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 1stelement. 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++ 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;
} |
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 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)))
|
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;
}
}
} |
// 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)); |
5
Time Complexity: O(N*log N)
Auxiliary Space: O(N)