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Maximum number of segments that can contain the given points

  • Last Updated : 01 Jun, 2021

Given an array arr[] containing N integers and two integers X and Y. Consider N line segments, where each line segment has starting and ending point as arr[i] – X and arr[i] + Y respectively. 
Given another array b[] of M points. The task is to assign these points to segments such that the number of segments that have been assigned a point is maximum. Note that a point can be assigned to at most 1 segment.
Examples: 

Input: arr[] = {1, 5}, b = {1, 1, 2}, X = 1, Y = 4 
Output:
Line Segments are [1-X, 1+Y] , [5-X, 5+Y] i.e. [0, 5] and [4, 9] 
The point 1 can be assigned to the first segment [0, 5] 
No points can be assigned to the second segment. 
So 2 can also be assigned to the first segment but it will not maximize the no. of segment. 
So the answer is 1.
Input: arr[] = {1, 2, 3, 4}, b = {1, 3, 5}, X = 0, Y = 0 
Output:

Approach: Sort both the input arrays. Now for every segment, we try to assign it the first unassigned point possible. If the current segment ends before the current point, it means that we won’t able to assign any point to it since all the points ahead of it are greater than the current point and the segment has already ended.
Below is the implementation of the above approach: 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the maximum number of segments
int countPoints(int n, int m, vector<int> a,
                vector<int> b, int x, int y)
{
    // Sort both the vectors
    sort(a.begin(), a.end());
    sort(b.begin(), b.end());
 
    // Initially pointing to the first element of b[]
    int j = 0;
    int count = 0;
    for (int i = 0; i < n; i++) {
 
        // Try to find a match in b[]
        while (j < m) {
 
            // The segment ends before b[j]
            if (a[i] + y < b[j])
                break;
 
            // The point lies within the segment
            if (b[j] >= a[i] - x && b[j] <= a[i] + y) {
                count++;
                j++;
                break;
            }
 
            // The segment starts after b[j]
            else
                j++;
        }
    }
 
    // Return the required count
    return count;
}
 
// Driver code
int main()
{
    int x = 1, y = 4;
    vector<int> a = { 1, 5 };
    int n = a.size();
    vector<int> b = { 1, 1, 2 };
    int m = a.size();
    cout << countPoints(n, m, a, b, x, y);
 
   return 0;
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Function to return the
// maximum number of segments
static int countPoints(int n, int m, int a[],
                        int[] b, int x, int y)
{
    // Sort both the vectors
    Arrays.sort(a);
    Arrays.sort(b);
 
    // Initially pointing to the first element of b[]
    int j = 0;
    int count = 0;
    for (int i = 0; i < n; i++)
    {
 
        // Try to find a match in b[]
        while (j < m)
        {
 
            // The segment ends before b[j]
            if (a[i] + y < b[j])
                break;
 
            // The point lies within the segment
            if (b[j] >= a[i] - x && b[j] <= a[i] + y)
            {
                count++;
                j++;
                break;
            }
 
            // The segment starts after b[j]
            else
                j++;
        }
    }
 
    // Return the required count
    return count;
}
 
// Driver code
public static void main(String args[])
{
    int x = 1, y = 4;
    int[] a = { 1, 5 };
    int n = a.length;
    int[] b = { 1, 1, 2 };
    int m = a.length;
    System.out.println(countPoints(n, m, a, b, x, y));
}
}
 
// This code is contributed by
// Surendra_Gangwar

Python3




# Python3 implementation of the approach
 
# Function to return the maximum
# number of segments
def countPoints(n, m, a, b, x, y):
 
    # Sort both the vectors
    a.sort()
    b.sort()
 
    # Initially pointing to the first
    # element of b[]
    j, count = 0, 0
    for i in range(0, n):
 
        # Try to find a match in b[]
        while j < m:
 
            # The segment ends before b[j]
            if a[i] + y < b[j]:
                break
 
            # The point lies within the segment
            if (b[j] >= a[i] - x and
                b[j] <= a[i] + y):
                count += 1
                j += 1
                break
 
