# Solve RMQ (Range Minimum Query) by finding LCA (Lowest Common Ancestor)¶

Given an array `A[0..N-1]`

.
For each query of the form `[L, R]`

we want to find the minimum in the array `A`

starting from position `L`

and ending with position `R`

.
We will assume that the array `A`

doesn't change in the process, i.e. this article describes a solution to the static RMQ problem

Here is a description of an asymptotically optimal solution. It stands apart from other solutions for the RMQ problem, since it is very different from them: it reduces the RMQ problem to the LCA problem, and then uses the Farach-Colton and Bender algorithm, which reduces the LCA problem back to a specialized RMQ problem and solves that.

## Algorithm¶

We construct a **Cartesian tree** from the array `A`

.
A Cartesian tree of an array `A`

is a binary tree with the min-heap property (the value of parent node has to be smaller or equal than the value of its children) such that the in-order traversal of the tree visits the nodes in the same order as they are in the array `A`

.

In other words, a Cartesian tree is a recursive data structure.
The array `A`

will be partitioned into 3 parts: the prefix of the array up to the minimum, the minimum, and the remaining suffix.
The root of the tree will be a node corresponding to the minimum element of the array `A`

, the left subtree will be the Cartesian tree of the prefix, and the right subtree will be a Cartesian tree of the suffix.

In the following image you can see one array of length 10 and the corresponding Cartesian tree.

The range minimum query `[l, r]`

is equivalent to the lowest common ancestor query `[l', r']`

, where `l'`

is the node corresponding to the element `A[l]`

and `r'`

the node corresponding to the element `A[r]`

.
Indeed the node corresponding to the smallest element in the range has to be an ancestor of all nodes in the range, therefor also from `l'`

and `r'`

.
This automatically follows from the min-heap property.
And is also has to be the lowest ancestor, because otherwise `l'`

and `r'`

would be both in the left or in the right subtree, which generates a contradiction since in such a case the minimum wouldn't even be in the range.

In the following image you can see the LCA queries for the RMQ queries `[1, 3]`

and `[5, 9]`

.
In the first query the LCA of the nodes `A[1]`

and `A[3]`

is the node corresponding to `A[2]`

which has the value 2, and in the second query the LCA of `A[5]`

and `A[9]`

is the node corresponding to `A[8]`

which has the value 3.

Such a tree can be built in $O(N)$ time and the Farach-Colton and Benders algorithm can preprocess the tree in $O(N)$ and find the LCA in $O(1)$.

## Construction of a Cartesian tree¶

We will build the Cartesian tree by adding the elements one after another.
In each step we maintain a valid Cartesian tree of all the processed elements.
It is easy to see, that adding an element `s[i]`

can only change the nodes in the most right path - starting at the root and repeatedly taking the right child - of the tree.
The subtree of the node with the smallest, but greater or equal than `s[i]`

, value becomes the left subtree of `s[i]`

, and the tree with root `s[i]`

will become the new right subtree of the node with the biggest, but smaller than `s[i]`

value.

This can be implemented by using a stack to store the indices of the most right nodes.

```
vector<int> parent(n, -1);
stack<int> s;
for (int i = 0; i < n; i++) {
int last = -1;
while (!s.empty() && A[s.top()] >= A[i]) {
last = s.top();
s.pop();
}
if (!s.empty())
parent[i] = s.top();
if (last >= 0)
parent[last] = i;
s.push(i);
}
```