Mice and Cheese

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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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

✓ Brute Force - Try All Combinations

⏱️ Time: O(C(n,k) × n) Space: O(k)

Try every possible way to select k cheeses for mouse 1, calculate total points for each combination, and return the maximum. This involves generating all C(n,k) combinations.

Greedy - Sort by Benefit Difference

⏱️ Time: O(n log n) Space: O(n)

Calculate the benefit difference (reward1[i] - reward2[i]) for each cheese. Sort by this difference in descending order and assign the top k cheeses to mouse 1, remaining to mouse 2.

Brute Force - Try All Combinations — Algorithm Steps

Generate all combinations of k cheeses from n total
For each combination, calculate total points
Return maximum points found

Visualization

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Step-by-Step Walkthrough

Generate Combinations

Create all possible ways to choose k cheeses for mouse 1

Calculate Points

For each combination, sum rewards from both mice

Find Maximum

Return the combination giving maximum points

Code -

solution.c — C

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

int popCount(int x) {
    int count = 0;
    while (x) {
        count += x & 1;
        x >>= 1;
    }
    return count;
}

int solution(int* reward1, int* reward2, int n, int k) {
    int maxPoints = 0;
    
    // Generate all combinations using bit manipulation
    for (int mask = 0; mask < (1 << n); mask++) {
        if (popCount(mask) == k) {
            int points = 0;
            for (int i = 0; i < n; i++) {
                if (mask & (1 << i)) {
                    points += reward1[i];
                } else {
                    points += reward2[i];
                }
            }
            if (points > maxPoints) {
                maxPoints = points;
            }
        }
    }
    
    return maxPoints;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[1000];
    
    // Parse reward1
    fgets(line, sizeof(line), stdin);
    int reward1[20], n1 = 0;
    parseArray(line, reward1, &n1);
    
    // Parse reward2
    fgets(line, sizeof(line), stdin);
    int reward2[20], n2 = 0;
    parseArray(line, reward2, &n2);
    
    // 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(C(n,k) × n)

C(n,k) combinations, each taking O(n) time to evaluate

✓ Linear Growth

Space Complexity

O(k)

Space to store current combination of size k

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Try All Combinations — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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