LeetCode #2611 — MEDIUM

Mice and Cheese

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

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], reward2 = [4,4,1,1], k = 2
Output: 15
Explanation: In this example, the first mouse eats the 2^nd (0-indexed) and the 3^rd types of cheese, and the second mouse eats the 0^th and the 1^st types of cheese.
The total points are 4 + 4 + 3 + 4 = 15.
It can be proven that 15 is the maximum total points that the mice can achieve.

Example 2:

Input: reward1 = [1,1], reward2 = [1,1], k = 2
Output: 2
Explanation: In this example, the first mouse eats the 0^th (0-indexed) and 1^st types of cheese, and the second mouse does not eat any cheese.
The total points are 1 + 1 = 2.
It can be proven that 2 is the maximum total points that the mice can achieve.

Constraints:

1 <= n == reward1.length == reward2.length <= 10⁵
1 <= reward1[i], reward2[i] <= 1000
0 <= k <= n

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[1,1,3,4]
[4,4,1,1]
2

Example 2

[1,1]
[1,1]
2

Core Insight

What unlocks the optimal approach

The intended solution uses greedy approach.
Imagine at first that the second mouse eats all the cheese, then we should choose k types of cheese with the maximum sum of - reward2[i] + reward1[i].

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2611: Mice and Cheese
class Solution {
    public int miceAndCheese(int[] reward1, int[] reward2, int k) {
        int n = reward1.length;
        Integer[] idx = new Integer[n];
        Arrays.setAll(idx, i -> i);
        Arrays.sort(idx, (i, j) -> reward1[j] - reward2[j] - (reward1[i] - reward2[i]));
        int ans = 0;
        for (int i = 0; i < k; ++i) {
            ans += reward1[idx[i]];
        }
        for (int i = k; i < n; ++i) {
            ans += reward2[idx[i]];
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2611: Mice and Cheese
func miceAndCheese(reward1 []int, reward2 []int, k int) (ans int) {
	n := len(reward1)
	idx := make([]int, n)
	for i := range idx {
		idx[i] = i
	}
	sort.Slice(idx, func(i, j int) bool {
		i, j = idx[i], idx[j]
		return reward1[j]-reward2[j] < reward1[i]-reward2[i]
	})
	for i := 0; i < k; i++ {
		ans += reward1[idx[i]]
	}
	for i := k; i < n; i++ {
		ans += reward2[idx[i]]
	}
	return
}

# Accepted solution for LeetCode #2611: Mice and Cheese
class Solution:
    def miceAndCheese(self, reward1: List[int], reward2: List[int], k: int) -> int:
        n = len(reward1)
        idx = sorted(range(n), key=lambda i: reward1[i] - reward2[i], reverse=True)
        return sum(reward1[i] for i in idx[:k]) + sum(reward2[i] for i in idx[k:])

// Accepted solution for LeetCode #2611: Mice and Cheese
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2611: Mice and Cheese
// class Solution {
//     public int miceAndCheese(int[] reward1, int[] reward2, int k) {
//         int n = reward1.length;
//         Integer[] idx = new Integer[n];
//         Arrays.setAll(idx, i -> i);
//         Arrays.sort(idx, (i, j) -> reward1[j] - reward2[j] - (reward1[i] - reward2[i]));
//         int ans = 0;
//         for (int i = 0; i < k; ++i) {
//             ans += reward1[idx[i]];
//         }
//         for (int i = k; i < n; ++i) {
//             ans += reward2[idx[i]];
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2611: Mice and Cheese
function miceAndCheese(reward1: number[], reward2: number[], k: number): number {
    const n = reward1.length;
    const idx = Array.from({ length: n }, (_, i) => i);
    idx.sort((i, j) => reward1[j] - reward2[j] - (reward1[i] - reward2[i]));
    let ans = 0;
    for (let i = 0; i < k; ++i) {
        ans += reward1[idx[i]];
    }
    for (let i = k; i < n; ++i) {
        ans += reward2[idx[i]];
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.