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:
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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].
- Line segment between point (A, B) and point (A, C) intersects.
- Line segment between point (A, B) and point (A, D) intersects.
- Line segment between point (A, B) and point (B, D) intersects.
- Line segment between point (A, B) and point (C, D) intersects.
- Line segment between point (A, B) and point (B, C) intersects.
- Line segment between point (A, C) and point (C, D) intersects.
- Line segment between point (A, C) and point (B, D) intersects
- Line segment between point (A, C) and point (A, D) intersects.
- Line segment between point (A, C) and point (B, C) intersects..
- Line segment between point (A, D) and point (B, D) intersects.
- Line segment between point (A, D) and point (C, D) intersects.
- Line segment between point (B, C) and point (B, D) intersects.
- Line segment between point (B, C) and point (C, D) intersects.
- 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:
- Store all pairs of coordinates in a data structure.
- 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:
- For every pair of coordinates, store these parameters (slope, intercept on the x-axis or y-axis) of a line.
- For lines parallel to X axis:
- Slope = 0, Intercept = Y[i]
- For lines parallel to Y axis:
- Slope = INF, Intercept = X[i]
- 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
- 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.
- 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 totalPairsOfIntersectineLines( int X[], int Y[], int N) { // Set to check the occurences // 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 occurence 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 etenstion 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 etenstion 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 etenstion 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 << totalPairsOfIntersectineLines(X, Y, N); return 0; } |
14
Time Complexity: O(N*N*logN)
Space Complexity: O(N)