Given segments and some points, for each point find the number of segments covering that point.
A segment (l, r) covers a point x if and only if l < = x < = r.
Examples:
Input: Segments = {{0, 3}, {1, 3}, {3, 8}},
Points = {-1, 3, 8}.
Output : {0, 3, 1}
Explanation :
- No segments passing through point -1
- All the segments passing through point 3
- Segment 3rd passing through point 8
Input: Segments = {{1, 3}, {2, 4}, {5, 7}},
Points = {0, 2, 5}.
Output: {0, 2, 1}
Explanation :
- No segments passing through point 0
- 1st and 2nd segment passing through point 2
- Segment 3rd passing through point 5
Approach:
- We can do this by using logic similar to prefix sum.
- Let’s represent a segment with (l, r). Form a vector of pairs, for each segment push two pairs in vector with values (l, +1) ans (r + 1, -1).
- Sort the points in ascending order, but we also need it’s position so mapped it with it’s position.
- Sort the segment vector in descending order because we iterate on it from back.
- Make a variable count of segments, which is initially zero.
- Then, we will iterate on the point and pop the pair from the segment vector until it’s first value is less than equal to current point and add it’s second value to the count.
- Finally, Store the values of count in an array to his respective position and print the array.
Below is the implementation of the above approach.
C++
// C++ program to find the number of // segments covering each points #include<bits/stdc++.h> using namespace std;
// Function to print an array void PrintArray( int n, int arr[])
{ for ( int i = 0; i < n; i++)
{
cout<<arr[i]<< " " ;
}
} // Function prints number of segments // covering by each points void NumberOfSegments(vector<pair< int , int > >segments,
vector< int >points, int s, int p)
{ vector< pair< int , int > >pts, seg;
// Pushing points and index in
// vector as a pairs
for ( int i = 0; i < p; i++)
{
pts.push_back({points[i], i});;
}
for ( int i = 0; i < s; i++)
{
// (l,+1)
seg.push_back({segments[i].first, 1});
// (r+1,-1)
seg.push_back({segments[i].second+1, -1});
}
// Sort the vectors
sort(seg.begin(), seg.end(),
greater<pair< int , int >>());
sort(pts.begin(),pts.end());
int count = 0;
int ans[p];
for ( int i = 0; i < p; i++)
{
int x = pts[i].first;
while (!seg.empty() &&
seg.back().first <= x)
{
count+= seg.back().second;
seg.pop_back();
}
ans[pts[i].second] = count;
}
// Print the answer
PrintArray(p, ans);
} //Driver code int main()
{ // Initializing vector of pairs
vector<pair< int , int >>seg;
// Push segments
seg.push_back({0, 3});
seg.push_back({1, 3});
seg.push_back({3, 8});
// Given points
vector< int >point{-1, 3, 7};
int s = seg.size();
int p = point.size();
NumberOfSegments(seg, point, s, p);
return 0;
} |
Java
// Java program to find the number of // segments covering each points import java.util.*;
import java.lang.*;
class GFG{
// Function to print an array static void PrintArray( int n, int arr[])
{ for ( int i = 0 ; i < n; i++)
{
System.out.print(arr[i] + " " );
}
} // Function prints number of segments // covering by each points static void NumberOfSegments(ArrayList< int []> segments,
int [] points, int s, int p)
{ ArrayList< int []> pts = new ArrayList<>(),
seg = new ArrayList<>();
// Pushing points and index in
// vector as a pairs
for ( int i = 0 ; i < p; i++)
{
pts.add( new int []{points[i], i});
}
for ( int i = 0 ; i < s; i++)
{
// (l,+1)
seg.add( new int []{segments.get(i)[ 0 ], 1 });
// (r+1,-1)
seg.add( new int []{segments.get(i)[ 1 ] + 1 , - 1 });
}
// Sort the vectors
Collections.sort(seg, (a, b) -> b[ 0 ] - a[ 0 ]);
Collections.sort(pts, (a, b) -> a[ 0 ] - b[ 0 ]);
int count = 0 ;
int [] ans = new int [p];
for ( int i = 0 ; i < p; i++)
{
int x = pts.get(i)[ 0 ];
while (seg.size() != 0 &&
seg.get(seg.size() - 1 )[ 0 ] <= x)
{
count += seg.get(seg.size() - 1 )[ 1 ];
seg.remove(seg.size() - 1 );
}
ans[pts.get(i)[ 1 ]] = count;
}
// Print the answer
PrintArray(p, ans);
} // Driver code public static void main(String[] args)
{ // Initializing vector of pairs
ArrayList< int []>seg = new ArrayList<>();
// Push segments
seg.add( new int []{ 0 , 3 });
seg.add( new int []{ 1 , 3 });
seg.add( new int []{ 3 , 8 });
// Given points
int [] point = {- 1 , 3 , 7 };
int s = seg.size();
int p = point.length;
