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 1st 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 2nd 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 Structuretypedef 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 < dint 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 linesint 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 Codeint 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 structureimport 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 methodfrom bisect import *# Defining function to count points of intersection of pairsdef 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 linesdef 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 Codeif __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 pairsfunction 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 linesfunction 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 Codelet lines = [[1, 5], [2, 1], [3, 7], [4, 1], [8, 2]];console.log(cntAllIntersections(lines, lines.length)); |
Output:
5
Time Complexity: O(N*log N)
Auxiliary Space: O(N)
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