Segment Tree

A Segment Tree maintains an array 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

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

which costs a space of $\mathcal{O}(n)$.

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.

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

Proof

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

y_combinator([&](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);

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.

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.

Proof

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

return y_combinator([&](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);