sort one array and other array following? - c++

here is the C++ sample
int a[1000] = {3,1,5,4}
int b[1000] = {7,9,11,3}
how do i make it so if i sort array a, array b also following array a
example
a[1000] = {1,3,4,5}
b[1000] = {9,7,3,11}
is it possible using sort function
sort(a,a+4)
but also sort array b aswell ?
edit: what if there are 3 arrays ?

Instead of using two arrays, can you use an array of pairs and then sort THAT using a special comparison functor rather than the default less-than operator?

The simplest way is to rearrange your data into an array-of-structs instead of a pair of arrays so that each datum is contiguous; then, you can use an appropriate comparator. For example:
struct CompareFirst
{
bool operator() (const std::pair<int,int>& lhs, const std::pair<int,int>& rhs)
{
return lhs.first < rhs.first;
}
};
// c[i].first contains a[i], c[i].second contains b[i] for all i
std::pair<int, int> c[1000];
std::sort(c, c+1000, CompareFirst());
If you can't refactor your data like that, then you need to define a custom class that acts as a RandomAccessIterator:
struct ParallalArraySortHelper
{
ParallelArraySortHelper(int *first, int *second)
: a(first), b(second)
{
}
int& operator[] (int index) { return a[index]; }
int operator[] const (int index) { return a[index]; }
ParallelArraySortHelper operator += (int distance)
{
a += distance;
b += distance;
return *this;
}
// etc.
// Rest of the RandomAccessIterator requirements left as an exercise
int *a;
int *b;
};
...
int a[1000] = {...};
int b[1000] = {...};
std::sort(ParallalArraySortHelper(a, b), ParallelArraySortHelper(a+1000, b+1000));

Generate an array the same size as the original, containing the indexes into the array: {0, 1, 2, 3}. Now use a custom comparator functor that compares the elements in an associated array rather than the indexes themselves.
template<typename T>
class CompareIndices
{
public:
CompareIndices(const T * array) : m_AssociatedArray(array) {}
bool operator() (int left, int right) const
{
return std::less(m_AssociatedArray[left], m_AssociatedArray[right]);
}
private:
const T * m_AssociatedArray;
};
std::sort(i, i+4, CompareIndices(a));
Once you have a sorted list of indices, you can apply it to the original array a, or any other b array you want.

Related

How to declare 2D dynamic array in C++

How to declare a 2D array in which the rows are dynamic but the rows length are fixed? Is there any way to do it without both row and rows length are dynamic?
P.S. I can't use STL containers or class string.
You can use std::vector and std::array for this.
std::vector is a dynamic-length array.
std::array is a fixed-length array, used as
array<int, 20> arr;
arr[10] = 42;
array<int, 20> anotherArr = arr; // copied here, as opposed to C arrays
int oldStyleArr[20];
oldStyleArr[10] = 42;
// int newStyleArr[20] = oldStyleArr; // error here
It is a convenient wrapper over the C-style array, and it provides value semantics and various conveniece methods like size().
So you can create array<vector<int>, 20> for an array of 20 dynamic vectors of ints or vector<array<int, 20>> for a dynamic vector of fixed-length arrays.
UPD: std::array works only with array bounds known at compile-time. If your array bounds are known only at runtime, you still can use std::vector's constructor from size and (optional) element:
int rowCount, columnCount;
cin >> rowCount >> columnCount;
using Row = vector<int>;
// create `vector` of `rowCount` rows,
// where each row is `vector` of `columnCount` ints
vector<Row> arr2d(rowCount, Row(columnCount));
However, that's not the most efficient solution because each row is allocated separately. You can solve this with a little wrapper over one-dimensional vector:
template<class T>
class Vector2D {
public:
Vector2D(int rows, int cols)
: data(rows*cols)
, rows(rows)
, cols(cols) {}
int rowCount() const { return rows; }
int columnCount() const { return cols; }
T& get(int r, int c) { return data[r*cols + c]; }
T const& get(int r, int c) const { return data[r*cols + c]; }
void addRow() {
data.resize(cols*(rows + 1));
}
// ...
private:
vector<T> data;
int rows;
int cols;
};
You can use std::vector for dynamic range arrays. You would want to instantiate a vector of arrays to achieve this.
What you are looking for is probably: std::vector<std::array<int, 20>> arr;
However, if STL containers are not an option you might need to implement the container yourself. A linked list is probably the easiest to implement.
Just declare a 1D array and calculate the index yourself like how compilers generate accesses to a multidimensional array. Each row has width items, so the [x][y] element will have the index x*width + y
template<typename T>
class twoD
{
const int width;
const int height;
T* data;
twoD(int w, int h) : width(w), height(h)
{
data = new T[width*height];
}
~twoD()
{
delete[] data;
}
T& at(int x, int y)
{
return data[x*width + y];
}
T const& at(int x, int y) const
{
return data[x*width + y];
}
}
twoD myArray(4, 8);
myArray.at(2, 3) = 5;
myArray.at(4, 5) = 6 - myArray.at(2, 3);
The same method can be used to access multidimensional arrays in any degree

