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

The optimal solution has a time complexity of O(n log n).

What data structures are used to solve Maximum Tastiness of Candy Basket?

This problem uses Array, Binary Search, Greedy, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum Tastiness of Candy Basket?

Insight 1: Sorting the prices allows us to efficiently find candies with a certain minimum difference between their prices. Insight 2: Binary search is a suitable approach to find the maximum tastiness because we can efficiently check if a given tastiness is achievable.

Which companies ask LeetCode 2517. Maximum Tastiness of Candy Basket in interviews?

This problem is frequently asked at PhonePe, eBay.

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Maximum Tastiness of Candy Basket - LeetCode 2517 Solution

Maximum Tastiness of Candy Basket - Complete Solution Guide

Maximum Tastiness of Candy Basket is LeetCode problem 2517, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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(|

Detailed Explanation

The problem asks us to find the maximum possible 'tastiness' of a candy basket containing 'k' distinct candies. The tastiness is defined as the smallest absolute difference between the prices of any two candies in the basket. We are given an array 'price' where each element represents the price of a candy, and an integer 'k' representing the number of candies we must choose for the basket. The goal is to maximize the minimum difference between candy prices in our chosen basket.

Solution Approach

The solution first sorts the prices of the candies. Then, it uses binary search on the possible range of tastiness values to find the maximum achievable tastiness. For each candidate tastiness value, the 'check' function verifies whether we can create a basket of 'k' candies with at least that much tastiness. The 'check' function uses a greedy approach: starting with the first candy, it greedily selects candies whose prices are at least 'tastiness' greater than the last selected candy's price. If it can select 'k' candies this way, then the given tastiness is achievable. Binary search efficiently narrows down the range of possible tastiness values to find the maximum possible value.

Step-by-Step Algorithm

Step 1: Sort the `price` array in ascending order. This allows us to easily find candies that are at least a certain tastiness apart.
Step 2: Define the search space for binary search. The lowest possible tastiness is 0, and the highest is the difference between the most expensive and the least expensive candy (price[-1] - price[0]).
Step 3: Perform binary search on the tastiness range [low, high].
Step 4: In each iteration of the binary search, calculate the middle value `mid = low + (high - low) // 2`. This `mid` value represents a potential tastiness.
Step 5: Call the `check` function with `mid` to determine if a basket of `k` candies can be formed with a minimum tastiness of `mid`.
Step 6: If `check(mid)` returns `true`, it means that a tastiness of `mid` or greater is possible. Update the answer (`ans = mid`) and move the `low` pointer to `mid + 1` to search for even higher tastiness values.
Step 7: If `check(mid)` returns `false`, it means that a tastiness of `mid` is not possible. Move the `high` pointer to `mid - 1` to search for lower tastiness values.
Step 8: Repeat steps 4-7 until `low > high`. The final value of `ans` will be the maximum possible tastiness.

Key Insights

Insight 1: Sorting the prices allows us to efficiently find candies with a certain minimum difference between their prices.
Insight 2: Binary search is a suitable approach to find the maximum tastiness because we can efficiently check if a given tastiness is achievable.
Insight 3: A greedy approach can be used to check if a given tastiness is achievable. Start with the lowest priced candy, then greedily pick candies as far apart as the given tastiness.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search, Greedy, Sorting.

Companies

Asked at: PhonePe, eBay.

Maximum Tastiness of Candy Basket | LeetCode 2517

Medium

Array Binary Search Greedy Sorting

Time: O(n log n)Space: O(1)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int maximumTastiness(int[] price, int k) {
        Arrays.sort(price);

        int low = 0;
        int high = price[price.length - 1] - price[0];
        int ans = 0;

        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (check(price, k, mid)) {
                ans = mid;
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }

        return ans;
    }

    private boolean check(int[] price, int k, int tastiness) {
        int count = 1;
        int last_price = price[0];
        for (int i = 1; i < price.length; i++) {
            if (price[i] >= last_price + tastiness) {
                count += 1;
                last_price = price[i];
            }
        }
        return count >= k;
    }
}

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Problem Info

Description

Problem 2517: Maximum Tastiness of Candy Basket - Detailed explanation and solution approaches.

