#2517

Medium

auto_awesomeBinary search over the valid answer space

LeetCode Problem Workspace

Maximum Tastiness of Candy Basket

Maximize the tastiness of a candy basket by choosing k candies from a list of candy prices.

Problem Statement

You are given an array of positive integers, price, where each element represents the price of a candy. You also have a positive integer, k, representing the number of candies you must select to form a basket. The goal is to maximize the tastiness of the basket, which is defined as the smallest absolute price difference between any two selected candies.

Return the maximum possible tastiness for a basket formed with exactly k candies. You can assume that at least k candies are available in the array.

Examples

Example 1

Input: price = [13,5,1,8,21,2], k = 3

Output: 8

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

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

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 Approach

Binary Search on the Answer

Since the tastiness of a basket is determined by the smallest price difference between any two selected candies, we can use binary search to find the maximum possible tastiness. We check for each mid value of tastiness if it's possible to select k candies where all price differences are greater than or equal to mid. Adjusting the binary search bounds allows us to converge on the maximum tastiness.

Greedy Selection of Candies

Once the binary search narrows down the maximum tastiness, we need to use a greedy approach to verify if it's possible to select k candies with the current mid value. This is done by iterating through the sorted list of candy prices and ensuring that the gap between any two selected candies is at least the current mid value.

Sorting for Efficient Checking

To facilitate the greedy selection, the array of candy prices is sorted before attempting to form the basket. Sorting helps by ensuring that we can easily check if the price differences between consecutive candies are large enough to satisfy the binary search criteria.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is primarily determined by the binary search, which runs in O(log(max_price - min_price)), combined with the greedy check, which takes O(n). Sorting the array beforehand contributes an additional O(n log n). Therefore, the overall time complexity is O(n log n + log(max_price - min_price) * n). The space complexity is O(n) due to the storage of the sorted array and auxiliary variables used in the greedy approach.

What Interviewers Usually Probe

Look for the candidate's understanding of binary search over a range of possible answers.
Check if the candidate effectively uses sorting and greedy methods to confirm the validity of a candidate solution.
Ensure the candidate can optimize the approach for both time and space, given the problem constraints.

Common Pitfalls or Variants

Common pitfalls

Failing to sort the array before applying the greedy approach can result in incorrect selections of candies.
Not properly adjusting the binary search bounds based on the feasibility of the current mid value.
Overcomplicating the problem by using brute-force methods instead of the efficient binary search and greedy solution.

Follow-up variants

What if the price array contains duplicate values? The approach remains the same, but you need to be mindful that some baskets may have a tastiness of 0.
What if k is always equal to the array length? The problem reduces to finding the smallest possible price difference between the candies.
How would the problem change if you were allowed to select fewer than k candies? This would require adjusting the feasibility check in the greedy approach.

FAQ

What is the time complexity of the Maximum Tastiness of Candy Basket problem?

The time complexity is O(n log n + log(max_price - min_price) * n), where n is the length of the price array and max_price - min_price is the range of possible tastiness values.

How do I solve the Maximum Tastiness of Candy Basket problem using binary search?

You can solve the problem by performing binary search over the possible tastiness values, checking for each mid value if it's possible to select k candies with that minimum price difference.

Can the tastiness of the basket ever be zero?

Yes, if there are multiple candies with the same price, the tastiness will be zero, as the absolute difference between any two candies is zero.

What should I do if the number of candies, k, is equal to the length of the price array?

In this case, the problem reduces to finding the smallest possible price difference between the candies, as you have to select all of them.

What is the main approach to solve the Maximum Tastiness of Candy Basket problem?

The main approach is binary search over the valid tastiness space, combined with a greedy selection of candies to verify the feasibility of each mid value.

terminal

Solution

Solution 1

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class Solution:
    def maximumTastiness(self, price: List[int], k: int) -> int:
        def check(x: int) -> bool:
            cnt, pre = 0, -x
            for cur in price:
                if cur - pre >= x:
                    pre = cur
                    cnt += 1
            return cnt >= k

        price.sort()
        l, r = 0, price[-1] - price[0]
        while l < r:
            mid = (l + r + 1) >> 1
            if check(mid):
                l = mid
            else:
                r = mid - 1
        return l

Continue Practicing

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