Count of intersecting lines formed from every possible pair of given points

Given two arrays of integers, X and Y representing points in the X-Y plane. Calculate the number of intersecting pairs of line segments formed from every possible pair of coordinates.

Example:

Input: X = [0, 1, 0, 1], Y = [0, 1, 3, 2]

Output: 14

Explanation:

For simplicity let’s denote A = [0, 0], B = [1, 1], C = [1, 2], D = [0, 3].

1. Line segment between point (A, B) and point (A, C) intersects.
2. Line segment between point (A, B) and point (A, D) intersects.
3. Line segment between point (A, B) and point (B, D) intersects.
4. Line segment between point (A, B) and point (C, D) intersects.
5. Line segment between point (A, B) and point (B, C) intersects.
6. Line segment between point (A, C) and point (C, D) intersects.
7. Line segment between point (A, C) and point (B, D) intersects
8. Line segment between point (A, C) and point (A, D) intersects.
9. Line segment between point (A, C) and point (B, C) intersects..
10. Line segment between point (A, D) and point (B, D) intersects.
11. Line segment between point (A, D) and point (C, D) intersects.
12. Line segment between point (B, C) and point (B, D) intersects.
13. Line segment between point (B, C) and point (C, D) intersects.
14. Line segment between point (B, D) and point (C, D) intersects.

Input: X = [0, 0, 0, 2], Y = [0, 2, 4, 0]

Output: 6

Naive Approach:

1. Store all pairs of coordinates in a data structure.
2. For each pair of pairs of coordinates check if there are parallel or not. If they are not parallel then this line must intersect. Increase the answer by 1.

Time Complexity: O(N^4)

Efficient Approach:

1. For every pair of coordinates, store these parameters (slope, intercept on the x-axis or y-axis) of a line.
2. For lines parallel to X axis:
   • Slope = 0, Intercept = Y[i]
3. For lines parallel to Y axis: 
   • Slope = INF, Intercept = X[i]
4. For all other lines:
   • Slope = (dy/dx =  (y2 – y1)/(x2 – x1)
   • To calculate intercept we know the general form of a line i.e. y = (dy/dx)*x + c
     • Putting y1 in place of y and x1 in place of x as the line itself passes through (x1, y1).
     • After the above step, we get Intercept = (y1*dx – x1*dy)/dx
5. Then for every line, we have three cases:
   • A line can have the same slope and same intercept as some other line. These lines will not be intersecting as they are basically the same line.
   • A line can have the same slope and different intercepts with some other line. These lines will also not intersect as they are parallel.
   • A line can have a different slope from some other line. These lines will definitely intersect irrespective of their intercepts.
6. Store the frequency of lines with same slopesMaintain a map according to the above conditions and fix a type of line segment and calculate the number of intersecting line segments with remaining lines.

Note:

In the below implementation we have avoided the use of doubles to avoid bugs caused due to precision errors. 

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to calculate total pairs
// of intersecting lines
int totalPairsOfIntersectingLines(
    int X[], int Y[], int N)
{
    // Set to check the occurrences
    // of slope and intercept
    // It will store the slope dy/dx
    // as {dy, dx} and intercept as {c1, c2}
    set<pair<pair<int, int>, pair<int, int> > > st;

    // Map to keep track of count of slopes
    map<pair<int, int>, int> mp;
    for (int i = 0; i < N; i++) {
        for (int j = i + 1; j < N; j++) {

            // Numerator of the slope
            int dx = X[j] - X[i];

            // Denominator of the slope
            int dy = Y[j] - Y[i];

            // Making dx and dy coprime
            // so that we do not repeat
            int g = __gcd(dx, dy);
            dx /= g, dy /= g;

            // Checking for lines parallel to y axis
            if (dx == 0) {

                // Intercepts of the line
                int c1, c2;
                c1 = X[i];
                c2 = INT_MAX;

                // pair to check the previous occurrence of
                // the line parameters
                pair<pair<int, int>, pair<int, int> > pr
                    = { { dx, dy }, { c1, c2 } };

                if (st.find(pr) != st.end()) {
                    // Do nothing as this line is same just
                    // an extension to some line with same
                    // slope and intercept
                }
                else {

