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# Custom Comparators and Coordinate Compression

Authors: Darren Yao, Siyong Huang, Michael Cao, Benjamin Qi, Nathan Chen

Using a custom comparator to sort custom objects or values in a non-default order, and compressing values from a large range to a smaller one.

### Prerequisites

Resources
IUSACO

partially based off this

CPH

short overview of what this module will cover

## Example - Wormhole Sort

Focus Problem – try your best to solve this problem before continuing!

View Internal Solution

We won't discuss the full solution here, as some of the concepts necessary for solving this problem will be introduced later in Silver. However, many solutions to this problem start by sorting the edges in nondecreasing order of weight. For example, the sample contains the following edges:

1 2 9
1 3 7
2 3 10
2 4 3

After sorting, it should look like

2 4 3
1 3 7
1 2 9
2 3 10

With C++, the easiest method is to use a vector of nested pairs:

#include <bits/stdc++.h>using namespace std;
#define f first#define s second
int main() {	int M = 4;	vector<pair<int, pair<int, int>>> v;	for (int i = 0; i < M; ++i) {

or a vector of array<int,3>s or vector<int>s:

int main() {	int M = 4;	vector<array<int, 3>> v;  // or vector<vector<int>>	for (int i = 0; i < M; ++i) {		int a, b, w;		cin >> a >> b >> w;		v.push_back({w, a, b});	}	sort(begin(v), end(v));	for (auto e : v) cout << e << " " << e << " " << e << "\n";}

In Python, we can use a list of lists.

But in Java, we can't sort an ArrayList of ArrayLists without writing some additional code. What should we do?

• If we only stored the edge weights and sorted them, we would have a sorted list of edge weights, but it would be impossible to tell which weights corresponded to which edges.
• However, if we create a class representing the edges and define a custom comparator to sort them by weight, we can sort the edges in ascending order while also keeping track of their endpoints.

## Classes

First, we need to define a class that represents what we want to sort. In our example we will define a class Edge that contains the two endpoints of the edge and the weight.

C++

### C++

A C++ struct is the same as a class in C++, but all members are public by default.

#include <bits/stdc++.h>using namespace std;
struct Edge {	int a, b, w;};
/* alternatively,class Edge {    public:

Java

import java.util.*;
public class Sol {	static class Edge {		int a, b, w;		public Edge(int _a, int _b, int _w) {			a = _a;			b = _b;			w = _w;		}

Python

class Edge:	def __init__(self, a, b, w):		self.a = a		self.b = b		self.w = w

v = []M = 4for i in range(M):	a, b, w = map(int, input().split())	v.append(Edge(a, b, w))for e in v:	print(e.a, e.b, e.w)

## Comparators

Normally, sorting functions rely on moving objects with a lower value in front of objects with a higher value if sorting in ascending order, and vice versa if in descending order. This is done through comparing two objects at a time.

C++

What a comparator does is compare two objects as follows, based on our comparison criteria:

• If object $x$ is less than object $y$, return true
• If object $x$ is greater than or equal to object $y$, return false

Essentially, the comparator determines whether object $x$ belongs to the left of object $y$ in a sorted ordering.

### Warning!

A comparator must return false for two equal objects (not doing so results in undefined behavior and potentially a verdict of wrong answer or runtime error).

In addition to returning the correct answer, comparators should also satisfy the following conditions:

• The function must be consistent with respect to reversing the order of the arguments: if $x \neq y$ and compare(x, y) is true, then compare(y, x) should be false and vice versa.
• The function must be transitive. If compare(x, y) is true and compare(y, z) is true, then compare(x, z) should also be true. If the first two compare functions both return false, the third must also return false.

This is the easiest to implement. However, it only works for objects (not primitives) and it doesn't allow you to define multiple ways to compare the same type of class.

In the context of Wormhole Sort (note the use of const Edge&):

#include <bits/stdc++.h>using namespace std;
struct Edge {	int a, b, w;	bool operator<(const Edge &y) { return w < y.w; }};
int main() {	int M = 4;

We can also overload the operator outside of the class:

struct Edge {	int a, b, w;};bool operator<(const Edge &x, const Edge &y) { return x.w < y.w; }

or within it using friend:

struct Edge {	int a, b, w;	friend bool operator<(const Edge &x, const Edge &y) { return x.w < y.w; }};

### Method 2 - Comparison Function

This works for both objects and primitives, and you can declare many different comparators for the same object.

#include <bits/stdc++.h>using namespace std;
struct Edge {	int a, b, w;};
bool cmp(const Edge &x, const Edge &y) { return x.w < y.w; }
int main() {

We can also use lambda expressions in C++11 or above:

sort(begin(v), end(v), [](const Edge &x, const Edge &y) { return x.w < y.w; });

Java

What a Comparator does is compare two objects as follows, based on our comparison criteria:

• If object $x$ is less than object $y$, return a negative integer.
• If object $x$ is greater than object $y$, return a positive integer.
• If object $x$ is equal to object $y$, return 0.

