LeetCode 2611: Mice and Cheese Solution

Master LeetCode problem 2611 (Mice and Cheese), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2611. Mice and Cheese

Medium

Array Greedy Sorting Heap (Priority Queue)

Problem Explanation

Explanation

To solve this problem, we can use dynamic programming. We can create a 3D dp array where dp[i][j][k] represents the maximum total points that can be achieved by the two mice if the first mouse eats j types of cheese from the first i types of cheese, and the second mouse eats k types of cheese from the first i types of cheese.

We can iterate over all types of cheese and for each type, we have two choices:

The first mouse eats the cheese: dp[i][j][k] = max(dp[i][j][k], dp[i-1][j-1][k] + reward1[i]).
The second mouse eats the cheese: dp[i][j][k] = max(dp[i][j][k], dp[i-1][j][k-1] + reward2[i]).

Finally, the answer will be the maximum value of dp[n][k][k].

Time Complexity

The time complexity of this approach is O(n^2 * k), where n is the number of types of cheese and k is the number of types of cheese the first mouse eats.

Space Complexity

The space complexity of this approach is O(n * k^2).

Solution Code

class Solution {
    public int maxPoints(int[] reward1, int[] reward2, int k) {
        int n = reward1.length;
        int[][][] dp = new int[n+1][k+1][k+1];

        for (int i = 1; i <= n; i++) {
            for (int j = 0; j <= k; j++) {
                for (int l = 0; l <= k; l++) {
                    dp[i][j][l] = Math.max(dp[i-1][j][l], (j > 0 ? dp[i-1][j-1][l] + reward1[i-1] : 0), (l > 0 ? dp[i-1][j][l-1] + reward2[i-1] : 0));
                }
            }
        }

        return dp[n][k][k];
    }
}

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