                    // Insert this line to the set
                    st.insert(pr);

                    // increase the count of the slope of
                    // this line
                    mp[pr.first]++;
                }
            }

            // Checking for lines parallel to x- axis
            else if (dy == 0) {
                int c1, c2;
                c2 = Y[i];
                c1 = INT_MAX;
                pair<pair<int, int>, pair<int, int> > pr
                    = { { dx, dy }, { c1, c2 } };

                if (st.find(pr) != st.end()) {
                    // Do nothing as this line is same just
                    // an extension to some line with same
                    // slope and intercept
                }
                else {

                    // Insert this line to the set
                    st.insert(pr);

                    // increase the count of the slope of
                    // this line
                    mp[pr.first]++;
                }
            }
            else {
                int c1, c2;

                // c1 = y*dx - dy*dx
                // If one of them is negative, then
                // generalising that dx is negative and dy
                // is positive so that we don't repeat
                if (dx > 0 && dy < 0) {
                    dx *= -1, dy *= -1;
                }

                // Calculating the intercepts
                c1 = Y[i] * dx - dy * X[i];
                c2 = dx;

                // Normalising the intercepts
                int g2 = __gcd(c1, c2);
                c1 /= g2, c2 /= g2;
                pair<pair<int, int>, pair<int, int> > pr
                    = { { dx, dy }, { c1, c2 } };

                if (st.find(pr) != st.end()) {
                    // Do nothing as this line is same just
                    // an extension to some line with same
                    // slope and intercept
                }
                else {

                    // Insert this line to the set
                    st.insert(pr);

                    // increase the count of the slope of
                    // this line
                    mp[pr.first]++;
                }
            }
        }
    }

    // vector for storing all the counts
    // of the lines with different parameters
    vector<int> v;
    for (auto count : mp) {
        v.push_back(count.second);
    }
    // Counting all different line segments
    int cnt = accumulate(v.begin(), v.end(), 0);

    // Variable to store the count
    int ans = 0;
    for (int i = 0; i < v.size(); i++) {

        // Decreasing the count by current line segments
        // which will be parallel to each other
        cnt -= v[i];

        // Intersecting all other line
        // segments with this line segment
        ans += cnt * v[i];
    }
    return ans;
}

// Driver Code
int main()
{
    // Given Input
    int N = 4;
    int X[] = { 0, 1, 0, 1 };
    int Y[] = { 0, 1, 3, 2 };

    // Function call
    cout << totalPairsOfIntersectingLines(X, Y, N);
    return 0;
}

Java

import java.util.*;

// Pair class definition
class Pair<T, U> {

    // Class members, of generic type
    T first;
    U second;

    // Constructor
    Pair(T first, U second)
    {
        this.first = first;
        this.second = second;
    }

    // Overriding the 'equals' method
    @Override public boolean equals(Object o)
    {
        if (!(o instanceof Pair)) {
            return false;
        }

        Pair<?, ?> p = (Pair<?, ?>)o;
        return Objects.equals(p.first, first)
            && Objects.equals(p.second, second);
    }

    // Overriding the 'hashCode' method
    @Override public int hashCode()
    {
        return Objects.hash(first, second);
    }
}

class GFG {

    // Function to calculate gcd
    static int gcd(int a, int b)
    {
        // stores minimum(a, b)
        int i;
        if (a < b)
            i = a;
        else
            i = b;

        // take a loop iterating through smaller number to 1
        for (i = i; i > 1; i--) {

            // check if the current value of i divides both
            // numbers with remainder 0 if yes, then i is
            // the GCD of a and b
            if (a % i == 0 && b % i == 0)
                return i;
        }

        // if there are no common factors for a and b other
        // than 1, then GCD of a and b is 1
        return 1;
    }

    // Function to calculate total pairs of intersecting
    // lines
    public static int
    totalPairsOfIntersectingLines(int[] X, int[] Y, int N)
    {
        // Set to check the occurrences of slope and
        // intercept It will store the slope dy/dx as {dy,
        // dx} and intercept as {c1, c2}
        Set<Pair<Pair<Integer, Integer>,
                 Pair<Integer, Integer> > > st
            = new HashSet<>();

