You are given an array nums of length n and a positive integer k.
A subarray of nums is called good if the absolute difference between its first and last element is exactly k, in other words, the subarray nums[i..j] is good if |nums[i] - nums[j]| == k.
Return the maximum sum of a good subarray of nums. If there are no good subarrays, return 0.
\n
Example 1:
\n\n\nInput: nums = [1,2,3,4,5,6], k = 1\nOutput: 11\nExplanation: The absolute difference between the first and last element must be 1 for a good subarray. All the good subarrays are: [1,2], [2,3], [3,4], [4,5], and [5,6]. The maximum subarray sum is 11 for the subarray [5,6].\n\n\n
Example 2:
\n\n\nInput: nums = [-1,3,2,4,5], k = 3\nOutput: 11\nExplanation: The absolute difference between the first and last element must be 3 for a good subarray. All the good subarrays are: [-1,3,2], and [2,4,5]. The maximum subarray sum is 11 for the subarray [2,4,5].\n\n\n
Example 3:
\n\n\nInput: nums = [-1,-2,-3,-4], k = 2\nOutput: -6\nExplanation: The absolute difference between the first and last element must be 2 for a good subarray. All the good subarrays are: [-1,-2,-3], and [-2,-3,-4]. The maximum subarray sum is -6 for the subarray [-1,-2,-3].\n\n\n
\n
Constraints:
\n\n2 <= nums.length <= 105-109 <= nums[i] <= 1091 <= k <= 109给你一个长度为 n 的数组 nums 和一个 正 整数 k 。
如果 nums 的一个子数组中，第一个元素和最后一个元素 差的绝对值恰好 为 k ，我们称这个子数组为 好 的。换句话说，如果子数组 nums[i..j] 满足 |nums[i] - nums[j]| == k ，那么它是一个好子数组。
请你返回 nums 中 好 子数组的 最大 和，如果没有好子数组，返回 0 。
\n\n
示例 1：
\n\n\n输入：nums = [1,2,3,4,5,6], k = 1\n输出：11\n解释：好子数组中第一个元素和最后一个元素的差的绝对值必须为 1 。好子数组有 [1,2] ，[2,3] ，[3,4] ，[4,5] 和 [5,6] 。最大子数组和为 11 ，对应的子数组为 [5,6] 。\n\n\n
示例 2：
\n\n\n输入：nums = [-1,3,2,4,5], k = 3\n输出：11\n解释：好子数组中第一个元素和最后一个元素的差的绝对值必须为 3 。好子数组有 [-1,3,2] 和 [2,4,5] 。最大子数组和为 11 ，对应的子数组为 [2,4,5] 。\n\n\n
示例 3：
\n\n\n输入：nums = [-1,-2,-3,-4], k = 2\n输出：-6\n解释：好子数组中第一个元素和最后一个元素的差的绝对值必须为 2 。好子数组有 [-1,-2,-3] 和 [-2,-3,-4] 。最大子数组和为 -6 ，对应的子数组为 [-1,-2,-3] 。\n\n\n
\n\n
提示：
\n\n2 <= nums.length <= 105-109 <= nums[i] <= 1091 <= k <= 109i store the element at index i + 1 as the key and the prefix sum till i as the value.",
"For each prefix sum ending at nums[i], try finding nums[i] - k and nums[i] + k in the HashMap and update the answer."
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