In addition to returning the correct number, comparators should also satisfy the following conditions:

• The function must be consistent with respect to reversing the order of the arguments: if compare(x, y) is positive, then compare(y, x) should be negative and vice versa.
• The function must be transitive. If compare(x, y) > 0 and compare(y, z) > 0, then compare(x, z) > 0. Same applies if the compare functions return negative numbers.
• Equality must be consistent. If compare(x, y) = 0, then compare(x, z) and compare(y, z) must both be positive, both negative, or both zero. Note that they don't have to be equal, they just need to have the same sign.

Java has default functions for comparing int, long, double types. The Integer.compare(), Long.compare(), and Double.compare() functions take two arguments $x$ and $y$ and compare them as described above.

There are two ways of implementing this in Java: Comparable, and Comparator. They essentially serve the same purpose, but Comparable is generally easier and shorter to code. Comparable is a function implemented within the class containing the custom object, while Comparator is its own class.

### Method 1 - Comparable

We'll need to put implements Comparable<Edge> into the heading of the class. Furthermore, we'll need to implement the compareTo method. Essentially, compareTo(x) is the compare function that we described above, with the object itself as the first argument, or compare(self, x).

When using Comparable, we can just call Arrays.sort(arr) or Collections.sort(list) on the array or list as usual.

import java.util.*;
public class Sol {	static class Edge implements Comparable<Edge> {		int a, b, w;		public Edge(int _a, int _b, int _w) {			a = _a;			b = _b;			w = _w;		}

### Method 2 - Comparator

If instead we choose to use Comparator, we'll need to declare a second class that implements Comparator<Edge>:

import java.util.*;
public class Sol {	static class Edge {		int a, b, w;		public Edge(int _a, int _b, int _w) {			a = _a;			b = _b;			w = _w;		}

When using Comparator, the syntax for using the built-in sorting function requires a second argument: Arrays.sort(arr, new Comp()), or Collections.sort(list, new Comp()).

Python

### Defining Less Than Operator

class Edge:	def __init__(self, a, b, w):		self.a = a		self.b = b		self.w = w
def __lt__(self, other):  # lt means less than		return self.w < other.w



### Key Function

This method maps an object to another comparable datatype with which to be sorted. This is the preferred method if you are only sorting something once. In this case we map edges to their weights.

class Edge:	def __init__(self, a, b, w):		self.a = a		self.b = b		self.w = w

v = []M = 4for i in range(M):	a, b, w = map(int, input().split())	v.append(Edge(a, b, w))v.sort(key=lambda x: x.w)for e in v:	print(e.a, e.b, e.w)

### Comparison Function

A comparison function in Python must satisfy the same properties as a comparator in Java. Note that old-style cmp functions are no longer supported, so the comparison function must be converted into a key function with cmp_to_key. Most of the time, it is better to use the key function, but in the rare case that the comparison function is not easily represented as a key function, we can use this.

from functools import cmp_to_key

class Edge:	def __init__(self, a, b, w):		self.a = a		self.b = b		self.w = w



## Variations

### Sorting in Decreasing Order of Weight

We can replace all occurrences of x.w < y.w with x.w > y.w in our C++ code. Similarly, we can replace all occurrences of Integer.compare(x, y) with -Integer.compare(x, y) in our Java code. In Python, we can pass the parameter reverse=True to the sort or sorted function.

### Sorting by Two Criteria

Now, suppose we wanted to sort a list of Edges in ascending order, primarily by weight and secondarily by first vertex (a). We can do this quite similarly to how we handled sorting by one criterion earlier. What the comparator function needs to do is to compare the weights if the weights are not equal, and otherwise compare first vertices.

C++

struct Edge {	int a, b, w;	bool operator<(const Edge &y) {		if (w != y.w) return w < y.w;		return a < y.a;	}};

Java

static class Edge implements Comparable<Edge> {	int a, b, w;	public Edge(int _a, int _b, int _w) {		a = _a;		b = _b;		w = _w;	}	public int compareTo(Edge y) {		if (w != y.w) return Integer.compare(w, y.w);		return Integer.compare(a, y.a);	}}

Python

In Python, tuples have a natural order based on their elements in order. We can take advantage of this to write a comparator:

class Edge:	def __init__(self, a, b, w):		self.a = a		self.b = b		self.w = w
def __lt__(self, other):  # lt means less than		return (self.w, self.a) < (other.w, other.a)

This also gives an easy way to write a key function to sort in this way:

edges: list[Edge]edges.sort(key=lambda edge: (edge.w, edge.a))

Sorting by an arbitrary number of criteria is done similarly.