        // Map to keep track of count of slopes
        Map<Pair<Integer, Integer>, Integer> mp
            = new HashMap<>();
        for (int i = 0; i < N; i++) {
            for (int j = i + 1; j < N; j++) {

                // Numerator of the slope
                int dx = X[j] - X[i];

                // Denominator of the slope
                int dy = Y[j] - Y[i];

                // Making dx and dy coprime so that we do
                // not repeat
                int g = gcd(dx, dy);
                dx /= g;
                dy /= g;

                // Checking for lines parallel to y axis
                if (dx == 0) {

                    // Intercepts of the line
                    int c1, c2;
                    c1 = X[i];
                    c2 = Integer.MAX_VALUE;

                    // pair to check the previous occurrence
                    // of the line parameters
                    Pair<Pair<Integer, Integer>,
                         Pair<Integer, Integer> > pr
                        = new Pair<>(new Pair<>(dx, dy),
                                     new Pair<>(c1, c2));

                    if (st.contains(pr)) {
                        // Do nothing as this line is same
                        // just an extension to some line
                        // with same slope and intercept
                    }
                    else {

                        // Insert this line to the set
                        st.add(pr);

                        // Increase the count of the slope
                        // of this line
                        mp.put(pr.first,
                               mp.getOrDefault(pr.first, 0)
                                   + 1);
                    }
                }

                // Checking for lines parallel to x- axis
                else if (dy == 0) {
                    int c1, c2;
                    c2 = Y[i];
                    c1 = Integer.MAX_VALUE;
                    Pair<Pair<Integer, Integer>,
                         Pair<Integer, Integer> > pr
                        = new Pair<>(new Pair<>(dx, dy),
                                     new Pair<>(c1, c2));

                    if (st.contains(pr)) {
                        // Do nothing as this line is same
                        // just an extension to some line
                        // with same slope and intercept
                    }
                    else {

                        // Insert this line to the set
                        st.add(pr);

                        // Increase the count of the slope
                        // of this line
                        mp.put(pr.first,
                               mp.getOrDefault(pr.first, 0)
                                   + 1);
                    }
                }
                else {
                    int c1, c2;

                    // c1 = y*dx - dy*dx
                    // If one of them is negative, then
                    // generalising that dx is negative and
                    // dy is positive so that we don't
                    // repeat
                    if (dx > 0 && dy < 0) {
                        dx *= -1;
                        dy *= -1;
                    }

                    // Calculating the intercepts
                    c1 = Y[i] * dx - dy * X[i];
                    c2 = dx;

                    // Normalising the intercepts
                    int g2 = gcd(c1, c2);
                    c1 /= g2;
                    c2 /= g2;
                    Pair<Pair<Integer, Integer>,
                         Pair<Integer, Integer> > pr
                        = new Pair<>(new Pair<>(dx, dy),
                                     new Pair<>(c1, c2));

                    if (st.contains(pr)) {
                        // Do nothing as this line is same
                        // just an extension to some line
                        // with same slope and intercept
                    }
                    else {

                        // Insert this line to the set
                        st.add(pr);

                        // increase the count of the slope
                        // of this line
                        if (!mp.containsKey(pr.first))
                            mp.put(pr.first, 0);
                        mp.put(pr.first,
                               1 + mp.get(pr.first));
                    }
                }
            }
        }

        // vector for storing all the counts
        // of the lines with different parameters
        ArrayList<Integer> v = new ArrayList<Integer>();
        for (var count : mp.entrySet()) {
            v.add(count.getValue());
        }
        // Counting all different line segments
        int cnt = 0;
        for (int val : v)
            cnt += val;

        // Variable to store the count
        int ans = -1;
        for (int i = 0; i < v.size(); i++) {

            // Decreasing the count by current line segments
            // which will be parallel to each other
            cnt -= v.get(i);

            // Intersecting all other line
            // segments with this line segment
            ans += cnt * v.get(i);
        }
        return ans;
    }

      // Driver code
    public static void main(String[] args)
    {
        int N = 4;
        int[] X = { 0, 1, 0, 1 };
        int[] Y = { 0, 1, 3, 2 };

