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: 1
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: 2
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 segmentsint 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 codeint 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 approachimport java.util.*;class GFG{// Function to return the // maximum number of segmentsstatic 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 codepublic 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), since no extra space has been taken.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
