Maximum number of elements without overlapping in a Line

0 0 5 minutes read

Given two arrays X and L of same size N. X_i represent the position in an infinite line. L_i 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

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 are 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 <bits/stdc++.h>
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

Javascript

<script>
 
// JavaScript program to find maximum number of
// elements without overlapping in a line
 
// Function to find maximum number of
// elements without overlapping in a line
function 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
    let ans = 2;
 
 
    for (let 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 endpolet of
            // segment covered by it
            x[i] = x[i] + l[i];
            ans++;
        }
    }
 
    // Return the required answer
    return ans;
}
 
// Driver code
 
let x = [1, 3, 4, 5, 8], l = [10, 1, 2, 2, 5];
 
let n = x.length;
 
// Function call
document.write(Segment(x, l, n));
 
// This code is contributed by _saurabh_jaiswal
 
</script>

Output

Time Complexity: O(N)
Auxiliary Space: O(1), no extra space is required, so it is a constant.

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 zambiatek!