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Java-based LeetCode algorithm problem solutions, regularly updated

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package g1401_1500.s1402_reducing_dishes;

// #Hard #Array #Dynamic_Programming #Sorting #Greedy
// #2022_03_25_Time_3_ms_(66.20%)_Space_41.3_MB_(80.80%)

import java.util.Arrays;

/**
 * 1402 - Reducing Dishes\.
 *
 * Hard
 *
 * A chef has collected data on the `satisfaction` level of his `n` dishes. Chef can cook any dish in 1 unit of time.
 *
 * **Like-time coefficient** of a dish is defined as the time taken to cook that dish including previous dishes multiplied by its satisfaction level i.e. `time[i] * satisfaction[i]`.
 *
 * Return _the maximum sum of **like-time coefficient** that the chef can obtain after dishes preparation_.
 *
 * Dishes can be prepared in **any** order and the chef can discard some dishes to get this maximum value.
 *
 * **Example 1:**
 *
 * **Input:** satisfaction = [-1,-8,0,5,-9]
 *
 * **Output:** 14
 *
 * **Explanation:** After Removing the second and last dish, the maximum total **like-time coefficient** will be equal to (-1\*1 + 0\*2 + 5\*3 = 14). Each dish is prepared in one unit of time.
 *
 * **Example 2:**
 *
 * **Input:** satisfaction = [4,3,2]
 *
 * **Output:** 20
 *
 * **Explanation:** Dishes can be prepared in any order, (2\*1 + 3\*2 + 4\*3 = 20)
 *
 * **Example 3:**
 *
 * **Input:** satisfaction = [-1,-4,-5]
 *
 * **Output:** 0
 *
 * **Explanation:** People do not like the dishes. No dish is prepared.
 *
 * **Constraints:**
 *
 * *   `n == satisfaction.length`
 * *   `1 <= n <= 500`
 * *   `-1000 <= satisfaction[i] <= 1000`
**/
public class Solution {
    public int maxSatisfaction(int[] satisfaction) {
        Arrays.sort(satisfaction);
        int sum = 0;
        int mulSum = 0;
        for (int i = 0; i < satisfaction.length; i++) {
            sum += satisfaction[i];
            mulSum += (i + 1) * satisfaction[i];
        }
        int maxVal = Math.max(0, mulSum);
        for (int j : satisfaction) {
            mulSum -= sum;
            sum -= j;
            maxVal = Math.max(maxVal, mulSum);
        }
        return maxVal;
    }
}