            # The segment starts after b[j]
            else:
                j += 1
 
    # Return the required count
    return count
 
# Driver code
if __name__ == "__main__":
 
    x, y = 1, 4
    a = [1, 5]
    n = len(a)
    b = [1, 1, 2]
    m = len(b)
    print(countPoints(n, m, a, b, x, y))
     
# This code is contributed by Rituraj Jain

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
    // Function to return the
    // maximum number of segments
    static int countPoints(int n, int m, int []a,
                            int []b, int x, int y)
    {
        // Sort both the vectors
        Array.Sort(a);
        Array.Sort(b);
     
        // Initially pointing to the
        // first element of b[]
        int j = 0;
        int count = 0;
        for (int i = 0; i < n; i++)
        {
     
            // Try to find a match in b[]
            while (j < m)
            {
     
                // The segment ends before b[j]
                if (a[i] + y < b[j])
                    break;
     
                // The point lies within the segment
                if (b[j] >= a[i] - x && b[j] <= a[i] + y)
                {
                    count++;
                    j++;
                    break;
                }
     
                // The segment starts after b[j]
                else
                    j++;
            }
        }
     
        // Return the required count
        return count;
    }
 
    // Driver code
    public static void Main()
    {
        int x = 1, y = 4;
        int[] a = {1, 5};
        int n = a.Length;
        int[] b = {1, 1, 2};
        int m = a.Length;
        Console.WriteLine(countPoints(n, m, a, b, x, y));
    }
}
 
// This code is contributed by Ryuga

PHP




<?php
// PHP implementation of the approach
 
// Function to return the maximum number of segments
function countPoints($n, $m, $a, $b, $x, $y)
{
    // Sort both the vectors
    sort($a);
    sort($b);
 
    // Initially pointing to the first element of b[]
    $j = 0;
    $count = 0;
    for ($i = 0; $i < $n; $i++)
    {
 
        // Try to find a match in b[]
        while ($j < $m)
        {
 
            // The segment ends before b[j]
            if ($a[$i] + $y < $b[$j])
                break;
 
            // The point lies within the segment
            if ($b[$j] >= $a[$i] - $x &&
                $b[$j] <= $a[$i] + $y)
            {
                $count++;
                $j++;
                break;
            }
 
            // The segment starts after b[j]
            else
                $j++;
        }
    }
 
    // Return the required count
    return $count;
}
 
    // Driver code
    $x = 1;
    $y = 4;
    $a = array( 1, 5 );
    $n = count($a);
    $b = array( 1, 1, 2 );
    $m = count($b);
    echo countPoints($n, $m, $a, $b, $x, $y);
 
// This code is contributed by Arnab Kundu
?>

Javascript




<script>
 
    // JavaScript implementation of the approach
     
    // Function to return the
    // maximum number of segments
    function countPoints(n, m, a, b, x, y)
    {
        // Sort both the vectors
        a.sort(function(a, b){return a - b});
        b.sort(function(a, b){return a - b});
      
        // Initially pointing to the
        // first element of b[]
        let j = 0;
        let count = 0;
        for (let i = 0; i < n; i++)
        {
      
            // Try to find a match in b[]
            while (j < m)
            {
      
                // The segment ends before b[j]
                if (a[i] + y < b[j])
                    break;
      
                // The point lies within the segment
                if (b[j] >= a[i] - x && b[j] <= a[i] + y)
                {
                    count++;
                    j++;
                    break;
                }
      
                // The segment starts after b[j]
                else
                    j++;
            }
        }
      
        // Return the required count
        return count;
    }
     
    let x = 1, y = 4;
    let a = [1, 5];
    let n = a.length;
    let b = [1, 1, 2];
    let m = a.length;
    document.write(countPoints(n, m, a, b, x, y));
     
</script>
Output: 
1

 

Time Complexity: O(N * log(N))
Auxiliary Space: O(1)

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