You are given an integer array nums of length n.
A trionic subarray is a contiguous subarray nums[l...r] (with 0 <= l < r < n) for which there exist indices l < p < q < r such that:
nums[l...p] is strictly increasing,nums[p...q] is strictly decreasing,nums[q...r] is strictly increasing.Return the maximum sum of any trionic subarray in nums.
\n
Example 1:
\n\nInput: nums = [0,-2,-1,-3,0,2,-1]
\n\nOutput: -4
\n\nExplanation:
\n\nPick l = 1, p = 2, q = 3, r = 5:
nums[l...p] = nums[1...2] = [-2, -1] is strictly increasing (-2 < -1).nums[p...q] = nums[2...3] = [-1, -3] is strictly decreasing (-1 > -3)nums[q...r] = nums[3...5] = [-3, 0, 2] is strictly increasing (-3 < 0 < 2).(-2) + (-1) + (-3) + 0 + 2 = -4.Example 2:
\n\nInput: nums = [1,4,2,7]
\n\nOutput: 14
\n\nExplanation:
\n\nPick l = 0, p = 1, q = 2, r = 3:
nums[l...p] = nums[0...1] = [1, 4] is strictly increasing (1 < 4).nums[p...q] = nums[1...2] = [4, 2] is strictly decreasing (4 > 2).nums[q...r] = nums[2...3] = [2, 7] is strictly increasing (2 < 7).1 + 4 + 2 + 7 = 14.\n
Constraints:
\n\n4 <= n = nums.length <= 105-109 <= nums[i] <= 109给你一个长度为 n 的整数数组 nums。
三段式子数组 是一个连续子数组 nums[l...r]（满足 0 <= l < r < n），并且存在下标 l < p < q < r，使得：
nums[l...p] 严格 递增，nums[p...q] 严格 递减，nums[q...r] 严格 递增。请你从数组 nums 的所有三段式子数组中找出和最大的那个，并返回其 最大 和。
\n\n
示例 1：
\n\n输入：nums = [0,-2,-1,-3,0,2,-1]
\n\n输出：-4
\n\n解释：
\n\n选择 l = 1, p = 2, q = 3, r = 5：
nums[l...p] = nums[1...2] = [-2, -1] 严格递增 (-2 < -1)。nums[p...q] = nums[2...3] = [-1, -3] 严格递减 (-1 > -3)。nums[q...r] = nums[3...5] = [-3, 0, 2] 严格递增 (-3 < 0 < 2)。(-2) + (-1) + (-3) + 0 + 2 = -4。示例 2:
\n\n输入: nums = [1,4,2,7]
\n\n输出: 14
\n\n解释:
\n\n选择 l = 0, p = 1, q = 2, r = 3：
nums[l...p] = nums[0...1] = [1, 4] 严格递增 (1 < 4)。nums[p...q] = nums[1...2] = [4, 2] 严格递减 (4 > 2)。nums[q...r] = nums[2...3] = [2, 7] 严格递增 (2 < 7)。1 + 4 + 2 + 7 = 14。\n\n
提示:
\n\n4 <= n = nums.length <= 105-109 <= nums[i] <= 109dp0...dp3 where dpk[i] is the max sum of a subarray ending at i after finishing k of the four phases (start -> inc -> dec -> inc)",
"Process each i>0",
"If nums[i]>nums[i‑1], set dp1[i]=max(dp1[i‑1]+nums[i], dp0[i‑1]+nums[i]), dp3[i]=max(dp3[i‑1]+nums[i], dp2[i‑1]+nums[i])",
"If nums[i], set dp2[i]=max(dp2[i‑1]+nums[i], dp1[i‑1]+nums[i])",
"Always carry over dp0[i]=dp0[i‑1]+nums[i] when nums[i]>nums[i‑1]",
"Return the maximum value in dp3"
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