Minimum Circles needed to be removed so that all remaining circles are non intersecting

Given n Circles present in x-y plane such that all the circles have their center aligned on the x-axis.
The task is to remove some of them, such that no two circles are intersecting. Find the minimum number of circles that need to be removed.
Note : Touching circles are also considered to be intersecting.
Given N and an array of pair of integers. Each pair contains two integers c and r each, denoting the circle with radius r and center (c, 0).

Examples:

Input : N=4, arr={(1, 1), (2, 1), (3, 1), (4, 1)}
Output : 2

Remove 2nd and 3rd circle to make the circles non-intersecting.

Input : N=4, arr={(1, 1), (4, 1), (5, 2), (7, 1)}
Output : 1

Approach:
Greedy strategy can be applied to solve the problem.

  • Find staring and ending points of the diameter of the circles.
  • Starting point would be equal to (c-r) and ending point would be equal to (c+r) where (c, 0) is the center of the particular circle and r is its radius.
  • Sort the {start, end} pair according to the value of end point. Less the value of end point, less is its index.
  • Start iterating the pairs and if the starting point of a circle is less then current end value, it means circles are intersecting hence increment the count. Else update the current end value.

Below is the implementation of the above approach:

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// C++ implementation of the above approach
#include <algorithm>
#include <iostream>
using namespace std;
  
struct circle {
    int start, end;
};
  
// Comparison function modified
// according to the end value
bool comp(circle a, circle b)
{
    if (a.end == b.end)
        return a.start < b.start;
    return a.end < b.end;
}
  
// Fucntion to return the count
// of non intersecting circles
void CountCircles(int c[], int r[], int n)
{
    // structure with start and
    // end of diameter of circles
    circle diameter[n];
  
    for (int i = 0; i < n; ++i) {
        diameter[i].start = c[i] - r[i];
  
        diameter[i].end = c[i] + r[i];
    }
  
    // sorting with smallest finish time first
    sort(diameter, diameter + n, comp);
  
    // count stores number of
    // circles to be removed
    int count = 0;
  
    // cur stores ending of first circle
    int cur = diameter[0].end;
    for (int i = 1; i < n; ++i) {
  
        // non intersecting circles
        if (diameter[i].start > cur) {
            cur = diameter[i].end;
        }
  
        // intersecting circles
        else
            count++;
    }
  
    cout << count << "\n";
}
  
// Driver Code
int main()
{
    // centers of circles
    int c[] = { 1, 2, 3, 4 };
    // radius of circles
    int r[] = { 1, 1, 1, 1 };
  
    // number of circles
    int n = sizeof(c) / sizeof(int);
  
    CountCircles(c, r, n);
  
    return 0;
}

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Output:

2

Time Complexity: O(N*log(N))
where N is the number of circles.



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