LeetCode #2517 — MEDIUM

Maximum Tastiness of Candy Basket

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an array of positive integers price where price[i] denotes the price of the i^th 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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Greedy

Example 1

[13,5,1,8,21,2]
3

Example 2

[1,3,1]
2

Example 3

[7,7,7,7]
2

Core Insight

What unlocks the optimal approach

The answer is binary searchable.
For some x, we can use a greedy strategy to check if it is possible to pick k distinct candies with tastiness being at least x.
Sort prices and iterate from left to right. For some price[i] check if the price difference between the last taken candy and price[i] is at least x. If so, add the candy i to the basket.
So, a candy basket with tastiness x can be achieved if the basket size is bigger than or equal to k.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2517: Maximum Tastiness of Candy Basket
class Solution {
    public int maximumTastiness(int[] price, int k) {
        Arrays.sort(price);
        int l = 0, r = price[price.length - 1] - price[0];
        while (l < r) {
            int mid = (l + r + 1) >> 1;
            if (check(price, k, mid)) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        return l;
    }

    private boolean check(int[] price, int k, int x) {
        int cnt = 0, pre = -x;
        for (int cur : price) {
            if (cur - pre >= x) {
                pre = cur;
                ++cnt;
            }
        }
        return cnt >= k;
    }
}

// Accepted solution for LeetCode #2517: Maximum Tastiness of Candy Basket
func maximumTastiness(price []int, k int) int {
	sort.Ints(price)
	return sort.Search(price[len(price)-1], func(x int) bool {
		cnt, pre := 0, -x
		for _, cur := range price {
			if cur-pre >= x {
				pre = cur
				cnt++
			}
		}
		return cnt < k
	}) - 1
}

# Accepted solution for LeetCode #2517: Maximum Tastiness of Candy Basket
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

// Accepted solution for LeetCode #2517: Maximum Tastiness of Candy Basket
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2517: Maximum Tastiness of Candy Basket
// class Solution {
//     public int maximumTastiness(int[] price, int k) {
//         Arrays.sort(price);
//         int l = 0, r = price[price.length - 1] - price[0];
//         while (l < r) {
//             int mid = (l + r + 1) >> 1;
//             if (check(price, k, mid)) {
//                 l = mid;
//             } else {
//                 r = mid - 1;
//             }
//         }
//         return l;
//     }
// 
//     private boolean check(int[] price, int k, int x) {
//         int cnt = 0, pre = -x;
//         for (int cur : price) {
//             if (cur - pre >= x) {
//                 pre = cur;
//                 ++cnt;
//             }
//         }
//         return cnt >= k;
//     }
// }

// Accepted solution for LeetCode #2517: Maximum Tastiness of Candy Basket
function maximumTastiness(price: number[], k: number): number {
    price.sort((a, b) => a - b);
    let l = 0;
    let r = price[price.length - 1] - price[0];
    const check = (x: number): boolean => {
        let [cnt, pre] = [0, -x];
        for (const cur of price) {
            if (cur - pre >= x) {
                pre = cur;
                ++cnt;
            }
        }
        return cnt >= k;
    };
    while (l < r) {
        const mid = (l + r + 1) >> 1;
        if (check(mid)) {
            l = mid;
        } else {
            r = mid - 1;
        }
    }
    return l;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.