# Minimum number of Circular obstacles required to obstruct the path in a Grid

Consider a grid of dimensions NxM and an array R consisting of available circular obstacles, the task is to find the minimum number of circular obstacles of given radiuses required to obstruct the path between source [0, 0] and destination [N-1, M-1]. If not possible print -1.
Note: The circular obstacles can overlap as shown in the image of example 1.

Examples:

Input: N = 4, M = 5, R[] = {1.0, 1.5, 1.25}
Output: 2

Input: N = 10, M = 12, R[] = {1.0, 1.25}
Output: -1

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

Approach:

• Find whether to put the obstacles row-wise or column-wise.
• Sort the radius in decreasing order.
• Since the obstacles cover an entire circle with radius R[i], therefore for a straight line, it covers the diameter.
• Decrease the val by 2 * Ri until it becomes zero using larger values in array R[].
• After using all the obstacles, when val &leq; 0 return the count of obstacles used and if the val > 0 after using all the obstacles print -1.

Below is the implementation of the above approach.

## CPP

 `// C++ program to find the minimum ` `// number of obstacles required ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find the minimum ` `// number of obstacles required ` `int` `solve(``int` `n, ``int` `m, ``int` `obstacles, ` `          ``double` `range[]) ` `{ ` `    ``// Find the minimum range required ` `    ``// to put obstacles ` `    ``double` `val = min(n, m); ` ` `  `    ``// Sorting the radius ` `    ``sort(range, range + obstacles); ` ` `  `    ``int` `c = 1; ` `    ``for` `(``int` `i = obstacles - 1; i >= 0; i--) { ` `        ``range[i] = 2 * range[i]; ` `        ``val -= range[i]; ` ` `  `        ``// If val is less than zero ` `        ``// then we have find the number of ` `        ``// obstacles required ` `        ``if` `(val <= 0) { ` `            ``return` `c; ` `        ``} ` `        ``else` `{ ` `            ``c++; ` `        ``} ` `    ``} ` ` `  `    ``if` `(val > 0) { ` `        ``return` `-1; ` `    ``} ` `} ` ` `  `// Driver function ` `int` `main() ` `{ ` `    ``int` `n = 4, m = 5, obstacles = 3; ` `    ``double` `range[] = { 1.0, 1.25, 1.15 }; ` `    ``cout << solve(n, m, obstacles, range) << ``"\n"``; ` `    ``return` `0; ` `} `

## Java

 `// Java program to find the minimum ` `// number of obstacles required ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to find the minimum ` `// number of obstacles required ` `static` `int` `solve(``int` `n, ``int` `m, ``int` `obstacles, ` `                ``double` `range[]) ` `{ ` `    ``// Find the minimum range required ` `    ``// to put obstacles ` `    ``double` `val = Math.min(n, m); ` ` `  `    ``// Sorting the radius ` `    ``Arrays.sort(range); ` ` `  `    ``int` `c = ``1``; ` `    ``for` `(``int` `i = obstacles - ``1``; i >= ``0``; i--) ` `    ``{ ` `        ``range[i] = ``2` `* range[i]; ` `        ``val -= range[i]; ` ` `  `        ``// If val is less than zero ` `        ``// then we have find the number of ` `        ``// obstacles required ` `        ``if` `(val <= ``0``) ` `        ``{ ` `            ``return` `c; ` `        ``} ` `        ``else` `        ``{ ` `            ``c++; ` `        ``} ` `    ``} ` ` `  `    ``if` `(val > ``0``)  ` `    ``{ ` `        ``return` `-``1``; ` `    ``} ` `    ``return` `0``; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `n = ``4``, m = ``5``, obstacles = ``3``; ` `    ``double` `range[] = { ``1.0``, ``1.25``, ``1.15` `}; ` `    ``System.out.print(solve(n, m, obstacles, range)+ ``"\n"``); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

## C#

 `// C# program to find the minimum ` `// number of obstacles required ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `    ``// Function to find the minimum ` `    ``// number of obstacles required ` `    ``static` `int` `solve(``int` `n, ``int` `m, ``int` `obstacles, ` `                    ``double` `[]range) ` `    ``{ ` `        ``// Find the minimum range required ` `        ``// to put obstacles ` `        ``double` `val = Math.Min(n, m); ` `     `  `        ``// Sorting the radius ` `        ``Array.Sort(range); ` `     `  `        ``int` `c = 1; ` `        ``for` `(``int` `i = obstacles - 1; i >= 0; i--) ` `        ``{ ` `            ``range[i] = 2 * range[i]; ` `            ``val -= range[i]; ` `     `  `            ``// If val is less than zero ` `            ``// then we have find the number of ` `            ``// obstacles required ` `            ``if` `(val <= 0) ` `            ``{ ` `                ``return` `c; ` `            ``} ` `            ``else` `            ``{ ` `                ``c++; ` `            ``} ` `        ``} ` `     `  `        ``if` `(val > 0)  ` `        ``{ ` `            ``return` `-1; ` `        ``} ` `        ``return` `0; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 4, m = 5, obstacles = 3; ` `        ``double` `[]range = { 1.0, 1.25, 1.15 }; ` `        ``Console.WriteLine(solve(n, m, obstacles, range)); ` `    ``} ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 program to find the minimum ` `# number of obstacles required ` ` `  `# Function to find the minimum ` `# number of obstacles required ` `def` `solve(n, m, obstacles,rangee): ` `     `  `    ``# Find the minimum rangee required ` `    ``# to put obstacles ` `    ``val ``=` `min``(n, m) ` ` `  `    ``# Sorting the radius ` `    ``rangee ``=` `sorted``(rangee) ` `    ``c ``=` `1` `    ``for` `i ``in` `range``(obstacles ``-` `1``, ``-``1``, ``-``1``): ` `        ``rangee[i] ``=` `2` `*` `rangee[i] ` `        ``val ``-``=` `rangee[i] ` `         `  `        ``# If val is less than zero ` `        ``# then we have find the number of ` `        ``# obstacles required ` `        ``if` `(val <``=` `0``): ` `            ``return` `c ` `        ``else``: ` `            ``c ``+``=` `1` ` `  `    ``if` `(val > ``0``): ` `        ``return` `-``1` ` `  `# Driver code ` `n ``=` `4` `m ``=` `5` `obstacles ``=` `3` `rangee ``=` `[``1.0``, ``1.25``, ``1.15``] ` `print``(solve(n, m, obstacles, rangee)) ` ` `  `# This code is contributed by mohit kumar 29 `

Output:

```2
```

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