You are given a 0-indexed integer array nums.
A subsequence of nums having length k and consisting of indices i0 < i1 < ... < ik-1 is balanced if the following holds:
nums[ij] - nums[ij-1] >= ij - ij-1, for every j in the range [1, k - 1].A subsequence of nums having length 1 is considered balanced.
Return an integer denoting the maximum possible sum of elements in a balanced subsequence of nums.
A subsequence of an array is a new non-empty array that is formed from the original array by deleting some (possibly none) of the elements without disturbing the relative positions of the remaining elements.
\n\n\n
Example 1:
\n\n\nInput: nums = [3,3,5,6]\nOutput: 14\nExplanation: In this example, the subsequence [3,5,6] consisting of indices 0, 2, and 3 can be selected.\nnums[2] - nums[0] >= 2 - 0.\nnums[3] - nums[2] >= 3 - 2.\nHence, it is a balanced subsequence, and its sum is the maximum among the balanced subsequences of nums.\nThe subsequence consisting of indices 1, 2, and 3 is also valid.\nIt can be shown that it is not possible to get a balanced subsequence with a sum greater than 14.\n\n
Example 2:
\n\n\nInput: nums = [5,-1,-3,8]\nOutput: 13\nExplanation: In this example, the subsequence [5,8] consisting of indices 0 and 3 can be selected.\nnums[3] - nums[0] >= 3 - 0.\nHence, it is a balanced subsequence, and its sum is the maximum among the balanced subsequences of nums.\nIt can be shown that it is not possible to get a balanced subsequence with a sum greater than 13.\n\n\n
Example 3:
\n\n\nInput: nums = [-2,-1]\nOutput: -1\nExplanation: In this example, the subsequence [-1] can be selected.\nIt is a balanced subsequence, and its sum is the maximum among the balanced subsequences of nums.\n\n\n
\n
Constraints:
\n\n1 <= nums.length <= 105-109 <= nums[i] <= 109给你一个下标从 0 开始的整数数组 nums 。
nums 一个长度为 k 的 子序列 指的是选出 k 个 下标 i0 < i1 < ... < ik-1 ，如果这个子序列满足以下条件，我们说它是 平衡的 ：
[1, k - 1] 内的所有 j ，nums[ij] - nums[ij-1] >= ij - ij-1 都成立。nums 长度为 1 的 子序列 是平衡的。
请你返回一个整数，表示 nums 平衡 子序列里面的 最大元素和 。
一个数组的 子序列 指的是从原数组中删除一些元素（也可能一个元素也不删除）后，剩余元素保持相对顺序得到的 非空 新数组。
\n\n\n\n
示例 1：
\n\n\n输入：nums = [3,3,5,6]\n输出：14\n解释：这个例子中，选择子序列 [3,5,6] ，下标为 0 ，2 和 3 的元素被选中。\nnums[2] - nums[0] >= 2 - 0 。\nnums[3] - nums[2] >= 3 - 2 。\n所以，这是一个平衡子序列，且它的和是所有平衡子序列里最大的。\n包含下标 1 ，2 和 3 的子序列也是一个平衡的子序列。\n最大平衡子序列和为 14 。\n\n
示例 2：
\n\n\n输入：nums = [5,-1,-3,8]\n输出：13\n解释：这个例子中，选择子序列 [5,8] ，下标为 0 和 3 的元素被选中。\nnums[3] - nums[0] >= 3 - 0 。\n所以，这是一个平衡子序列，且它的和是所有平衡子序列里最大的。\n最大平衡子序列和为 13 。\n\n\n
示例 3：
\n\n\n输入：nums = [-2,-1]\n输出：-1\n解释：这个例子中，选择子序列 [-1] 。\n这是一个平衡子序列，而且它的和是 nums 所有平衡子序列里最大的。\n\n\n
\n\n
提示：
\n\n1 <= nums.length <= 105-109 <= nums[i] <= 109dp[x] represent the maximum sum of a balanced subsequence ending at x.",
"Rewriting the formula nums[ij] - nums[ij-1] >= ij - ij-1 gives nums[ij] - ij >= nums[ij-1] - ij-1.",
"So, for some index x, we need to find an index y, y < x, such that dp[x] = nums[x] + dp[y] is maximized, and nums[x] - x >= nums[y] - y.",
"There are many ways to achieve this. One method involves sorting the values of nums[x] - x for all indices x and using a segment/Fenwick tree with coordinate compression.",
"Hence, using a dictionary or map, let's call it dict, where dict[nums[x] - x] represents the position of the value, nums[x] - x, in the segment tree.",
"The tree is initialized with zeros initially.",
"For indices x in order from [0, n - 1], dp[x] = max(nums[x], nums[x] + the maximum query from the tree in the range [0, dict[nums[x] - x]]), and if dp[x] is greater than the value in the tree at position dict[nums[x] - x], we update the value in the tree.",
"The answer to the problem is the maximum value in dp."
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