有序三元组中的最大值 I

You are given a 0-indexed integer array nums.

Return the maximum value over all triplets of indices (i, j, k) such that i < j < k. If all such triplets have a negative value, return 0.

The value of a triplet of indices (i, j, k) is equal to (nums[i] - nums[j]) * nums[k].

 

Example 1:

Input: nums = [12,6,1,2,7]
Output: 77
Explanation: The value of the triplet (0, 2, 4) is (nums[0] - nums[2]) * nums[4] = 77.
It can be shown that there are no ordered triplets of indices with a value greater than 77. 

Example 2:

Input: nums = [1,10,3,4,19]
Output: 133
Explanation: The value of the triplet (1, 2, 4) is (nums[1] - nums[2]) * nums[4] = 133.
It can be shown that there are no ordered triplets of indices with a value greater than 133.

Example 3:

Input: nums = [1,2,3]
Output: 0
Explanation: The only ordered triplet of indices (0, 1, 2) has a negative value of (nums[0] - nums[1]) * nums[2] = -3. Hence, the answer would be 0.

 

Constraints:

• 3 <= nums.length <= 100
• 1 <= nums[i] <= 106

/*
 * 思路：使用 Java Stream API 实现上述逻辑。
 * 尽管 Stream API 主要用于简化集合处理，但我们可以将其用于生成 i, j, k 的所有可能组合。
 */

import java.util.stream.IntStream;

public class Solution {
    public int maximumTripletValue(int[] nums) {
        int n = nums.length;
        return IntStream.range(0, n - 2)
            .boxed()
            .flatMap(i -> IntStream.range(i + 1, n - 1)
                .boxed()
                .flatMap(j -> IntStream.range(j + 1, n)
                    .mapToObj(k -> new int[]{i, j, k})))
            .mapToInt(triplet -> (nums[triplet[0]] - nums[triplet[1]]) * nums[triplet[2]])
            .filter(value -> value > 0)
            .max()
            .orElse(0);
    }
}

该题解采用了前缀最大值和后缀最大值的方法。首先，通过一次从后向前的遍历构建一个后缀最大值数组`suf_max`，该数组在每个位置i存储了从i到数组末尾的最大值。随后，在一次从前向后的遍历中，使用一个变量`pre_max`来记录从数组开始到当前位置的最大值，然后对于每个位置j，计算`(pre_max - nums[j]) * suf_max[j + 1]`，并更新结果`ans`为所有可能组合中的最大值。这种方法有效地利用了动态规划的思想，通过记录前面的计算结果来避免重复计算，从而优化了时间复杂度。

O(n)

O(n)