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 in example 1.

Examples:

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

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

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 <= 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,``range``):` `    `  `    ``# Find the minimum range required` `    ``# to put obstacles` `    ``val ``=` `min``(n, m)`   `    ``# Sorting the radius` `    ``range` `=` `sorted``(``range``)` `    ``c ``=` `1` `    ``for` `i ``in` `range``(obstacles ``-` `1``, ``-``1``, ``-``1``):` `        ``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 ``+``=` `1`   `    ``if` `(val > ``0``):` `        ``return` `-``1`   `# Driver code` `n ``=` `4` `m ``=` `5` `obstacles ``=` `3` `range` `=` `[``1.0``, ``1.25``, ``1.15``]` `print``(solve(n, m, obstacles, ``range``))`   `# This code is contributed by mohit kumar 29`

Javascript

 ``

Output:

`2`

Time Complexity: O(N * log(N)), where N is the number of given obstacles.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

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