Maximum Balanced Subsequence Sum
<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>.</p>

<p>A <strong>subsequence</strong> of <code>nums</code> having length <code>k</code> and consisting of <strong>indices</strong> <code>i<sub>0</sub>&nbsp;&lt;&nbsp;i<sub>1</sub> &lt;&nbsp;... &lt; i<sub>k-1</sub></code> is <strong>balanced</strong> if the following holds:</p>

<ul>
	<li><code>nums[i<sub>j</sub>] - nums[i<sub>j-1</sub>] &gt;= i<sub>j</sub> - i<sub>j-1</sub></code>, for every <code>j</code> in the range <code>[1, k - 1]</code>.</li>
</ul>

<p>A <strong>subsequence</strong> of <code>nums</code> having length <code>1</code> is considered balanced.</p>

<p>Return <em>an integer denoting the <strong>maximum</strong> possible <strong>sum of elements</strong> in a <strong>balanced</strong> subsequence of </em><code>nums</code>.</p>

<p>A <strong>subsequence</strong> of an array is a new <strong>non-empty</strong> array that is formed from the original array by deleting some (<strong>possibly none</strong>) of the elements without disturbing the relative positions of the remaining elements.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,3,5,6]
<strong>Output:</strong> 14
<strong>Explanation:</strong> In this example, the subsequence [3,5,6] consisting of indices 0, 2, and 3 can be selected.
nums[2] - nums[0] &gt;= 2 - 0.
nums[3] - nums[2] &gt;= 3 - 2.
Hence, it is a balanced subsequence, and its sum is the maximum among the balanced subsequences of nums.
The subsequence consisting of indices 1, 2, and 3 is also valid.
It can be shown that it is not possible to get a balanced subsequence with a sum greater than 14.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [5,-1,-3,8]
<strong>Output:</strong> 13
<strong>Explanation:</strong> In this example, the subsequence [5,8] consisting of indices 0 and 3 can be selected.
nums[3] - nums[0] &gt;= 3 - 0.
Hence, it is a balanced subsequence, and its sum is the maximum among the balanced subsequences of nums.
It can be shown that it is not possible to get a balanced subsequence with a sum greater than 13.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [-2,-1]
<strong>Output:</strong> -1
<strong>Explanation:</strong> In this example, the subsequence [-1] can be selected.
It is a balanced subsequence, and its sum is the maximum among the balanced subsequences of nums.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>
Binary Indexed Tree; Segment Tree; Array; Binary Search; Dynamic Programming
const inf int = 1e18

type BinaryIndexedTree struct {
	n int
	c []int
}

func NewBinaryIndexedTree(n int) BinaryIndexedTree {
	c := make([]int, n+1)
	for i := range c {
		c[i] = -inf
	}
	return BinaryIndexedTree{n: n, c: c}
}

func (bit *BinaryIndexedTree) update(x, v int) {
	for x <= bit.n {
		bit.c[x] = max(bit.c[x], v)
		x += x & -x
	}
}

func (bit *BinaryIndexedTree) query(x int) int {
	mx := -inf
	for x > 0 {
		mx = max(mx, bit.c[x])
		x -= x & -x
	}
	return mx
}

func maxBalancedSubsequenceSum(nums []int) int64 {
	n := len(nums)
	arr := make([]int, n)
	for i, x := range nums {
		arr[i] = x - i
	}
	sort.Ints(arr)
	m := 0
	for i, x := range arr {
		if i == 0 || x != arr[i-1] {
			arr[m] = x
			m++
		}
	}
	arr = arr[:m]
	tree := NewBinaryIndexedTree(m)
	for i, x := range nums {
		j := sort.SearchInts(arr, x-i) + 1
		v := max(tree.query(j), 0) + x
		tree.update(j, v)
	}
	return int64(tree.query(m))
}