Given two arrays X and L of same size N. Xi represent 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
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:
// 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 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 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# 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 |
<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> |
4
Time Complexity: O(N)
Auxiliary Space: O(1), no extra space is required, so it is a constant.