs at least two candies.
All candies have the same tastiness value
How to Handle:
The algorithm should still correctly identify the two candies with the maximum possible tastiness within the limit.
Large number of candies, potentially exceeding memory limits.
How to Handle:
The solution should be optimized for space complexity, possibly using a streaming or iterative approach if the entire candies array cannot fit in memory.
tastiness values that cause integer overflow when summed
How to Handle:
Use a data type with a larger range (e.g., long) to store the sum of tastiness values to prevent overflow.
k (the maximum number of candies) is zero or negative
How to Handle:
Return 0 as no candies can be selected, leading to zero tastiness.
k is larger than the number of candies available
How to Handle:
Adjust k to be the number of candies available to avoid out-of-bounds errors.
Candies with zero or negative tastiness values.
How to Handle:
The algorithm should handle these correctly; negative values will reduce the overall tastiness, and zero values will not contribute.
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