Sorting a list by a struct member in c++

I have a list filled with this struct:
struct singlePaymentStruct
{
std::string payer;
int payment;
double amount;
std::time_t timeRec;
singlePaymentStruct() {
payer="Empty";
payment=0;
amount=0;
timeRec = time(0);
}
};
I want to be able to sort this list by any of the fields. How exactly do I do this?
I didn't quite understand how sort method works with something more complex than just a list of records...
Solution found:
singlePaymentList.sort( []( const singlePaymentStruct &a, const singlePaymentStruct &b)
{return a.payer > b.payer;}
);
1.overloading operator<
you can do this by overloading the < operator
struct Foo{
int bar;
bool operator<(Foo &x){
return bar < x.bar;
}
};
2.using lambda expressions
(what is lambda expression?)
Foo array[10];
std::sort(array,array + 10,[](Foo const &l, Foo const &r) {
return l.bar < r.bar; });
3.using custom compare functions
If the possible fields to be used for sorting are known prior, it may be easier to read to implement custom compare functions specifically for the sorting.
struct Foo {
int bar;
SpecialType daa; // Assume daa.IsLessThan() available.
static bool lessBar(const Foo& l, const Foo& r) {
return l.bar < r.bar;
}
static bool lessDaa(const Foo& l, const Foo& r) {
return l.daa.IsLessThan(r.daa);
}
};
Foo array1[10]; // To be sorted by Foo::bar
Foo array2[10]; // To be sorted by Foo::daa
std::sort(array1, array1+10, Foo::lessBar);
std::sort(array2, array2+10, Foo::lessDaa);
std::sort accepts a third optional parameter that is a comparator function. This function should behave as < between elements (i.e. return true when the first is "less than" the second.
For example to sort an std::vector of your structures on increasing payment value what you can do is:
std::sort(data.begin(), data.end(),
[](const singlePaymentStruct& a, const singlePaymentStruct& b) {
return a.payment < b.payment;
});
let the array be struct singlePaymentStruct a[N]
sort(a,a+N,cmp);
bool cmp(struct singlePaymentStruct x, struct singlePaymentStruct y)
{
return x.field < y.field ; //or anything you want to do and return boolean
}
How it works under the hood?
Simply put basically it uses some sorting algoritm like quicksort or mergesort.
Why do we specify comparator functor ?
Well we need that comparator functor to decide the ordering of elements.
The basic thing is in any sorting algortihm the basic operation is comparison..and if we can specify that we are basically controlling the sorting operation.
Hope now you get the pieces together. That's why cmp() takes two values which it will compare and based on which order them.