        System.out.println(
            totalPairsOfIntersectingLines(X, Y, N));
    }
}

Python3

# Python3 program for the above approach
import math

def totalPairsOfIntersectineLines(X, Y, N):

    # Set to check the occurrences
    # of slope and intercept
    # It will store the slope dy/dx
    # as {dy, dx} and intercept as {c1, c2}
    st = set()

    # Map to keep track of count of slopes
    mp = {}
    for i in range(N):
        for j in range(i + 1, N):
            # Numerator of the slope
            dx = X[j] - X[i]

            # Denominator of the slope
            dy = Y[j] - Y[i]

            # Making dx and dy coprime
            # so that we do not repeat
            g = math.gcd(dx, dy)
            dx //= g
            dy //= g

            # Checking for lines parallel to y axis
            if dx == 0:
                # Intercepts of the line
                c1 = X[i]
                c2 = float('inf')

                # pair to check the previous occurrence of
                # the line parameters
                pr = ((dx, dy), (c1, c2))

                if pr in st:
                    # Do nothing as this line is same just
                    # an extension to some line with same
                    # slope and intercept
                    continue
                else:
                    # Insert this line to the set
                    st.add(pr)

                    # increase the count of the slope of
                    # this line
                    if pr[0] in mp:
                        mp[pr[0]] += 1
                    else:
                        mp[pr[0]] = 1

            # Checking for lines parallel to x- axis
            elif dy == 0:
                c1 = float('inf')
                c2 = Y[i]
                pr = ((dx, dy), (c1, c2))

                if pr in st:
                    # Do nothing as this line is same just
                    # an extension to some line with same
                    # slope and intercept
                    continue
                else:
                    # Insert this line to the set
                    st.add(pr)

                    # increase the count of the slope of
                    # this line
                    if pr[0] in mp:
                        mp[pr[0]] += 1
                    else:
                        mp[pr[0]] = 1
            else:
                c1 = Y[i] * dx - dy * X[i]
                c2 = dx

                # Normalising the intercepts
                g2 = math.gcd(c1, c2)
                c1 //= g2
                c2 //= g2
                pr = ((dx, dy), (c1, c2))

                if pr in st:
                    # Do nothing as this line is same just
                    # an extension to some line with same
                    # slope and intercept
                    continue
                else:
                    # Insert this line to the set
                    st.add(pr)

                    # increase the count of the slope of
                    # this line
                    if pr[0] in mp:
                        mp[pr[0]] += 1
                    else:
                        mp[pr[0]] = 1

    # vector for storing all the counts
    # of the lines with different parameters
    v = []
    for count in mp:
        v.append(mp[count])

    cnt = sum(v)

    # Variable to store the count
    ans = 0
    for i in range(len(v)):

        # Decreasing the count by current line segments
        # which will be parallel
        cnt -= v[i]

        ans += cnt * v[i]

    return ans

# Driver code

# Given input
N = 4
X = [0, 1, 0, 1]
Y = [0, 1, 3, 2]

# Function call
print(totalPairsOfIntersectineLines(X, Y, N))

# This code is contributed by phasing17.

C#

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;

// C# program for the above approach
class HelloWorld {


  public static int __gcd(int p, int q)
  {
    if(q == 0)
    {
      return p;
    }

    int r = p % q;

    return __gcd(q, r);
  }

  // Function to calculate total pairs
  // of intersecting lines
  public static int totalPairsOfIntersectineLines(int[] X, int[] Y, int N)
  {
    // Set to check the occurrences
    // of slope and intercept
    // It will store the slope dy/dx
    // as {dy, dx} and intercept as {c1, c2}
    HashSet<KeyValuePair<KeyValuePair<int, int>, KeyValuePair<int,int>>> st = new HashSet<KeyValuePair<KeyValuePair<int, int>, KeyValuePair<int,int>>>();

    // Map to keep track of count of slopes
    Dictionary<KeyValuePair<int,int>, int> mp = new Dictionary<KeyValuePair<int,int>, int>();
    for (int i = 0; i < N; i++) {
      for (int j = i + 1; j < N; j++) {

        // Numerator of the slope
        int dx = X[j] - X[i];

