What is the time complexity of Reducing Dishes?

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

What data structures are used to solve Reducing Dishes?

This problem uses Array, Dynamic Programming, Greedy, Sorting. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Reducing Dishes?

Insight 1: Sorting the satisfaction values is crucial. Preparing the most satisfying dishes later in the process contributes more to the overall 'like-time coefficient' sum because they are multiplied by a larger 'time' value. Insight 2: The problem can be solved using a greedy approach. After sorting, we iterate backwards, adding dishes as long as the cumulative sum of satisfaction values is positive. If adding a dish makes the suffix sum non-positive, it won't contribute to the overall maximum.

Which companies ask LeetCode 1402. Reducing Dishes in interviews?

This problem is frequently asked at Sony.

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Reducing Dishes - LeetCode 1402 Solution

Reducing Dishes - Complete Solution Guide

Reducing Dishes is LeetCode problem 1402, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

A chef has collected data on the satisfaction level of his n dishes. Chef can cook any dish in 1 unit of time. Like-time coefficient of a dish is defined as the time taken to cook that dish including previous dishes multiplied by its satisfaction level i.e. time[i] * satisfaction[i] . Return the maximum sum of like-time coefficient that the chef can obtain after preparing some amount of dishes. Dishes can be prepared in any order and the chef can discard some dishes to get this maximum value. Ex

Detailed Explanation

The problem asks us to maximize the 'like-time coefficient' of dishes a chef prepares. Each dish has a satisfaction value. The like-time coefficient for a dish is the product of its satisfaction and the time it takes to prepare (including all preceding dishes). We can choose to prepare any subset of the dishes and in any order, and the goal is to maximize the sum of like-time coefficients for the chosen dishes. The input is a list of integers representing the satisfaction values of the dishes. The output is a single integer representing the maximum possible like-time coefficient sum.

Solution Approach

The provided solution uses a greedy approach combined with sorting. First, the satisfaction array is sorted in ascending order. Then, it iterates through the sorted array from right to left (from the most satisfying to the least). In each iteration, it maintains a running 'suffix_sum' of the satisfaction values seen so far. If the 'suffix_sum' is positive, it means adding this dish (and all dishes to its right) contributes to a higher total coefficient, so the 'suffix_sum' is added to the 'max_coeff'. If the 'suffix_sum' becomes non-positive, it indicates that adding any further dishes to the left would only decrease the total coefficient, so the loop breaks.

Step-by-Step Algorithm

Step 1: Sort the `satisfaction` array in ascending order.
Step 2: Initialize `max_coeff` and `suffix_sum` to 0.
Step 3: Iterate through the sorted `satisfaction` array from right to left (from the largest element to the smallest).
Step 4: In each iteration, add the current element's satisfaction value to `suffix_sum`.
Step 5: Check if `suffix_sum` is greater than 0. If it is, add `suffix_sum` to `max_coeff`.
Step 6: If `suffix_sum` is not greater than 0 (i.e., less than or equal to 0), break out of the loop because adding further smaller values will never increase the total coefficient.
Step 7: After the loop finishes, return `max_coeff`.

Key Insights

Insight 1: Sorting the satisfaction values is crucial. Preparing the most satisfying dishes later in the process contributes more to the overall 'like-time coefficient' sum because they are multiplied by a larger 'time' value.
Insight 2: The problem can be solved using a greedy approach. After sorting, we iterate backwards, adding dishes as long as the cumulative sum of satisfaction values is positive. If adding a dish makes the suffix sum non-positive, it won't contribute to the overall maximum.
Insight 3: We don't need to consider all possible orderings and subsets explicitly. Sorting and then greedily adding dishes from the end based on the suffix sum guarantees the optimal solution.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Dynamic Programming, Greedy, Sorting.

Companies

Asked at: Sony.

