Segment Tree

The Segment Tree is a data structure that maintains a sequence of $n=2^k$ numbers $a_0,a_1,\dots ,a_{n-1}$ by maintaining $s(0,2^k),s(0,2^{k-1}),s(2^{k-1},2^k),\dots ,s(0,1),s(1,2),\dots ,s(2^k-1,2^k)$, where

$$s(l,r)=\sum _{i=l}^{r-1}{a}_i$$

This requires $\mathcal{O}(n)$ space.

Proof

$$\begin{array}{rl}\sum _{i=0}^k{2}^i& =2^{k+1}-1\\ & =2n-1\\ & \in \mathcal{O}(n)\end{array}$$

Modify

Modify updates $a_i$ to $x$ in $\mathcal{O}(\log n)$ time and $\mathcal{O}(\log n)$ space.

Algorithm

Updating the maintained segments containing the given element yields an algorithm that solves the problem in $\mathcal{O}(\log n)$ time and $\mathcal{O}(\log n)$ space.

void modify(int i, int x) {
	[&](this auto &&self, int o, int s, int t) -> void {
		if (s + 1 == t) {
			sum[o] = x;
			return;
		}
 
		int mid = std::midpoint(s, t);
		if (i < mid) {
			self(o << 1, s, mid);
		} else {
			self(o << 1 | 1, mid, t);
		}
		sum[o] = sum[o << 1] + sum[o << 1 | 1];
	} (1, 0, n);
}

Proof

It is easy to prove that exactly $1$ segment is visited in each layer.

Range Sum Query

Range Sum Query computes ${\sum }_{i=l}^{r-1}{a}_i$ in $\mathcal{O}(\log n)$ time and $\mathcal{O}(\log n)$ space.

Algorithm

Decomposing the query interval into maintained segments yields an algorithm that solves the problem in $\mathcal{O}(\log n)$ time and $\mathcal{O}(\log n)$ space.

int range_sum_query(int l, int r) {
	return [&](this auto &&self, int o, int s, int t) -> int {
		if (s >= r || t <= l) {
			return 0;
		}
		if (l <= s && t <= r) {
			return sum[o];
		}
 
		int mid = std::midpoint(s, t);
		return self(o << 1, s, mid) + self(o << 1 | 1, mid, t);
	} (1, 0, n);
}

Proof

It is easy to prove that at most $4$ segments are visited in each layer.