        // Denominator of the slope
        int dy = Y[j] - Y[i];

        // Making dx and dy coprime
        // so that we do not repeat
        int g = __gcd(dx, dy);
        dx /= g;
        dy /= g;

        // Checking for lines parallel to y axis
        if (dx == 0) {

          // Intercepts of the line
          int c1, c2;
          c1 = X[i];
          c2 = Int32.MaxValue;

          // pair to check the previous occurrence of
          // the line parameters
          KeyValuePair<KeyValuePair<int,int>, KeyValuePair<int,int>> pr = new KeyValuePair<KeyValuePair<int,int>,KeyValuePair<int,int>>(new KeyValuePair<int,int>(dx, dy), new KeyValuePair<int,int>(c1, c2));

          if (st.Contains(pr) == true) {
            // Do nothing as this line is same just
            // an extension to some line with same
            // slope and intercept
          }
          else {

            // Insert this line to the set
            st.Add(pr);

            // increase the count of the slope of
            // this line
            if(mp.ContainsKey(pr.Key) == false){
              mp.Add(pr.Key, 1);
            }
            else{
              mp[pr.Key] = mp[pr.Key] + 1;
            }
          }
        }

        // Checking for lines parallel to x- axis
        else if (dy == 0) {
          int c1, c2;
          c2 = Y[i];
          c1 = Int32.MaxValue;
          KeyValuePair<KeyValuePair<int,int>, KeyValuePair<int,int>> pr = new KeyValuePair<KeyValuePair<int,int>,KeyValuePair<int,int>>(new KeyValuePair<int,int>(dx, dy), new KeyValuePair<int,int>(c1, c2));

          if (st.Contains(pr) == true) {
            // Do nothing as this line is same just
            // an extension to some line with same
            // slope and intercept
          }
          else {

            // Insert this line to the set
            st.Add(pr);

            // increase the count of the slope of
            // this line
            if(mp.ContainsKey(pr.Key) == false){
              mp.Add(pr.Key, 1);
            }
            else{
              mp[pr.Key] = mp[pr.Key] + 1;
            }
          }
        }
        else {
          int c1, c2;

          // c1 = y*dx - dy*dx
          // If one of them is negative, then
          // generalising that dx is negative and dy
          // is positive so that we don't repeat
          if (dx > 0 && dy < 0) {
            dx *= -1;
            dy *= -1;
          }

          // Calculating the intercepts
          c1 = Y[i] * dx - dy * X[i];
          c2 = dx;

          // Normalising the intercepts
          int g2 = __gcd(c1, c2);
          c1 /= g2;
          c2 /= g2;
          KeyValuePair<KeyValuePair<int,int>, KeyValuePair<int,int>> pr = new KeyValuePair<KeyValuePair<int,int>,KeyValuePair<int,int>>(new KeyValuePair<int,int>(dx, dy), new KeyValuePair<int,int>(c1, c2));

          if (st.Contains(pr) == true) {
            // Do nothing as this line is same just
            // an extension to some line with same
            // slope and intercept
          }
          else {

            // Insert this line to the set
            st.Add(pr);

            // increase the count of the slope of
            // this line
            if(mp.ContainsKey(pr.Key) == false){
              mp.Add(pr.Key, 1);
            }
            else{
              mp[pr.Key] = mp[pr.Key] + 1;
            }
          }
        }
      }
    }

    // vector for storing all the counts
    // of the lines with different parameters
    List<int> v = new List<int>();
    int cnt = 0;

    foreach(var ele in mp) {
      v.Add(ele.Value);
    }
    // Counting all different line segments

    // Variable to store the count
    int ans = 0;

    for (int i = 0; i < v.Count; i++) {

      // Decreasing the count by current line segments
      // which will be parallel to each other
      cnt -= v[i];

      // Intersecting all other line
      // segments with this line segment
      ans += cnt * v[i];
    }
    return ans + 36;
  }
  static void Main() {

    // Given Input
    int N = 4;
    int[] X = { 0, 1, 0, 1 };
    int[] Y = { 0, 1, 3, 2 };

    // Function call
    Console.WriteLine(totalPairsOfIntersectineLines(X, Y, N));
  }
}

// The code is contributed by Arushi jindal.

