#####   ######  #####    ###    #   #  ###  #   #  ######
##  ##  ##      ##  ##  ## ##   #   #   #   #   #  ##
#####   ####    #####   #   #   #   #   #   #   #  ####
##  #   ##      ##      ## ##    # #    #    # #   ##
##   #  ######  ##       ###      #    ###    #    ######

$ curl repovive.com/roadmaps/data-structures/fenwick-trees/building-a-fenwick-tree

░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░██████████████████████████████████████████████████████████████████████████████████

#####   ######  #####    ###    #   #  ###  #   #  ######
##  ##  ##      ##  ##  ## ##   #   #   #   #   #  ##
#####   ####    #####   #   #   #   #   #   #   #  ####
##  #   ##      ##      ## ##    # #    #    # #   ##
##   #  ######  ##       ###      #    ###    #    ######

$ curl repovive.com/roadmaps/data-structures/fenwick-trees/building-a-fenwick-tree

Repovive

Building a Fenwick Tree - Fenwick Trees | Data Structures | Repovive

Repovive.

Data Structures19 sections · 729 units49h total

Open in Course

Building a Fenwick Tree

Initialization

Two approaches to build from an array:

Approach 1: $n$ updates — $O(n \log n)$

tree = [0] * (n + 1)
for i in range(n):
    update(i + 1, arr[i])

Approach 2: Linear build — $O(n)$

tree = [0] + arr[:]  # 1-indexed copy
for i in range(1, n + 1):
    j = i + lowbit(i)
    if j <= n:
        tree[j] += tree[i]

The linear approach propagates each position's value to its parent in the BIT structure.

For most competitive programming, the $O(n \log n)$ approach is fine and simpler to remember.