Reducing Dishes | LeetCode 1402

Hard

Array Dynamic Programming Greedy Sorting

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

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int maxSatisfaction(int[] satisfaction) {
        Arrays.sort(satisfaction);
        int n = satisfaction.length;
        int max_coeff = 0;
        int suffix_sum = 0;

        for (int i = n - 1; i >= 0; i--) {
            suffix_sum += satisfaction[i];

            if (suffix_sum <= 0) {
                break;
            }

            max_coeff += suffix_sum;
        }

        return max_coeff;
    }
}

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Problem Info

Description

Problem 1402: Reducing Dishes - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Greedy Sorting

Companies

Sony

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1402

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

7 steps

1
Step 1: Sort the `satisfaction` array in ascending order.
2
Step 2: Initialize `max_coeff` and `suffix_sum` to 0.
3
Step 3: Iterate through the sorted `satisfaction` array from right to left (from the largest element to the smallest).
4
Step 4: In each iteration, add the current element's satisfaction value to `suffix_sum`.
5
Step 5: Check if `suffix_sum` is greater than 0. If it is, add `suffix_sum` to `max_coeff`.
6
Step 6: If `suffix_sum` is not greater than 0 (i.e., less than or equal to 0), break out of the loop because adding further smaller values will never increase the total coefficient.
7
Step 7: After the loop finishes, return `max_coeff`.

💡

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?

The sorting algorithm typically has a time complexity of O(n log n). The loop that iterates through the sorted array takes O(n) time. Since O(n log n) dominates O(n), the overall time complexity is O(n log n). The space complexity is O(1) because we are only using a few variables to store the running sum and the maximum coefficient, and these don't depend on the size of the input.

📖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

Dynamic Programming

A dynamic programming approach could be used, where dp[i][j] represents the maximum like-time coefficient using the first 'i' dishes and selecting 'j' of them. However, this is generally less efficient.

Time: O(n^2), where n is the number of dishes.Space: O(n^2), where n is the number of dishes.

Trade-offs: The dynamic programming approach has higher time and space complexity than the greedy solution. The greedy approach is preferred due to its simplicity and efficiency. DP might be needed if there were additional constraints that invalidated the greedy approach, but here the simpler greedy one works perfectly.

✨

Key Insights

1Insight 1: Sorting the satisfaction values is crucial. Preparing the most satisfying dishes later in the process contributes more to the overall 'like-time coefficient' sum because they are multiplied by a larger 'time' value.
2Insight 2: The problem can be solved using a greedy approach. After sorting, we iterate backwards, adding dishes as long as the cumulative sum of satisfaction values is positive. If adding a dish makes the suffix sum non-positive, it won't contribute to the overall maximum.
3Insight 3: We don't need to consider all possible orderings and subsets explicitly. Sorting and then greedily adding dishes from the end based on the suffix sum guarantees the optimal solution.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

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: All satisfaction values are negative. In this case, the algorithm correctly returns 0 because no dishes should be prepared.
Edge case 2: All satisfaction values are positive. The algorithm correctly prepares all dishes in sorted order, maximizing the like-time coefficient.
Edge case 3: Empty satisfaction array. Although the problem states n >= 1, it's a good habit to consider it. Most implementations would handle an empty array without error and return 0. Adding an explicit check for empty array at the beginning would improve robustness.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the satisfaction values. The order of dishes matters significantly; the most satisfying ones should be prepared later.

✗Common mistake 2: Trying to consider all possible subsets of dishes. This leads to exponential time complexity and is not necessary due to the greedy approach.

✗Common mistake 3: Incorrectly implementing the 'suffix_sum' logic. It's important to iterate backwards and stop the loop when 'suffix_sum' becomes non-positive.

✅

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: Task scheduling with varying rewards. We can think of satisfaction as the reward for completing a task and the time as the order in which tasks are completed. The goal is to maximize the total reward, considering that tasks done later are worth more.

🏦Real-world scenario 2: Investment strategies where different investments have varying returns. The time can represent the stage of investment, and the goal is to maximize the overall return, knowing that later investments have a greater multiplier effect. Ordering the investments (sorting by potential return) and choosing which ones to pursue (greedy selection) can be guided by this algorithm.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ 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: October 21, 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 14

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

@CodeNinjaJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Reducing Dishes"?

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

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