Data Modelling & AI Data Structure & Algorithm

Count number of intersections points for given lines between (i, 0) and (j, 1)

23 June 2025

0

Given an array, lines[] of N pairs of the form (i, j) where (i, j) represents a line segment from coordinate (i, 0) to (j, 1), the task is to find the count of points of intersection of the given lines.

Example:

Input: lines[] = {{1, 2}, {2, 1}}
Output: 1
Explanation: For the given two pairs, the line form (1, 0) to (2, 1) intersect with the line from (2, 0) to (1, 1) at point (1.5, 0.5). Hence the total count of points of intersection is 1.

Input: lines[] = {{1, 5}, {2, 1}, {3, 7}, {4, 1}, {8, 2}}
Output: 5

Approach: The given problem can be solved using a Greedy approach using the policy-based data structure. It can be observed that for lines represented b two pairs (a, b) and (c, d) to intersect either (a > c and b < d) or (a < c and b > d) must hold true.

Therefore using this observation, the given array of pairs can be sorted in decreasing order of the 1^st element. While traversing the array, insert the value of the second element into the policy-based data structure and find the count of elements smaller than the second element of the inserted pair using the order_of_key function and maintain the sum of count in a variable. Similarly, calculate for the cases after sorting the given array of pairs in decreasing order of their 2^nd element.

Below is the implementation of the above approach:

C++

// C++ Program of the above approach
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
using namespace std;
 
// Defining Policy Based Data Structure
typedef tree<int, null_type,
             less_equal<int>, rb_tree_tag,
             tree_order_statistics_node_update>
    ordered_multiset;
 
// Function to count points
// of intersection of pairs
// (a, b) and (c, d)
// such that a > c and b < d
int cntIntersections(
    vector<pair<int, int> > lines,
    int N)
{
    // Stores the count
    // of intersection points
    int cnt = 0;
 
    // Initializing Ordered Multiset
    ordered_multiset s;
 
    // Loop to iterate the array
    for (int i = 0; i < N; i++) {
 
        // Add the count of integers
        // smaller than lines[i].second
        // in the total count
        cnt += s.order_of_key(lines[i].second);
 
        // Insert lines[i].second into s
        s.insert(lines[i].second);
    }
 
    // Return Count
    return cnt;
}
 
// Function to find the
// total count of points of
// intersections of all the given lines
int cntAllIntersections(
    vector<pair<int, int> > lines,
    int N)
{
    // Sort the array in decreasing
    // order of 1st element
    sort(lines.begin(), lines.end(),
         greater<pair<int, int> >());
 
    // Stores the total count
    int totalCnt = 0;
 
    // Function call for cases
    // with a > c and b < d
    totalCnt += cntIntersections(lines, N);
 
    // Swap all the pairs of the array in order
    // to calculate cases with a < c and b > d
    for (int i = 0; i < N; i++) {
        swap(lines[i].first, lines[i].second);
    }
 
    // Function call for cases
    // with a < c and b > d
    totalCnt += cntIntersections(lines, N);
 
    // Return Answer
    return totalCnt;
}
 
// Driver Code
int main()
{
    vector<pair<int, int> > lines{
        {1, 5}, {2, 1}, {3, 7}, {4, 1}, {8, 2}
    };
 
    cout << cntAllIntersections(lines,
                                lines.size());
 
    return 0;
}

Java

import java.io.*;
import java.lang.*;
import java.util.*;
// Importing the policy-based data structure
import java.util.TreeSet;
import java.util.function.*;
 
class Main {
    // Function to count points
    // of intersection of pairs
    // (a, b) and (c, d)
    // such that a > c and b < d
    static int cntIntersections(Pair[] lines, int N)
    {
        // Stores the count
        // of intersection points
        int cnt = 0;
 
        // Initializing TreeSet
        TreeSet<Integer> s = new TreeSet<Integer>();
 
        // Loop to iterate the array
        for (int i = 0; i < N; i++) {
 
            // Add the count of integers
            // smaller than lines[i].second
            // in the total count
            cnt += s.headSet(lines[i].b, true).size();
 
            // Insert lines[i].second into s
            s.add(lines[i].b);
        }
 
        // Return Count
        return cnt;
    }
 
    // Function to find the
    // total count of points of
    // intersections of all the given lines
    static int cntAllIntersections(Pair[] lines, int N)
    {
        // Sort the array in decreasing
        // order of 1st element
        Arrays.sort(lines, new Comparator<Pair>() {
            public int compare(Pair p1, Pair p2)
            {
                if (p1.a == p2.a) {
                    return p1.b - p2.b;
                }
                return p2.a - p1.a;
            }
        });
 
        // Stores the total count
        int totalCnt = 0;
 
        // Function call for cases
        // with a > c and b < d
        totalCnt += cntIntersections(lines, N);
 
        // Swap all the pairs of the array in order
        // to calculate cases with a < c and b > d
        for (int i = 0; i < N; i++) {
            int temp = lines[i].a;
            lines[i].a = lines[i].b;
            lines[i].b = temp;
        }
 
        // Function call for cases
        // with a < c and b > d
        totalCnt += cntIntersections(lines, N);
 
        // Return Answer
        return totalCnt;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        Pair[] lines = { new Pair(1, 5), new Pair(2, 1),
                         new Pair(3, 7), new Pair(4, 1),
                         new Pair(8, 2) };
 
        System.out.println(
            cntAllIntersections(lines, lines.length));
    }
 
    // Pair class to represent a pair of integers
    static class Pair {
        int a, b;
 
