# Maximum number of segments that can contain the given points

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.

Input: arr[] = {1, 2, 3, 4}, b = {1, 3, 5}, X = 0, Y = 0
Output: 2

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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 ` `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(a.begin(), a.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

 `= ``\$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 ` `?> `

Output:

```1
```

Time Complexity: O(N * log(N))

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