Topics

Array Binary Search Greedy Sorting

Companies

PhonePeeBay

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2517

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Sort the `price` array in ascending order. This allows us to easily find candies that are at least a certain tastiness apart.
2
Step 2: Define the search space for binary search. The lowest possible tastiness is 0, and the highest is the difference between the most expensive and the least expensive candy (price[-1] - price[0]).
3
Step 3: Perform binary search on the tastiness range [low, high].
4
Step 4: In each iteration of the binary search, calculate the middle value `mid = low + (high - low) // 2`. This `mid` value represents a potential tastiness.
5
Step 5: Call the `check` function with `mid` to determine if a basket of `k` candies can be formed with a minimum tastiness of `mid`.
6
Step 6: If `check(mid)` returns `true`, it means that a tastiness of `mid` or greater is possible. Update the answer (`ans = mid`) and move the `low` pointer to `mid + 1` to search for even higher tastiness values.
7
Step 7: If `check(mid)` returns `false`, it means that a tastiness of `mid` is not possible. Move the `high` pointer to `mid - 1` to search for lower tastiness values.
8
Step 8: Repeat steps 4-7 until `low > high`. The final value of `ans` will be the maximum possible tastiness.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

Sorting the array dominates the time complexity. Binary search performs a logarithmic number of checks, and the `check` function iterates through the sorted array once in each iteration. The additional space required is constant.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Generate all possible combinations of 'k' candies and calculate the tastiness for each combination. Then, find the maximum tastiness among all combinations.

Time: O(C(n, k) * k^2), where n is the number of candies and k is the number of candies in the basket. C(n, k) is the binomial coefficient representing the number of combinations of k items from n items. k^2 is the complexity to compute the min difference.Space: O(k), for storing the selected basket.

Trade-offs: This is extremely inefficient for larger values of 'n' and 'k' due to the exponential number of combinations. It is not practical for the given problem constraints.

✨

Key Insights

1Insight 1: Sorting the prices allows us to efficiently find candies with a certain minimum difference between their prices.
2Insight 2: Binary search is a suitable approach to find the maximum tastiness because we can efficiently check if a given tastiness is achievable.
3Insight 3: A greedy approach can be used to check if a given tastiness is achievable. Start with the lowest priced candy, then greedily pick candies as far apart as the given tastiness.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: `k = 2`. The code correctly handles the case where we need to pick just two candies. The check function still works and the binary search will yield the largest difference between any two candies in the sorted array that allows picking k candies.
Edge case 2: All prices are the same (e.g., `price = [7, 7, 7, 7]`). In this case, the tastiness will be 0, which the code will correctly identify because the `check` function will always return true for a tastiness of 0, and the binary search will converge to 0.
Edge case 3: k = price.length. The algorithm still works correctly, essentially finding the smallest difference between adjacent sorted prices that would allow picking all k candies.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the prices first. Without sorting, it's difficult to efficiently check if a certain tastiness can be achieved.

✗Common mistake 2: Using the wrong range for binary search. The range should be from 0 to the difference between the maximum and minimum prices, inclusive. Using the wrong range may lead to incorrect results.

✗Common mistake 3: Not handling the case when the same candy can exist with duplicate prices. The provided solution works because it correctly counts the number of distinct positions available.

✗Common mistake 4: Overflow issues when calculating mid = (low + high) / 2. The code uses mid = low + (high - low) // 2 to avoid potential overflow problems when low and high are very large.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation. Imagine you're allocating servers with different processing power (prices) to 'k' different tasks. You want to ensure that the difference in processing power between any two servers allocated to tasks is maximized. This is similar to maximizing the tastiness of the candy basket.

🏦Real-world scenario 2: Selecting stocks. Suppose you want to select 'k' stocks from a list of stocks with different prices. You want to maximize the minimum difference in price between any two stocks in your selection. This could be done to diversify the stocks in your portfolio and reduce the risk of having stocks with similar price fluctuations.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 25, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Maximum Tastiness of Candy Basket"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?