No. The operation has an inverse, and binary-indexed blocks decompose any prefix.

Each index participates in a ladder of increasingly large blocks. An update climbs to every block containing the point; a query descends through disjoint blocks covering the prefix.

What the function promises: Fenwick(n) supports zero-based point additions and inclusive range sums. Invalid indices raise ValueError.

class Fenwick:
    def __init__(self, n):
        self.n, self.tree = n, [0] * (n + 1)
    def add(self, index, delta):
        if not 0 <= index < self.n:
            raise ValueError("Invalid index")
        i = index + 1
        while i <= self.n:
            self.tree[i] += delta
            i += i & -i
    def prefix(self, end):
        total = 0
        i = min(end + 1, self.n)
        while i > 0:
            total += self.tree[i]
            i -= i & -i
        return total
    def range_sum(self, left, right):
        return self.prefix(right) - self.prefix(left - 1)

bit = Fenwick(4)
for i, value in enumerate([2,1,3,4]): bit.add(i, value)
assert bit.range_sum(1,3) == 8
bit.add(1,4)
assert bit.range_sum(1,3) == 12

Work and storage: O(log n) per update/query and O(n) storage.

When to recognize it: The operation has an inverse, and binary-indexed blocks decompose any prefix.

Check before applying it: Fenwick trees are specialized; arbitrary range minimum with updates lacks the same subtraction trick.

Store boundary changes in a Fenwick tree; a point value is their prefix sum.