        Pair(int a, int b)
        {
            this.a = a;
            this.b = b;
        }
    }
}

Python3

# Python3 implementation of the above approach
 
# Importing in-built module for sorting and bisect_left method
from bisect import *
 
# Defining function to count points of intersection of pairs
def cntIntersections(lines, N):
    # Stores the count of intersection points
    cnt = 0
 
    # Initializing list to store ending points of lines
    s = []
 
    # Loop to iterate the array
    for i in range(N):
        # Add the count of integers smaller than lines[i][1] in the total count
        cnt += bisect_left(s, lines[i][1])
        # Insert lines[i][1] into s
        s.append(lines[i][1])
        s.sort()
         
    # Return Count
    return cnt
 
# Function to find the total count of points of intersections of all the given lines
def cntAllIntersections(lines, N):
    # Sort the array in decreasing order of 1st element
    lines = sorted(lines, reverse=True)
 
    # Stores the total count
    totalCnt = 0
 
    # Function call for cases with a > c and b < d
    totalCnt += cntIntersections(lines, N)
 
    # Swap all the pairs of the array in order to calculate cases with a < c and b > d
    lines = [(b, a) for (a, b) in lines]
 
    # Function call for cases with a < c and b > d
    totalCnt += cntIntersections(lines, N)
 
    # Return Answer
    return totalCnt
 
# Driver Code
if __name__ == '__main__':
    lines = [(1, 5), (2, 1), (3, 7), (4, 1), (8, 2)]
    print(cntAllIntersections(lines, len(lines)))

C#

using System;
using System.Collections.Generic;
 
public class Program {
    // Function to count points
    // of intersection of pairs
    // (a, b) and (c, d)
    // such that a > c and b < d
    static int cntIntersections(Pair[] lines, int N)
    {
        // Stores the count
        // of intersection points
        int cnt = 0;
        // Initializing SortedSet
        SortedSet<int> s = new SortedSet<int>();
 
        // Loop to iterate the array
        for (int i = 0; i < N; i++) {
            // Add the count of integers
            // smaller than lines[i].second
            // in the total count
            cnt += s.GetViewBetween(int.MinValue,
                                    lines[i].b)
                       .Count;
 
            // Insert lines[i].second into s
            s.Add(lines[i].b);
        }
 
        // Return Count
        return cnt;
    }
 
    // Function to find the
    // total count of points of
    // intersections of all the given lines
    static int cntAllIntersections(Pair[] lines, int N)
    {
        // Sort the array in decreasing
        // order of 1st element
        Array.Sort(lines, new PairComparer());
 
        // Stores the total count
        int totalCnt = 0;
 
        // Function call for cases
        // with a > c and b < d
        totalCnt += cntIntersections(lines, N);
 
        // Swap all the pairs of the array in order
        // to calculate cases with a < c and b > d
        for (int i = 0; i < N; i++) {
            int temp = lines[i].a;
            lines[i].a = lines[i].b;
            lines[i].b = temp;
        }
 
        // Function call for cases
        // with a < c and b > d
        totalCnt += cntIntersections(lines, N);
 
        // Return Answer
        return totalCnt;
    }
 
    // Driver Code
    public static void Main()
    {
        Pair[] lines = { new Pair(1, 5), new Pair(2, 1),
                         new Pair(3, 7), new Pair(4, 1),
                         new Pair(8, 2) };
 
        Console.WriteLine(
            cntAllIntersections(lines, lines.Length));
    }
 
    // Pair class to represent a pair of integers
    public class Pair {
        public int a, b;
 
        public Pair(int a, int b)
        {
            this.a = a;
            this.b = b;
        }
    }
 
    // PairComparer class to compare pairs in decreasing
    // order
    public class PairComparer : IComparer<Pair> {
        public int Compare(Pair p1, Pair p2)
        {
            if (p1.a == p2.a) {
                return p1.b - p2.b;
            }
            return p2.a - p1.a;
        }
    }
}

Javascript

// Defining function to count points of intersection of pairs
function cntIntersections(lines, N) {
    // Stores the count of intersection points
    let cnt = 0;
 
    // Initializing list to store ending points of lines
    let s = [];
 
    // Loop to iterate the array
    for (let i = 0; i < N; i++) {
        // Add the count of integers smaller than lines[i][1] in the total count
        cnt += s.filter(x => x < lines[i][1]).length;
        // Insert lines[i][1] into s
        s.push(lines[i][1]);
        s.sort((a, b) => a - b);
    }
 
    // Return Count
    return cnt;
}
 
// Function to find the total count of points of intersections of all the given lines
function cntAllIntersections(lines, N) {
    // Sort the array in decreasing order of 1st element
    lines = lines.sort((a, b) => b[0] - a[0]);
 
    // Stores the total count
    let totalCnt = 0;
 
    // Function call for cases with a > c and b < d
    totalCnt += cntIntersections(lines, N);
 
    // Swap all the pairs of the array in order to calculate cases with a < c and b > d
    lines = lines.map(x => [x[1], x[0]]);
 
    // Function call for cases with a < c and b > d
    totalCnt += cntIntersections(lines, N);
 
    // Return Answer
    return totalCnt;
}
 
// Driver Code
let lines = [[1, 5], [2, 1], [3, 7], [4, 1], [8, 2]];
console.log(cntAllIntersections(lines, lines.length));

Output:

Time Complexity: O(N*log N)
Auxiliary Space: O(N)

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Count number of intersections points for given lines between (i, 0) and (j, 1)

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