NumberOfSegments(seg, point, s, p);
} } // This code is contributed by offbeat |
Python3
# Python3 program to find the number # of segments covering each point # Function to print an array def PrintArray(n, arr):
for i in range (n):
print (arr[i], end = " " )
# Function prints number of segments # covering by each points def NumberOfSegments(segments, points, s, p):
pts = []
seg = []
# Pushing points and index in
# vector as a pairs
for i in range (p):
pts.append([points[i], i])
for i in range (s):
# (l, +1)
seg.append([segments[i][ 0 ], 1 ])
# (r+1, -1)
seg.append([segments[i][ 1 ] + 1 , - 1 ])
# Sort the vectors
seg.sort(reverse = True )
pts.sort(reverse = False )
count = 0
ans = [ 0 for i in range (p)]
for i in range (p):
x = pts[i][ 0 ]
while ( len (seg) ! = 0 and
seg[ len (seg) - 1 ][ 0 ] < = x):
count + = seg[ len (seg) - 1 ][ 1 ]
seg.remove(seg[ len (seg) - 1 ])
ans[pts[i][ 1 ]] = count
# Print the answer
PrintArray(p, ans)
# Driver code if __name__ = = '__main__' :
# Initializing vector of pairs
seg = []
# Push segments
seg.append([ 0 , 3 ])
seg.append([ 1 , 3 ])
seg.append([ 3 , 8 ])
# Given points
point = [ - 1 , 3 , 7 ]
s = len (seg)
p = len (point)
NumberOfSegments(seg, point, s, p)
# This code is contributed by Bhupendra_Singh |
C#
// C# program to find the number of // segments covering each points using System;
using System.Collections.Generic;
namespace ConsoleApp1
{ class Program
{
// Function to print an array
static void PrintArray( int [] arr)
{
foreach ( int x in arr)
{
Console.Write(x + " " );
}
}
// Function to find the number of segments covering each points
static void NumberOfSegments(List<Tuple< int , int >> segments, int [] points)
{
List<Tuple< int , int >> pts = new List<Tuple< int , int >>();
List<Tuple< int , int >> seg = new List<Tuple< int , int >>();
// Pushing points and index in
// list as a tuples
for ( int i = 0; i < points.Length; i++)
{
pts.Add(Tuple.Create(points[i], i));
}
for ( int i = 0; i < segments.Count; i++)
{
// (l,+1)
seg.Add(Tuple.Create(segments[i].Item1, 1));
// (r+1,-1)
seg.Add(Tuple.Create(segments[i].Item2 + 1, -1));
}
// Sort the lists
seg.Sort((x, y) => y.Item1.CompareTo(x.Item1));
pts.Sort();
int count = 0;
int [] ans = new int [points.Length];
for ( int i = 0; i < points.Length; i++)
{
int x = pts[i].Item1;
while (seg.Count > 0 && seg[seg.Count - 1].Item1 <= x)
{
count += seg[seg.Count - 1].Item2;
seg.RemoveAt(seg.Count - 1);
}
ans[pts[i].Item2] = count;
}
// Print the answer
PrintArray(ans);
}
static void Main( string [] args)
{
// Initializing list of tuples
List<Tuple< int , int >> seg = new List<Tuple< int , int >>();
// Push segments
seg.Add(Tuple.Create(0, 3));
seg.Add(Tuple.Create(1, 3));
seg.Add(Tuple.Create(3, 8));
// Given points
int [] point = { -1, 3, 7 };
NumberOfSegments(seg, point);
Console.ReadLine();
}
}
} |
Javascript
<script> // JavaScript program to find the number of // segments covering each points // Function to print an array function PrintArray(n,arr)
{ for (let i = 0; i < n; i++)
{
document.write(arr[i], " " );
}
} // Function prints number of segments // covering by each points function NumberOfSegments(segments,points,s,p)
{ let pts = []; let seg = []; // Pushing points and index in // vector as a pairs for (let i = 0; i < p; i++)
{ pts.push([points[i], i]);
} for (let i = 0; i < s; i++)
{ // (l,+1)
seg.push([segments[i][0], 1]);
// (r+1,-1)
seg.push([segments[i][1]+1, -1]);
} // Sort the vectors seg.sort((a,b) => b[0]-a[0]); pts.sort((a,b) => a[0]-b[0]); let count = 0; let ans = new Array(p);
for (let i = 0; i < p; i++)
{ let x = pts[i][0];
while (seg.length>0 && seg[seg.length-1][0] <= x)
{
count+= seg[seg.length-1][1];
seg.pop();
}
ans[pts[i][1]] = count;
} // Print the answer PrintArray(p, ans); } // Driver code // Initializing vector of pairs let seg = []; // Push segments seg.push([0, 3]); seg.push([1, 3]); seg.push([3, 8]); // Given points let point = [-1, 3, 7]; let s = seg.length; let p = point.length; NumberOfSegments(seg, point, s, p); // This code is contributed by shinjanpatra. </script> |
Output:
0 3 1
Time Complexity: O(s*log(s) + p*log(p)), where s is the number of segments and p is the number of points.
Auxiliary Space: O(s + p).
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