<p>You are given two <strong>0-indexed</strong> integer arrays <code>nums</code> and <code>removeQueries</code>, both of length <code>n</code>. For the <code>i<sup>th</sup></code> query, the element in <code>nums</code> at the index <code>removeQueries[i]</code> is removed, splitting <code>nums</code> into different segments.</p>

<p>A <strong>segment</strong> is a contiguous sequence of <strong>positive</strong> integers in <code>nums</code>. A <strong>segment sum</strong> is the sum of every element in a segment.</p>

<p>Return<em> an integer array </em><code>answer</code><em>, of length </em><code>n</code><em>, where </em><code>answer[i]</code><em> is the <strong>maximum</strong> segment sum after applying the </em><code>i<sup>th</sup></code> <em>removal.</em></p>

<p><strong>Note:</strong> The same index will <strong>not</strong> be removed more than once.</p>

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

<pre>
<strong>Input:</strong> nums = [1,2,5,6,1], removeQueries = [0,3,2,4,1]
<strong>Output:</strong> [14,7,2,2,0]
<strong>Explanation:</strong> Using 0 to indicate a removed element, the answer is as follows:
Query 1: Remove the 0th element, nums becomes [0,2,5,6,1] and the maximum segment sum is 14 for segment [2,5,6,1].
Query 2: Remove the 3rd element, nums becomes [0,2,5,0,1] and the maximum segment sum is 7 for segment [2,5].
Query 3: Remove the 2nd element, nums becomes [0,2,0,0,1] and the maximum segment sum is 2 for segment [2]. 
Query 4: Remove the 4th element, nums becomes [0,2,0,0,0] and the maximum segment sum is 2 for segment [2]. 
Query 5: Remove the 1st element, nums becomes [0,0,0,0,0] and the maximum segment sum is 0, since there are no segments.
Finally, we return [14,7,2,2,0].</pre>

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

<pre>
<strong>Input:</strong> nums = [3,2,11,1], removeQueries = [3,2,1,0]
<strong>Output:</strong> [16,5,3,0]
<strong>Explanation:</strong> Using 0 to indicate a removed element, the answer is as follows:
Query 1: Remove the 3rd element, nums becomes [3,2,11,0] and the maximum segment sum is 16 for segment [3,2,11].
Query 2: Remove the 2nd element, nums becomes [3,2,0,0] and the maximum segment sum is 5 for segment [3,2].
Query 3: Remove the 1st element, nums becomes [3,0,0,0] and the maximum segment sum is 3 for segment [3].
Query 4: Remove the 0th element, nums becomes [0,0,0,0] and the maximum segment sum is 0, since there are no segments.
Finally, we return [16,5,3,0].
</pre>

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

<ul>
	<li><code>n == nums.length == removeQueries.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>0 &lt;= removeQueries[i] &lt; n</code></li>
	<li>All the values of <code>removeQueries</code> are <strong>unique</strong>.</li>
</ul>

# from sortedcontainers import SortedSet
# import bisect

# class Solution:
#     def maximumSegmentSum(self, nums: List[int], removeQueries: List[int]) -> List[int]:

#         # Two soluton SortedList (Multiset/Treeset) and REVERSE UNION FIND **

#         n = len(nums)
#         s = SortedSet([0])  # Use SortedSet to maintain sorted order with unique elements
#         in_segments = SortedSet([(0, n - 1)])
        
#         # Compute prefix sums
#         pre = [0] * n
#         pre[0] = nums[0]
#         for i in range(1, n):
#             pre[i] = pre[i - 1] + nums[i]
        
#         # Insert the sum of the whole array initially
#         s.add(pre[n - 1])
        
#         ans = []
        
#         for p in removeQueries:
#             # Find the segment containing index p
#             idx = bisect.bisect_right(in_segments, (p, float('inf'))) - 1
#             it = in_segments[idx]
            
#             # Calculate segment sums
#             L = (pre[p - 1] if p > 0 else 0) - (pre[it[0] - 1] if it[0] > 0 else 0)
#             R = pre[it[1]] - pre[p]
#             T = pre[it[1]] - (pre[it[0] - 1] if it[0] > 0 else 0)
            
#             # Remove the total segment sum
#             s.remove(T)
            
#             # Update segments and their sums
#             if it[0] <= p - 1:
#                 in_segments.add((it[0], p - 1))
#                 s.add(L)
            
#             if it[1] >= p + 1:
#                 in_segments.add((p + 1, it[1]))
#                 s.add(R)
            
#             in_segments.remove(it)
            
#             # Record the maximum sum
#             ans.append(s[-1])
        
#         return ans

# ********************* REVERSE DSU Solution *********************
class UnionFind:
    def __init__(self):
        self.parent = {}
        self.sum = {}
        self.maxSumSegment = 0
        
    def addAndMergeAdjSegment(self, u, value):
        self.parent[u] = u
        self.sum[u] = value
        self.maxSumSegment = max(self.maxSumSegment, value)
        if u-1 in self.parent:
            self.union(u, u-1)
        if u+1 in self.parent:
            self.union(u, u+1)
        
    def find(self, u):
        if u != self.parent[u]:
            self.parent[u] = self.find(self.parent[u])  # Path compression
        return self.parent[u]
    
    def union(self, u, v):
        pu = self.find(u)
        pv = self.find(v)
        if pu == pv: return
        self.sum[pu] += self.sum[pv]
        self.parent[pv] = pu
        self.maxSumSegment = max(self.maxSumSegment, self.sum[pu])

class Solution:
    def maximumSegmentSum(self, nums: List[int], removeQueries: List[int]) -> List[int]:
        n = len(nums)
        uf = UnionFind()
        
        ans = [0] * n
        for i in range(n-1, -1, -1):
            ans[i] = uf.maxSumSegment
            q = removeQueries[i]
            uf.addAndMergeAdjSegment(q, nums[q])
        return ans
943