<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>

from typing import List
import bisect
import math

class FenwickTree:
    def __init__(self, size):
        self.N = size + 1  # 1-based indexing
        self.tree = [-math.inf] * self.N

    def update(self, idx, value):
        while idx < self.N:
            if value > self.tree[idx]:
                self.tree[idx] = value
            else:
                break  # No need to update further if current value is not greater
            idx += idx & -idx

    def query(self, idx):
        res = -math.inf
        while idx > 0:
            if self.tree[idx] > res:
                res = self.tree[idx]
            idx -= idx & -idx
        return res

class Solution:
    def maxBalancedSubsequenceSum(self, nums: List[int]) -> int:
        if not nums:
            return 0

        n = len(nums)
        dp = [0] * n

        # Compute key[j] = nums[j] - j
        keys = [nums[j] - j for j in range(n)]

        # Sort unique keys
        sorted_keys = sorted(set(keys))
        key_to_idx = {key: idx + 1 for idx, key in enumerate(sorted_keys)}  # 1-based indexing

        # Initialize Fenwick Tree
        ft = FenwickTree(len(sorted_keys))

        max_sum = -math.inf

        for i in range(n):
            threshold = nums[i] - i

            # Find the rightmost index where key[j] <= threshold
            idx = bisect.bisect_right(sorted_keys, threshold)

            if idx > 0:
                max_dp = ft.query(idx)
                if max_dp != -math.inf:
                    dp[i] = max(nums[i], nums[i] + max_dp)  # Corrected line
                else:
                    dp[i] = nums[i]
            else:
                dp[i] = nums[i]

            # Update Fenwick Tree with dp[i] at key[j] = nums[i] - i
            key = nums[i] - i
            key_idx = key_to_idx[key]
            ft.update(key_idx, dp[i])

            # Update the global maximum
            if dp[i] > max_sum:
                max_sum = dp[i]

        return max_sum

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