Hello everyone,

finding the diameter is one of the most frequent ways to solve problems about trees. In this tutorial we will see how to find a diameter and some of its properties, and we will use them to solve some problems of increasing difficulty. I tried to put a lot of examples to make the understanding easier.

The first part of the tutorial is quite basic, so feel free to skip it and jump to the problems if you already know the concepts.

Target: rating $$$[1400, 2100]$$$ on CF

Prerequisites: basic graph theory, greedy

## The diameter

Given an unweighted tree, let's define $$$\text{dist}(a, b) =$$$ the number of edges in the simple path $$$a \rightarrow b$$$.

A diameter of the tree is a simple path $$$a \rightarrow b$$$ that maximizes $$$\text{dist}(a, b)$$$ over all pairs of nodes. If there are multiple diameters, let's pick any of them.

The same definition is valid for a weighted tree with nonnegative weights (with $$$\text{dist}(a, b) =$$$ the sum of the weights of the edges in the simple path $$$a \rightarrow b$$$).

#### Finding a diameter

Given a tree with $$$n$$$ nodes are multiple ways to find a diameter. Here is one of the simplest ways:

*Run a DFS from any node $$$p$$$. Let $$$a$$$ be a node whose distance from node $$$p$$$ is maximized. Run another DFS from node $$$a$$$. Let $$$b$$$ be a node whose distance from node $$$a$$$ is maximized. $$$a \rightarrow b$$$ is a diameter.*

#### Tree = edges of a diameter + forest

Before proving the previous algorithm, let's analyze the structure of the tree (we will mention the diameter, but we will not use the fact that $$$a \rightarrow b$$$ is actually a diameter before proving it).

We started a DFS from node $$$p = 11$$$, and we got that node $$$a = 1$$$ is the farthest from $$$p$$$, and node $$$b = 7$$$ is the farthest from $$$a$$$.

Let's represent the diameter on a line. If you remove the edges of the diameter, you get a forest (i.e., several trees). Let's root each tree at the node in the diameter. What's the height (i.e., the maximum distance from the root to any node) of each component?

Let $$$q$$$ be the root of the component of $$$p$$$. Let's consider any component whose root $$$d$$$ is between $$$a$$$ and $$$q$$$ (included), and one of its nodes $$$c$$$.

We get

$$$\text{dist}(p, a) \geq \text{dist}(p, c) \implies \text{dist}(p, a) - \text{dist}(p, d) \geq \text{dist}(p, c) - \text{dist}(p, d) \implies \text{dist}(c, a) \geq \text{dist}(c, d)$$$.

In other words, since \text{dist}(a, q) \geq \text{dist}(q, b), the height of each component with root in the left half of the diameter is at most the distance of the root of the component from the left end of the diameter.

You can prove the same statement for the right end of the diameter, using that $$$b$$$ is the farthest node from $$$a$$$.

#### Farthest node for each node

For each node $$$i$$$, let's find a node $$$j$$$ such that $$$\text{dist}(i, j)$$$ is maximum.

_Claim: $$$j = a$$$ or $$$j = b$$$ always works.

Proof:

- if $$$i$$$ and $$$j$$$ are in the same subtree

#### Proof that $$$a \rightarrow b$$$ is a diameter

For each $$$x$$$, let $$$f(x)$$$ be the number of inversions if you consider only the elements from $$$1$$$ to $$$x$$$ in the permutation.

First, let's put $$$x$$$ at the end of the permutation: this requires $$$x - \text{pos}(x)$$$ moves. That's optimal (the actual proof is similar to Proof 1; in an intuitive way, if you put the last element to the end of the array, it doesn't interfere anymore with the other swaps).