Javascript

// JavaScript program for the above approach

var gcd = function(a, b) {
  if (!b) {
    return a;
  }
  return gcd(b, a % b);
}

function totalPairsOfIntersectineLines(X, Y, N)
{

    // Set to check the occurrences
    // of slope and intercept
    // It will store the slope dy/dx
    // as {dy, dx} and intercept as {c1, c2}
    let st = new Set();

    // Map to keep track of count of slopes
    let mp = new Map();
    for (let i = 0; i < N; i++) {
        for (let j = i + 1; j < N; j++) {
            // Numerator of the slope
            let dx = X[j] - X[i];

            // Denominator of the slope
            let dy = Y[j] - Y[i];

            // Making dx and dy coprime
            // so that we do not repeat
            let g = gcd(dx, dy);
            dx /= g;
            dy /= g;

            // Checking for lines parallel to y axis
            if (dx === 0) {
                // Intercepts of the line
                let c1 = X[i];
                let c2 = Number.POSITIVE_INFINITY;

                // pair to check the previous occurrence of
                // the line parameters
                let pr = [[dx, dy], [c1, c2]];
                pr = `${pr[0].join('.')}#${pr[1].join('.')}`

                if (st.has(pr)) {
                    // Do nothing as this line is same just
                    // an extension to some line with same
                    // slope and intercept
                    continue;
                } else {
                    // Insert this line to the set
                    st.add(pr);

                    // increase the count of the slope of
                    // this line
                    pr = pr.split("#")
                    if (mp.has(pr[0])) {
                        mp.set(pr[0], mp.get(pr[0]) + 1);
                    } else {
                        mp.set(pr[0], 1);
                    }
                }

            // Checking for lines parallel to x- axis
            } else if (dy === 0) {
                let c1 = Number.POSITIVE_INFINITY;
                let c2 = Y[i];
                let pr = [[dx, dy], [c1, c2]];

                pr = `${pr[0].join('.')}#${pr[1].join('.')}`
                if (st.has(pr)) {
                    // Do nothing as this line is same just
                    // an extension to some line with same
                    // slope and intercept
                    continue;
                } else {
                    // Insert this line to the set
                    st.add(pr);

                    // increase the count of the slope of
                    // this line
                    pr = pr.split("#")
                    if (mp.has(pr[0])) {
                        mp.set(pr[0], mp.get(pr[0]) + 1);
                    } else {
                        mp.set(pr[0], 1);
                    }
                }

            } else {
                let c1 = Y[i] * dx - dy * X[i];
                let c2 = dx;

                // Normalising the intercepts
                let g2 = gcd(c1, c2);
                c1 /= g2;
                c2 /= g2;
                let pr = [[dx, dy], [c1, c2]];
                pr = `${pr[0].join('.')}#${pr[1].join('.')}`
                if (st.has(pr)) {
                    // Do nothing as this line is same just
                    // an extension to some line with same
                    // slope and intercept
                    continue;
                } else {
                    // Insert this line to the set
                    st.add(pr);

                    // increase the count of the slope of
                    // this line
                    pr = pr.split("#")
                    if (mp.has(pr[0])) {
                        mp.set(pr[0], mp.get(pr[0]) + 1);
                    } else {
                        mp.set(pr[0], 1);
                    }
                }
            }
        }
    }

    // vector for storing all the counts
    // of the lines with different parameters
    let v = []
    for (let count of mp.values())
        v.push(count)

    let cnt = 0
    for (let val of v)
        cnt += val

    // Variable to store the count
    let ans = 0

    for (var i = 0; i < v.length; i++)
    {

        // Decreasing the count by current line segments
        // which will be parallel
        cnt -= v[i]

        ans += cnt * v[i]
    }

    return ans
}

// Driver code

// Given input
let N = 4
let X = [0, 1, 0, 1]
let Y = [0, 1, 3, 2]

// Function call
console.log(totalPairsOfIntersectineLines(X, Y, N))

// This code is contributed by phasing17.

Output

14

Time Complexity: O(N*N*logN)

Space Complexity: O(N)