# 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:

## C++

 `// C++ implementation of the above approach ` `#include ` `#include ` `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; ` `} ` ` `  `// Function 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.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; ` `} `

## Java

 `// Java implementation of the above approach ` `import` `java.util.Arrays; ` `import` `java.util.Comparator; ` ` `  `public` `class` `MinimumCirclesTobeRemoved { ` ` `  `    ``private` `class` `Circle ``implements` `Comparator{ ` `        ``int` `start; ` `        ``int` `end; ` `         `  `        ``// Comparison function modified ` `        ``// according to the end value ` `        ``public` `int` `compare(Circle a , Circle b){ ` `            ``if``(a.end == b.end){ ` `                ``return` `(a.start - b.start); ` `            ``} ` `            ``return` `a.end - b.end; ` `        ``} ` `    ``} ` `     `  `    ``// Function to return the count ` `    ``// of non intersecting circles ` `    ``public` `void` `CountCircles(``int``[] c, ``int``[] r, ``int` `n){ ` `     `  `        ``// structure with start and ` `        ``// end of diameter of circles ` `        ``Circle diameter[] = ``new` `Circle[n]; ` ` `  `        ``for``(``int` `i = ``0``; i < n; i++) ` `        ``{ ` `            ``diameter[i] = ``new` `Circle(); ` `            ``diameter[i].start = (c[i] - r[i]); ` `            ``diameter[i].end = (c[i] + r[i]); ` `        ``} ` `         `  `        ``// sorting with smallest finish time first ` `        ``Arrays.sort(diameter, ``new` `Circle()); ` `         `  `        ``// count stores number of ` `        ``// circles to be removed ` `        ``int` `count = ``0``; ` `         `  `        ``// cur stores ending of first circle ` `        ``int` `curr = diameter[``0``].end; ` ` `  `        ``for``(``int` `i = ``1``; i < n; i++) ` `        ``{ ` `             `  `            ``// non intersecting circles ` `            ``if``(diameter[i].start > curr) ` `            ``{ ` `                ``curr = diameter[i].end; ` `            ``} ` `            ``else` `            ``{ ` `                ``count++; ` `            ``} ` `        ``} ` `        ``System.out.println(count); ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``MinimumCirclesTobeRemoved a = ``new` `MinimumCirclesTobeRemoved(); ` `         `  `        ``// centers of circles ` `        ``int``[] c = ``new` `int``[]{``1``, ``2``, ``3``, ``4``}; ` `         `  `        ``// radius of circles ` `        ``int``[] r = ``new` `int``[]{``1``, ``1``, ``1``, ``1``}; ` `        ``a.CountCircles(c, r, c.length); ` `    ``} ` `} ` ` `  `// This code is contributed by parshavnahta97 `

Output:

```2
```

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

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