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g2601_2700.s2611_mice_and_cheese.Solution Maven / Gradle / Ivy

package g2601_2700.s2611_mice_and_cheese;

// #Medium #Array #Sorting #Greedy #Heap_Priority_Queue
// #2023_08_30_Time_11_ms_(99.56%)_Space_55_MB_(81.66%)

import java.util.Arrays;

/**
 * 2611 - Mice and Cheese\.
 *
 * Medium
 *
 * 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 2nd (0-indexed) and the 3rd types of cheese, and the second mouse eats the 0th and the 1st 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 0th (0-indexed) and 1st 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`
**/
public class Solution {
    public int miceAndCheese(
            int[] firstReward, int[] seondReward, int numberOfTypesOfCheeseForFirstMouse) {
        int maximumPoints = 0;
        final int totalTypesOfCheese = firstReward.length;
        int[] differenceInRewards = new int[totalTypesOfCheese];
        for (int i = 0; i < totalTypesOfCheese; ++i) {
            differenceInRewards[i] = firstReward[i] - seondReward[i];
            maximumPoints += seondReward[i];
        }
        Arrays.sort(differenceInRewards);
        for (int i = totalTypesOfCheese - 1;
                i > totalTypesOfCheese - numberOfTypesOfCheeseForFirstMouse - 1;
                --i) {
            maximumPoints += differenceInRewards[i];
        }
        return maximumPoints;
    }
}