C++

Java

With Java, we can implement a comparator for arrays of arbitrary length (although this might be more confusing than creating a separate class).

import java.util.*;
public class Sol {	static class Comp implements Comparator<int[]> {		public int compare(int[] a, int[] b) {			for (int i = 0; i < a.length; ++i)				if (a[i] != b[i]) return Integer.compare(a[i], b[i]);			return 0;		}	}

Python

## Coordinate Compression

Coordinate compression describes the process of mapping each value in a list to its index if that list was sorted. For example, the list $\{7, 3, 4, 1\}$ would be compressed to $\{3, 1, 2, 0\}$. Notice that $1$ is the least value in the first list, so it becomes $0$, and $7$ is the greatest value, so it becomes $3$, the largest index in the list.

When we have values from a large range, but we only care about their relative order (for example, if we have to know if one value is above another), coordinate compression is a simple way to help with implementation. For example, if we have a set of integers ranging from $0$ to $10^9$, we can't use them as array indices because we'd have to create an array of size $10^9$, which would surely cause a Memory Limit Exceeded verdict. However, if there are only $N \leq 10^6$ such integers, we can coordinate-compress their values, which guarantees that the values will all be in the range from $0$ to $N-1$, which can be used as array indices.

Focus Problem – try your best to solve this problem before continuing!

View Internal Solution

### Example 1

A good example of coordinate compression in action is in the solution of USACO Rectangular Pasture. Again, we won't delve into the full solution but instead discuss how coordinate compression is applied. Since the solution uses 2D-prefix sums (another Silver topic), it is helpful if all point coordinates are coordinate-compressed to the range $0$ to $N-1$ so they can be used as array indices. Without coordinate compression, creating a large enough array would result in a Memory Limit Exceeded verdict.

Below you will find the solution to Rectangular Pasture, which uses coordinate compression at the beginning. Observe how a custom comparator is used to sort the points:

C++

 Code Snippet: Solution code (Click to expand)
typedef pair<int, int> Point;bool ycomp(Point p, Point q) { return p.second < q.second; }
Code Snippet: Solution code (Click to expand)
sort(P, P + N);	for (int i = 0; i < N; i++) P[i].first = i + 1;	sort(P, P + N, ycomp);	for (int i = 0; i < N; i++) P[i].second = i + 1;
Code Snippet: Solution code (Click to expand)

Java

 Code Snippet: Solution code (Click to expand)
int[] xs = new int[n];		int[] ys = new int[n];
Integer[] cows = new Integer[n];		for (int j = 0; j < n; j++) {			xs[j] = in.nextInt();			ys[j] = in.nextInt();			cows[j] = j;

The solution uses a lambda function as the custom comparator, which our guide didn't discuss, but it should be apparent which coordinate (x or y) that the comparator is sorting by.

The solution to Rectangular Pasture directly replaces coordinates with their compressed values, and forgets the real values of the coordinates because they are unnecessary. However, there may be problems for which we need to also remember the original values of coordinates that we compress.

### Example 2

Focus Problem – try your best to solve this problem before continuing!

This problem will require prefix sums and coordinate compression. However, the implementation of coordinate compression in this solution will also require remembering values in addition to compressing them (as opposed to just replacing the original values, as in the last problem). If you just want to focus on the implementation of coordinate compression and how it can switch between compressed indices and original values, see the contracted code below. indices is a list of values that need to be compressed. After it gets sorted and has duplicate values removed, it is ready to use. The method getCompressedIndex takes in an original value, and binary searches for its position in indices to get its corresponding compressed index. To go from a compressed index to an original value, the code can just access that index in indices.

We also provide a more detailed explanation:

Detailed Explanation

C++

 Code Snippet: Solution code (Click to expand)
// finds the "compressed index" of a special index (a.k.a. its position in the// sorted list)int getCompressedIndex(int a) {	return lower_bound(indices.begin(), indices.end(), a) - indices.begin();}
Code Snippet: Solution code (Click to expand)	//========= COORDINATE COMPRESSION =======

Java

 Code Snippet: Solution code (Click to expand)	// finds the "compressed index" of a special index (a.k.a. its position in	// the sorted list)	static int getCompressedIndex(int a) {		return Collections.binarySearch(indices, a);	} 	Code Snippet: Solution code (Click to expand)		//========= COORDINATE COMPRESSION =======		TreeSet<Integer> temp = new TreeSet<Integer>(indices);		// Since temp is a set, all duplicate elements are removed

## Problems

### Pro Tip

Many of the problems below may use other Silver concepts, such as prefix sums.

StatusSourceProblem NameDifficultyTags
CSESEasy
Show TagsPrefix Sums, Sorting
SilverEasy
Show TagsPrefix Sums, Sorting
SilverNormal
Show TagsPrefix Sums, Sorting
SilverNormal
Show TagsSorting
SilverNormal
Show TagsSorting
GoldNormal
Show TagsPrefix Sums, Sorting
CFNormal
Show TagsSorting
CFNormal
Show TagsPrefix Sums, Sorting
CFNormal
Show TagsSorting
SilverHard
Show TagsSorting
SilverHard
Show TagsSorting
SilverVery Hard
Show TagsSorting

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