Multidimensional array: operator overloading

I have a class with a multidimensional array:
it is possible to create a one, two, ..., n dimensional array with this class
if the array has n dimensions, i want to use n operator[] to get an object:
example:
A a({2,2,2,2}];
a[0][1][1][0] = 5;
but array is not a vector of pointer which lead to other vectors etc...
so i want the operator[] to return a class object until the last dimension, then return a integer
This is a strongly simplified code, but it shows my problem:
The error i receive: "[Error] cannot convert 'A::B' to 'int' in initialization"
#include <cstddef> // nullptr_t, ptrdiff_t, size_t
#include <iostream> // cin, cout...
class A {
private:
static int* a;
public:
static int dimensions;
A(int i=0) {
dimensions = i;
a = new int[5];
for(int j=0; j<5; j++) a[j]=j;
};
class B{
public:
B operator[](std::ptrdiff_t);
};
class C: public B{
public:
int& operator[](std::ptrdiff_t);
};
B operator[](std::ptrdiff_t);
};
//int A::count = 0;
A::B A::operator[] (std::ptrdiff_t i) {
B res;
if (dimensions <= 1){
res = C();
}
else{
res = B();
}
dimensions--;
return res;
}
A::B A::B::operator[] (std::ptrdiff_t i){
B res;
if (dimensions <=1){
res = B();
}
else{
res = C();
}
dimensions--;
return res;
}
int& A::C::operator[](std::ptrdiff_t i){
return *(a+i);
}
int main(){
A* obj = new A(5);
int res = obj[1][1][1][1][1];
std::cout<< res << std::endl;
}
The operator[] is evaluated from left to right in obj[1][1]...[1], so obj[1] returns a B object. Suppose now you just have int res = obj[1], then you'll assign to a B object (or C object in the case of multiple invocations of []) an int, but there is no conversion from B or C to int. You probably need to write a conversion operator, like
operator int()
{
// convert to int here
}
for A, B and C, as overloaded operators are not inherited.
I got rid of your compiling error just by writing such operators for A and B (of course I have linking errors since there are un-defined functions).
Also, note that if you want to write something like obj[1][1]...[1] = 10, you need to overload operator=, as again there is no implicit conversion from int to A or your proxy objects.
Hope this makes sense.
PS: see also #Oncaphillis' comment!
vsoftco is totally right, you need to implement an overload operator if you want to actually access your elements. This is necessary if you want it to be dynamic, which is how you describe it. I actually thought this was an interesting problem, so I implemented what you described as a template. I think it works, but a few things might be slightly off. Here's the code:
template<typename T>
class nDimArray {
using thisT = nDimArray<T>;
T m_value;
std::vector<thisT*> m_children;
public:
nDimArray(std::vector<T> sizes) {
assert(sizes.size() != 0);
int thisSize = sizes[sizes.size() - 1];
sizes.pop_back();
m_children.resize(thisSize);
if(sizes.size() == 0) {
//initialize elements
for(auto &c : m_children) {
c = new nDimArray(T(0));
}
} else {
//initialize children
for(auto &c : m_children) {
c = new nDimArray(sizes);
}
}
}
~nDimArray() {
for(auto &c : m_children) {
delete c;
}
}
nDimArray<T> &operator[](const unsigned int index) {
assert(!isElement());
assert(index < m_children.size());
return *m_children[index];
}
//icky dynamic cast operators
operator T() {
assert(isElement());
return m_value;
}
T &operator=(T value) {
assert(isElement());
m_value = value;
return m_value;
}
private:
nDimArray(T value) {
m_value = value;
}
bool isElement() const {
return m_children.size() == 0;
}
//no implementation yet
nDimArray(const nDimArray&);
nDimArray&operator=(const nDimArray&);
};
The basic idea is that this class can either act as an array of arrays, or an element. That means that in fact an array of arrays COULD be an array of elements! When you want to get a value, it tries to cast it to an element, and if that doesn't work, it just throws an assertion error.
Hopefully it makes sense, and of course if you have any questions ask away! In fact, I hope you do ask because the scope of the problem you describe is greater than you probably think it is.
It could be fun to use a Russian-doll style template class for this.
// general template where 'd' indicates the number of dimensions of the container
// and 'n' indicates the length of each dimension
// with a bit more template magic, we could probably support each
// dimension being able to have it's own size
template<size_t d, size_t n>
class foo
{
private:
foo<d-1, n> data[n];
public:
foo<d-1, n>& operator[](std::ptrdiff_t x)
{
return data[x];
}
};
// a specialization for one dimension. n can still specify the length
template<size_t n>
class foo<1, n>
{
private:
int data[n];
public:
int& operator[](std::ptrdiff_t x)
{
return data[x];
}
};
int main(int argc, char** argv)
{
foo<3, 10> myFoo;
for(int i=0; i<10; ++i)
for(int j=0; j<10; ++j)
for(int k=0; k<10; ++k)
myFoo[i][j][k] = i*10000 + j*100 + k;
return myFoo[9][9][9]; // would be 090909 in this case
}
Each dimension keeps an array of previous-dimension elements. Dimension 1 uses the base specialization that tracks a 1D int array. Dimension 2 would then keep an array of one-dimentional arrays, D3 would have an array of two-dimensional arrays, etc. Then access looks the same as native multi-dimensional arrays. I'm using arrays inside the class in my example. This makes all the memory contiguous for the n-dimensional arrays, and doesn't require dynamic allocations inside the class. However, you could provide the same functionality with dynamic allocation as well.

