What is the time complexity of Mice and Cheese?

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

What data structures are used to solve Mice and Cheese?

This problem uses Array, Greedy, Sorting, Heap (Priority Queue). Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Mice and Cheese?

Insight 1: The key is to prioritize cheeses where the difference between reward1 and reward2 is the highest. By giving the first mouse these cheeses, we maximize the overall score. Insight 2: Sorting is crucial for identifying the 'k' cheeses with the greatest difference in rewards.

Which companies ask LeetCode 2611. Mice and Cheese in interviews?

This problem is frequently asked at DoorDash.

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Mice and Cheese - LeetCode 2611 Solution

Mice and Cheese - Complete Solution Guide

Mice and Cheese is LeetCode problem 2611, 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

There are two mice and n different types of cheese, each type of cheese should be eaten by exactly one mouse. A point of the cheese with index i ( 0-indexed ) is: reward1[i] if the first mouse eats it. reward2[i] if the second mouse eats it. You are given a positive integer array reward1 , a positive integer array reward2 , and a non-negative integer k . Return the maximum points the mice can achieve if the first mouse eats exactly k types of cheese. Example 1: Input: reward1 = [1,1,3,4], reward

Detailed Explanation

The problem requires us to maximize the total reward obtained by two mice eating 'n' different types of cheese. Each cheese type 'i' has reward1[i] if eaten by the first mouse and reward2[i] if eaten by the second mouse. We are given 'k', the number of cheese types the first mouse must eat. The goal is to return the maximum possible total reward achievable under this constraint.

Solution Approach

The provided solution calculates the difference between reward1 and reward2 for each cheese type. It then sorts these differences in descending order. It begins by assuming the second mouse eats all the cheese, calculating the initial total reward. Finally, it iterates through the 'k' largest differences, adding these differences to the total reward. This represents giving the 'k' cheeses with the biggest reward advantage to the first mouse, thereby maximizing the total score.

Step-by-Step Algorithm

Step 1: Calculate the difference between reward1[i] and reward2[i] for each cheese type 'i'. Store these differences in a new array called 'diffs'.
Step 2: Sort the 'diffs' array in descending order. This places the cheese types with the greatest advantage for the first mouse at the beginning of the array.
Step 3: Initialize 'total_score' to the sum of all rewards in the 'reward2' array. This is the score obtained if the second mouse eats all the cheese.
Step 4: Iterate through the first 'k' elements of the sorted 'diffs' array (or all elements if k > n). Add each of these differences to the 'total_score'. This adjusts the score as the first mouse eats the 'k' cheese types with the greatest reward advantage.
Step 5: Return the final 'total_score'.

Key Insights

Insight 1: The key is to prioritize cheeses where the difference between reward1 and reward2 is the highest. By giving the first mouse these cheeses, we maximize the overall score.
Insight 2: Sorting is crucial for identifying the 'k' cheeses with the greatest difference in rewards.
Insight 3: Starting with the assumption that the second mouse eats all the cheese and then strategically allocating cheeses to the first mouse simplifies the problem.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Greedy, Sorting, Heap (Priority Queue).

Companies

Asked at: DoorDash.

Mice and Cheese | LeetCode 2611

Medium

Array Greedy Sorting Heap (Priority Queue)

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

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int miceAndCheese(int[] reward1, int[] reward2, int k) {
        int n = reward1.length;
        Integer[] diffs = new Integer[n];
        for (int i = 0; i < n; i++) {
            diffs[i] = reward1[i] - reward2[i];
        }
        
        Arrays.sort(diffs, (a, b) -> b - a);
        
        // Start with the total score if the second mouse eats all the cheese.
        int total_score = 0;
        for (int i = 0; i < n; i++) {
            total_score += reward2[i];
        }
        
        // For the k cheeses that the first mouse must eat, we choose the ones
        // that provide the maximum additional points compared to the second mouse.
        // This is equivalent to picking the k largest differences.
        // The change in score for these k cheeses is the sum of their differences.
        if (k > 0) {
            for (int i = 0; i < k; i++) {
                total_score += diffs[i];
            }
        }
            
        return total_score;
    }
}

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Problem Info

Description

Problem 2611: Mice and Cheese - Detailed explanation and solution approaches.

