Finding repeated points in 2D plane - c++

Question

I am trying to find repeated points from a given 10 points, where each points has x and y values. I have written the below code but can not get correct results. The output should be {3,5},{4,2},{2,4},{7,8}
#include <iostream>
#include<stdlib.h>
using namespace std;
struct point
{
int x;
int y;
};
void distinctPoints(point arr[], int size)
{
cout<<"Repeated Points"<<endl;
cout<<"x, y"<<endl;
for(int i = 0; i< size; i++)
for(int j = i+1; j< size; j++)
{
if ((arr[i].x==arr[j].x) && (arr[i].y==arr[j].y))
{
cout<<arr[j].x <<", "<<arr[j].y<<endl;
break;
}
}
}
int main()
{ int size=10;
point points[size]={{3,5},{4,2},{2,4},{3,5},{7,8},{7,8},{4,2},{7,8},{3,5},{2,4}};
distinctPoints(points, size);
return 0;
}

Answer 1

Your approach (once corrected, as VHS's answer did) could be fine for a small number of points, but, with a bigger set of data, an O(N2) algorithm could be too inefficient.
You can take advantage of the average costant time that takes inserting an element in a std::unordered_set, even if you are required to write a comparison function and an hash function for your class.
The algorithm presented below uses two unordered_set's:
uniques ends up storing all the elements that are present in the source container, without repetitions.
repeated stores a unique instance of only the elements that are present multiple times.
An element is copied to the output only if it's already present in uniques, but not in repeated.
#include <iostream>
#include <vector>
#include <unordered_set>
#include <algorithm>
#include <iterator>
struct point
{
int x, y;
bool operator== (point const& b) const
{
return x == b.x && y == b.y;
}
};
namespace std {
template<> struct hash<point>
{
std::size_t operator() (point const& p) const
{
return (std::hash<int>{}(p.x) << 1) ^ std::hash<int>{}(p.y);
}
};
}
std::ostream& operator<< (std::ostream& os, point const& pt)
{
return os << '(' << pt.x << ", " << pt.y << ')';
}
template<class InputIt, class OutputIt>
OutputIt copy_repeated_values(InputIt first, InputIt last, OutputIt dest)
{
using value_type = typename InputIt::value_type;
std::unordered_set<value_type> uniques, repeated;
return std::copy_if(
first, last, dest, [&] (value_type const& value) {
return
not uniques.insert(value).second &&
repeated.insert(value).second;
}
);
}
int main()
{
std::vector<point> points {
{3,5}, {4,2}, {2,4}, {3,5}, {7,8}, {7,8}, {4,2}, {7,8}, {3,5}, {2,4}
};
copy_repeated_values(
std::begin(points), std::end(points),
std::ostream_iterator<point>(std::cout, " ")
);
std::cout << '\n';
}
The output is:
(3, 5) (7, 8) (4, 2) (2, 4)

Answer 2

I have tweaked your distinctPoints method so that it doesn't print the duplicates multiple times even if the dups appear more than twice. See the following edits:
void distinctPoints(point arr[], int size)
{
point dups[size];
cout<<"Distinct Points"<<endl;
cout<<"x, y"<<endl;
for(int i = 0; i < size; i++)
for(int j = 0; j < size; j++) {
if ((arr[i].x==arr[j].x) && (arr[i].y==arr[j].y)) {
if(j < i) {
break;
}
else if( j == i) {
continue;
}
else {
cout<<arr[i].x <<", "<<arr[i].y<<endl;
break;
}
}
}
}

Answer 3

This should do what you are trying to achieve, I am making use of set and maps in c++ which takes care of unique entries.
The map keeps track of already visited points.
#include <iostream>
#include<stdlib.h>
#include <set>
#include <map>
using namespace std;
struct point
{
int x;
int y;
};
map<pair<int, int>, int> mapCountOfPoints;
set<pair<int, int> > disPoints;
void distinctPoints(point arr[], int size)
{
for(int i=0; i<size; i++) {
pair<int, int> temp = make_pair(arr[i].x, arr[i].y);
if(mapCountOfPoints.find(temp) != mapCountOfPoints.end()) {
disPoints.insert(temp);
} else {
mapCountOfPoints[temp] = 1;
}
}
// Now we will iterate over the set to get the distinct set of points
for(set<pair<int, int>>::iterator it=disPoints.begin(); it!=disPoints.end(); it++) {
cout<<it->first<<" "<<it->second<<endl;
}
}
int main()
{ int size=10;
point points[size]={{3,5},{4,2},{2,4},{3,5},{7,8},{7,8},{4,2},{7,8},{3,5},{2,4}};
distinctPoints(points, size);
return 0;
}
Hope this helps!