Mice and Cheese - Practice Coding Problems

Mice and Cheese - Problem

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.

Input & Output

Example 1 — Basic Case

$ Input: reward1 = [1,1,3,4], reward2 = [4,4,1,1], k = 2

› Output: 15

💡 Note: Mouse 1 should eat cheeses 2 and 3 (benefit differences +2 and +3). Mouse 2 gets cheeses 0 and 1. Total: 3 + 4 + 4 + 4 = 15

Example 2 — All to Mouse 1

$ Input: reward1 = [1,1], reward2 = [1,1], k = 2

› Output: 2

💡 Note: Both cheeses have equal benefit for both mice, so mouse 1 eats all. Total: 1 + 1 = 2

Example 3 — Mouse 2 Better

$ Input: reward1 = [1,1,1], reward2 = [5,5,5], k = 1

› Output: 11

💡 Note: Mouse 1 must eat exactly 1 cheese (any gives same result). Mouse 2 gets the other 2. Total: 1 + 5 + 5 = 11

Constraints

1 ≤ reward1.length, reward2.length ≤ 10⁵
1 ≤ reward1[i], reward2[i] ≤ 1000
0 ≤ k ≤ reward1.length

Visualization

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Understanding the Visualization

Input

Two reward arrays and constraint k

Process

Calculate benefit differences and sort

Output

Maximum total points achievable

Key Takeaway

🎯 Key Insight: Assign cheeses to mouse 1 based on maximum benefit difference (reward1[i] - reward2[i])

Asked in

G Google 25 a Amazon 18 f Facebook 15

The key insight is to assign cheeses based on benefit difference (reward1[i] - reward2[i]). Best approach is greedy sorting: sort by benefit difference and give the top k to mouse 1. Time: O(n log n), Space: O(n)

Common Approaches

Approach	Time	Space	Notes
✓ Greedy - Sort by Benefit Difference	O(n log n)	O(n)	Sort cheeses by benefit difference and greedily assign the best k to mouse 1
Brute Force - Try All Combinations	O(C(n,k) × n)	O(k)	Generate all possible combinations of k cheeses for the first mouse

Greedy - Sort by Benefit Difference — Algorithm Steps

Calculate benefit difference for each cheese
Sort indices by benefit difference (descending)
Assign top k cheeses to mouse 1, rest to mouse 2

Visualization

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Step-by-Step Walkthrough

Calculate Differences

Compute reward1[i] - reward2[i] for each cheese

Sort by Difference

Sort cheeses by benefit difference in descending order

Assign Greedily

Give top k cheeses to mouse 1, rest to mouse 2

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct {
    int diff;
    int index;
} Pair;

int compare(const void* a, const void* b) {
    Pair* pa = (Pair*)a;
    Pair* pb = (Pair*)b;
    return pb->diff - pa->diff; // descending order
}

int solution(int* reward1, int* reward2, int n, int k) {
    // Calculate benefit difference for each cheese
    Pair* differences = (Pair*)malloc(n * sizeof(Pair));
    for (int i = 0; i < n; i++) {
        differences[i].diff = reward1[i] - reward2[i];
        differences[i].index = i;
    }
    
    // Sort by benefit difference in descending order
    qsort(differences, n, sizeof(Pair), compare);
    
    // Assign top k cheeses to mouse 1, rest to mouse 2
    int totalPoints = 0;
    int* mouse1Indices = (int*)calloc(n, sizeof(int));
    
    for (int i = 0; i < k; i++) {
        mouse1Indices[differences[i].index] = 1;
    }
    
    for (int i = 0; i < n; i++) {
        if (mouse1Indices[i]) {
            totalPoints += reward1[i];
        } else {
            totalPoints += reward2[i];
        }
    }
    
    free(differences);
    free(mouse1Indices);
    return totalPoints;
}

int main() {
    char line[1000];
    
    // Parse reward1
    fgets(line, sizeof(line), stdin);
    int reward1[100], n1 = 0;
    char* token = strtok(line + 1, ",]");
    while (token) {
        reward1[n1++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    // Parse reward2
    fgets(line, sizeof(line), stdin);
    int reward2[100], n2 = 0;
    token = strtok(line + 1, ",]");
    while (token) {
        reward2[n2++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    // Parse k
    int k;
    scanf("%d", &k);
    
    int result = solution(reward1, reward2, n1, k);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n log n)

Sorting n elements takes O(n log n), other operations are O(n)

⚡ Linearithmic

Space Complexity

O(n)

Array to store indices and their benefit differences

⚡ Linearithmic Space

23.0K Views

Medium Frequency

~15 min Avg. Time

890 Likes

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Greedy - Sort by Benefit Difference — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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