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Maximum number of elements without overlapping in a Line
• Last Updated : 02 Aug, 2019

Given two arrays X and L of same size N. Xi represents the position in an infinite line. Li represents the range up to which ith element can cover on both sides. The task is to select the maximum number of elements such that no two selected elements overlap if they cover the right or the left side segment.

Note: Array X is sorted.

Examples:

Input : x[] = {10, 15, 19, 20} , L[] = {4, 1, 3, 1}
Output : 4
Suppose, first element covers left side segment [6, 10]
second element covers left side segment 14, 15]
Third element covers left side segment [16, 19]
Fourth element covers right side segment [20, 21]

Input : x[] = {1, 3, 4, 5, 8}, L[] = {10, 1, 2, 2, 5}
Output : 4

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

Approach:
This problem can be solved greedily. We can always make the first element cover the left segment and the last element cover the right segment. For the other elements first, try to give left segment if possible otherwise try to give the right segment. If none of them possible then leave the element.

Below is the implementation of the above approach:

C++

 `// CPP program to find maximum number of ``// elements without overlapping in a line``#include ``using` `namespace` `std;`` ` `// Function to find maximum number of ``// elements without overlapping in a line``int` `Segment(``int` `x[], ``int` `l[], ``int` `n)``{``    ``// If n = 1, then answer is one``    ``if` `(n == 1)``        ``return` `1;``     ` `    ``// We can always make 1st element to cover ``    ``// left segment and nth the right segment``    ``int` `ans = 2;``         ` `         ` `    ``for` `(``int` `i = 1; i < n - 1; i++)``    ``{``        ``// If left segment for ith element doesnt overlap``        ``// with i - 1 th element then do left``        ``if` `(x[i] - l[i] > x[i - 1])``            ``ans++;`` ` `        ``// else try towards right if possible``        ``else` `if` `(x[i] + l[i] < x[i + 1])``        ``{``            ``// update x[i] to right endpoint of ``            ``// segment covered by it``            ``x[i] = x[i] + l[i];``            ``ans++;``        ``}``    ``}``     ` `    ``// Return the required answer``    ``return` `ans;``}`` ` `// Driver code``int` `main()``{``    ``int` `x[] = {1, 3, 4, 5, 8}, l[] = {10, 1, 2, 2, 5};``     ` `    ``int` `n = ``sizeof``(x) / ``sizeof``(x[0]);`` ` `    ``// Function call``    ``cout << Segment(x, l, n);`` ` `    ``return` `0;``}`

Java

 `// Java program to find maximum number of ``// elements without overlapping in a line``import` `java.util.*;`` ` `class` `GFG``{`` ` `// Function to find maximum number of ``// elements without overlapping in a line``static` `int` `Segment(``int` `x[], ``int` `l[], ``int` `n)``{``    ``// If n = 1, then answer is one``    ``if` `(n == ``1``)``        ``return` `1``;``     ` `    ``// We can always make 1st element to cover ``    ``// left segment and nth the right segment``    ``int` `ans = ``2``;``         ` `    ``for` `(``int` `i = ``1``; i < n - ``1``; i++)``    ``{``        ``// If left segment for ith element ``        ``// doesn't overlap with i - 1 th``        ``// element then do left``        ``if` `(x[i] - l[i] > x[i - ``1``])``            ``ans++;`` ` `        ``// else try towards right if possible``        ``else` `if` `(x[i] + l[i] < x[i + ``1``])``        ``{``            ``// update x[i] to right endpoint of ``            ``// segment covered by it``            ``x[i] = x[i] + l[i];``            ``ans++;``        ``}``    ``}``     ` `    ``// Return the required answer``    ``return` `ans;``}`` ` `// Driver code``public` `static` `void` `main(String[] args) ``{``    ``int` `x[] = {``1``, ``3``, ``4``, ``5``, ``8``},``        ``l[] = {``10``, ``1``, ``2``, ``2``, ``5``};``     ` `    ``int` `n = x.length;`` ` `    ``// Function call``    ``System.out.println(Segment(x, l, n));``}``}`` ` `// This code is contributed by 29AjayKumar`

Python3

 `# Python3 program to find maximum number of``# elements without overlapping in a line`` ` `# Function to find maximum number of``# elements without overlapping in a line``def` `Segment(x, l, n):``     ` `    ``# If n = 1, then answer is one``    ``if` `(n ``=``=` `1``):``        ``return` `1`` ` `    ``# We can always make 1st element to cover``    ``# left segment and nth the right segment``    ``ans ``=` `2``     ` `    ``for` `i ``in` `range``(``1``, n ``-` `1``):``         ` `        ``# If left segment for ith element doesnt overlap``        ``# with i - 1 th element then do left``        ``if` `(x[i] ``-` `l[i] > x[i ``-` `1``]):``            ``ans ``+``=` `1`` ` `        ``# else try towards right if possible``        ``elif` `(x[i] ``+` `l[i] < x[i ``+` `1``]):``             ` `            ``# update x[i] to right endpoof``            ``# segment covered by it``            ``x[i] ``=` `x[i] ``+` `l[i]``            ``ans ``+``=` `1`` ` `    ``# Return the required answer``    ``return` `ans`` ` `# Driver code``x ``=` `[``1``, ``3``, ``4``, ``5``, ``8``]``l ``=` `[``10``, ``1``, ``2``, ``2``, ``5``]`` ` `n ``=` `len``(x)`` ` `# Function call``print``(Segment(x, l, n))`` ` `# This code is contributed ``# by Mohit Kumar`

C#

 `// C# program to find maximum number of ``// elements without overlapping in a line``using` `System;``     ` `class` `GFG``{`` ` `// Function to find maximum number of ``// elements without overlapping in a line``static` `int` `Segment(``int` `[]x, ``int` `[]l, ``int` `n)``{``    ``// If n = 1, then answer is one``    ``if` `(n == 1)``        ``return` `1;``     ` `    ``// We can always make 1st element to cover ``    ``// left segment and nth the right segment``    ``int` `ans = 2;``         ` `    ``for` `(``int` `i = 1; i < n - 1; i++)``    ``{``        ``// If left segment for ith element ``        ``// doesn't overlap with i - 1 th``        ``// element then do left``        ``if` `(x[i] - l[i] > x[i - 1])``            ``ans++;`` ` `        ``// else try towards right if possible``        ``else` `if` `(x[i] + l[i] < x[i + 1])``        ``{``            ``// update x[i] to right endpoint of ``            ``// segment covered by it``            ``x[i] = x[i] + l[i];``            ``ans++;``        ``}``    ``}``     ` `    ``// Return the required answer``    ``return` `ans;``}`` ` `// Driver code``public` `static` `void` `Main(String[] args) ``{``    ``int` `[]x = {1, 3, 4, 5, 8};``    ``int` `[]l = {10, 1, 2, 2, 5};``     ` `    ``int` `n = x.Length;`` ` `    ``// Function call``    ``Console.WriteLine(Segment(x, l, n));``}``}`` ` `// This code is contributed by PrinciRaj1992`

Output:
` 4 `

Time Complexity: O(N)

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