use n_th element in a container, but with another key

I have two vectors. One that actually holds the data (let's say floats) and one that holds the indices. I want to pass at nth_element the indices vector, but I want the comparison to be done by the vector that actually holds the data. I was thinking about a functor, but this provides only the () operator I guess. I achieved that by making the data vector a global one, but of course that's not desired.
std::vector<float> v; // data vector (global)
bool myfunction (int i,int j) { return (v[i]<v[j]); }
int find_median(std::vector<int> &v_i)
{
size_t n = v_i.size() / 2;
nth_element(v_i.begin(), v_i.begin()+n, v_i.end(), myfunction);
return v_i[n];
}
You may use a functor like:
class comp_with_indirection
{
public:
explicit comp_with_indirection(const std::vector<float>& floats) :
floats(floats)
{}
bool operator() (int lhs, int rhs) const { return floats[lhs] < floats[rhs]; }
private:
const std::vector<float>& floats;
};
And then you may use it like:
int find_median(const std::vector<float>& v_f, std::vector<int> &v_i)
{
assert(!v_i.empty());
assert(v_i.size() <= v_f.size());
const size_t n = v_i.size() / 2;
std::nth_element(v_i.begin(), v_i.begin() + n, v_i.end(), comp_with_indirection(v_f));
return v_i[n];
}
Note: with C++11, you may use lambda instead of named functor class.
int find_median(const std::vector<float>& v_f, std::vector<int> &v_i)
{
assert(!v_i.empty());
assert(v_i.size() <= v_f.size());
const size_t n = v_i.size() / 2;
std::nth_element(
v_i.begin(), v_i.begin() + n, v_i.end(),
[&v_f](int lhs, int rhs) {
return v_f[lhs] < v_f[rhs];
});
return v_i[n];
}

Return one dimension of an array?

How would I only return one dimension of an array, while ignoring the other?
Such as:
int map[4][8];
int MapManager::getMapX(int x)
{
return map[x];
}
The return type should be int* or int[], not int:
int* MapManager::getMapX(int x) {
return map[x];
}
Other than that, you are fine.
With a 2 dimensional array you will only be able to directly return a one dimensional array for the inner array by converting to a pointer, such as
int map[4][8];
int* MapManager::getMapX(int x)
{
return map[x];
}
For the other 'dimension', you will need to have another external array to copy to:
int* MapManager::getMapY(int y, int *outArray, int numElements)
{
for(int i = 0; i < numElements; i++) {
outArray[i] = map[i][y];
}
}
This array needs to be allocated with the correct size. (8 in this case).
The reason for this is that the array elements for the y 'column' are not contiguous in memory, but spread across several arrays (which are the 'x' rows). C arrays rely on this contiguous concept for accessing the elements.
Since none of the other answers here return an actual array (pointers are not arrays), I thought I might show how to really return an array. Or the closest thing possible, which is a reference to an array.
typedef int row_type[8];
row_type& MapManager::getMapX(int x) {
return map[x];
}
What's the point of this? The pointer works the same!
No, it doesn't. The pointer loses type information, namely, the size. You can make begin() and end() functions work with arrays, but you can't with pointers:
// pointer version
int* MapManager::getMapX_ptr(int x) {
return map[x];
}
row_type& row = getMapX(0);
// row is of type int(&)[8]
// notice the size is not lost in the type!
std::sort(begin(row), end(row));
// compiles fine! end() correctly finds the end of the row
int* ptr = getMapX_ptr(0);
// notice how ptr has no size information at all
std::sort(begin(ptr), end(ptr));
// can't possible be made to work!
You can't write end for int*.
template <typename T, std::size_t N>
T* end(T(&arr)[N]) {
return &arr[0] + N;
}
template <typename T>
T* end(T* ptr) {
// what here?
// guess?
// pick a random number?
}
You could create a special class that provides a strided view of an array. That is to say, it skips over values. Here's the beginnings of something like that:
template<typename T>
class StridedView
{
public:
StridedView(T * b, int stride, int count)
:begin_(b), stride_(stride), count_(count)
{}
template<int N>
StridedView(T (&arr)[N])
:begin_(arr), stride_(1), count_(N)
{}
T & operator[](int index)
{
return *(begin_ + index * stride_);
}
const T & operator[](int index) const
{
return *(begin_ + index * stride_);
}
int size() const { return count_; }
private:
T * begin_;
int stride_;
int count_;
};
Then you could have functions that can get you a row or a column as appropriate:
template<typename T, int R, int C>
StridedView<T> GetRow(T (&arr)[R][C], int row)
{
T * begin = (*arr) + (row * C);
return StridedView<T>(begin, 1, C);
}
template<typename T, int R, int C>
StridedView<T> GetColumn(T (&arr)[R][C], int column)
{
T * begin = (*arr) + column;
return StridedView<T>(begin, C, R);
}
Demo