Topics

Array Greedy Sorting Heap (Priority Queue)

Companies

DoorDash

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2611

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

5 steps

1
Step 1: Calculate the difference between reward1[i] and reward2[i] for each cheese type 'i'. Store these differences in a new array called 'diffs'.
2
Step 2: Sort the 'diffs' array in descending order. This places the cheese types with the greatest advantage for the first mouse at the beginning of the array.
3
Step 3: Initialize 'total_score' to the sum of all rewards in the 'reward2' array. This is the score obtained if the second mouse eats all the cheese.
4
Step 4: Iterate through the first 'k' elements of the sorted 'diffs' array (or all elements if k > n). Add each of these differences to the 'total_score'. This adjusts the score as the first mouse eats the 'k' cheese types with the greatest reward advantage.
5
Step 5: Return the final 'total_score'.

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The dominant operation in terms of time complexity is the sorting of the 'diffs' array, which takes O(n log n) time. The other operations, such as calculating differences and summing rewards, take O(n) time. Space complexity is O(n) because we create a new array, 'diffs', of size 'n' to store the differences between rewards. The C code modifies this analysis slightly as it allocates memory explicitly, but the Big-O complexity remains the same.

📖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

Heap (Priority Queue)

Instead of sorting the entire 'diffs' array, we can use a max-heap (priority queue) to find the 'k' largest differences. We maintain a heap of size 'k' and iterate through the 'diffs' array. If a difference is greater than the smallest element in the heap, we replace the smallest element with the current difference. Finally, we sum the elements in the heap.

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

Trade-offs: This approach is more efficient than sorting when 'k' is significantly smaller than 'n' (k << n), as O(n log k) can be less than O(n log n). However, if 'k' is close to 'n', sorting is generally faster.

✨

Key Insights

1Insight 1: The key is to prioritize cheeses where the difference between reward1 and reward2 is the highest. By giving the first mouse these cheeses, we maximize the overall score.
2Insight 2: Sorting is crucial for identifying the 'k' cheeses with the greatest difference in rewards.
3Insight 3: Starting with the assumption that the second mouse eats all the cheese and then strategically allocating cheeses to the first mouse simplifies the problem.

📚

Key Terms Explained

For Students

Array

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

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 = 0. The first mouse eats no cheese. The solution correctly calculates the total score by assuming the second mouse eats all the cheese.
Edge case 2: k = n. The first mouse eats all the cheese. The solution sorts the differences and adds them, effectively calculating the total score if only the first mouse eats any cheese.
Edge case 3: Empty arrays. While not explicitly covered in the constraints, if `reward1` and `reward2` are empty, the code will throw an error since n is used without initialization and dereferencing a null array would cause problems. An explicit size check at the beginning would be beneficial to handle this case gracefully (returning 0, for example).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the 'diffs'. The difference should be calculated as reward1[i] - reward2[i] to correctly reflect the advantage of giving the cheese to the first mouse.

✗Common mistake 2: Not sorting the 'diffs' array or sorting it in the wrong order. Sorting is essential to prioritize the cheeses that provide the greatest reward advantage to the first mouse.

✗Common mistake 3: Incorrectly initializing the 'total_score'. It's important to start with the total reward when the second mouse eats all the cheese and then adjust the score based on the 'k' largest differences.

✅

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 allocating resources to different projects, where each project yields different profits depending on who is assigned to it. The mice and cheese problem is analogous to this scenario, where resources are cheeses, projects are mice, and profits are rewards. Goal: Maximize total profit under allocation constraints.

🏦Real-world scenario 2: Task assignment. Consider assigning tasks to two workers, where each task has a different efficiency rating for each worker. Assigning k tasks to worker 1 such that the total efficiency across both workers is maximized directly reflects the mouse and cheese problem.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: August 31, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

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

@InterviewProJul 7

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

@CodeNinjaJun 30

❓

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.

What are common greedy problem patterns?

Activity selection (sort by end time), interval scheduling (sort by start/end), Huffman coding (always combine smallest), fractional knapsack (sort by value/weight ratio), and minimum spanning tree. Look for problems where local decisions don't affect future options.

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 "Mice and Cheese"?

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 greedy techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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