For example, if you have the permutation $$$[2, 3, 7, 8, 6, 9, 1, 4, 5]$$$ and you move the $$$9$$$ to the end, you get $$$[2, 3, 7, 8, 6, 1, 4, 5, 9]$$$ and now you need to sort $$$[2, 3, 7, 8, 6, 1, 4, 5]$$$. Hence, $$$f(x) = f(x-1) + x - \text{pos}(x)$$$. For each $$$x$$$, $$$x - \text{pos}(x)$$$ is actually the number of pairs $$$(i, j)$$$ ($$$1 \leq i < j \leq x$$$) such that $$$x = a_i > a_j$$$. So $$$f(x)$$$ is equal to the number of inversions.

#### Counting inversions in $$$O(n \log n)$$$

You can use a Fenwick tree (or a segment tree). There are other solutions (for example, using divide & conquer + merge sort), but they are usually harder to generalize.

For each $$$j$$$, calculate the number of $$$i < j$$$ such that $$$a_i > a_j$$$.

The Fenwick tree should contain the frequency of each value in $$$[1, n]$$$ in the prefix $$$[1, j - 1]$$$ of the array.

So, for each $$$j$$$, the queries look like

- $$$res := res + \text{range_sum}(a_j + 1, n)$$$
- add $$$1$$$ in the position $$$a_j$$$ of the Fenwick tree

#### Observations / slight variations of the problem

By using a Fenwick tree, you are actually calculating the number of inversions for each prefix of the array.

You can calculate the number of swaps required to sort an array (not necessarily a permutation, but for now let's assume that its elements are distinct) by compressing the values of the array. For example, the array $$$[13, 18, 34, 38, 28, 41, 5, 29, 30]$$$ becomes $$$[2, 3, 7, 8, 6, 9, 1, 4, 5]$$$.

You can also calculate the number of swaps required to get an array $$$b$$$ (for now let's assume that its elements are distinct) starting from $$$a$$$, by renaming the values. For example,

$$$a = [2, 3, 7, 8, 6, 9, 1, 4, 5], b = [9, 8, 5, 2, 1, 4, 7, 3, 6]$$$

is equivalent to

$$$a = [4, 8, 7, 2, 9, 1, 5, 6, 3], b = [1, 2, 3, 4, 5, 6, 7, 8, 9]$$$

$$$a^{-1}$$$ (a permutation such that $$$(a^{-1})_{a_x} = x$$$, i.e. $$$(a^{-1})_x$$$ is equal to the position of $$$x$$$ in $$$a$$$) has the same number of inversions as $$$a$$$. For example, $$$[2, 3, 7, 8, 6, 9, 1, 4, 5]$$$ and $$$[7, 1, 2, 8, 9, 5, 3, 4, 6]$$$ have both $$$16$$$ inversions. Sketch of a proof: note that, when you swap two elements in adjacent positions in $$$a$$$, you are swapping two adjacent values in $$$a^{-1}$$$, and the number of inversions in $$$a^{-1}$$$ also increases by $$$1$$$ or decreases by $$$1$$$ (like in Proof 1).

## 1430E - String Reversal (rating: 1900)

**Hint 1**

**Hint 2**

**Hint 3**

**Solution**

## 103148B - Luna Likes Love (EGOI 2021/2)

**Hint 1**

**Hint 2**

**Hint 3**

**Hint 4**

**Solution**

## arc088_e (rating: 2231)

**Hint 1**

**Hint 2**

**Hint 3**

**Hint 4**

**Solution**

## arc097_e (rating: 2247)

**Hint 1**

**Hint 2**

**Hint 3**

**Hint 4**

**Solution**

## Other problems

IOI 2019/1

arc120_c (suggested by Ghassane)

Hackerearth — Swapping numbers (Inferno03)

Hackerearth — Make the strings equal (Inferno03)

1526D - Kill Anton (somil_jain_120)

JOI 2021/3 (Final Round) (you can submit here)

## Conclusions

We've seen that a lot of problems where you have to swap adjacent elements can be tackled with greedy observations, such as looking at the optimal relative positions of the values in the final array; then, a lot of these problems can be reduced to "find the number of inversions" or similar.

Of course, suggestions/corrections are welcome. In particular, please share in the comments other problems where you have to swap adjacent elements.

I hope you enjoyed the blog!