Finding repeated points in 2D plane

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

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 2

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 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!