Given two integers **L** and **W** representing the dimensions of a grid, and two arrays **X[]** and **Y[]** of length **N** denoting the number of towers present on the grid at positions **(X[i], Y[i]), **where** (0 <= i <= N – 1). **The task is to find the largest unbounded area in the field which is not defended by the towers.

**Examples:**

Input:L = 15, W = 8, N = 3 X[] = {3, 11, 8} Y[] = {8, 2, 6}Output:12Explanation:

The coordinates of the towers are (3, 8), (11, 2) and (8, 6).

Observe that the largest area of the grid which is not guarded by any of the towers is within the cells (4, 3), (7, 3), (7, 5) and (4, 5). Hence, the area for that part is 12.

Input:L = 3, W = 3, N = 1 X[] = {1} Y[] = {1}Output:4Explanation:

Observe that the largest area of the grid which is not guarded by any of the towers is 4.

**Naive Approach: **Follow the steps below to solve the problem:

- Initialize a
**matrix**of dimensions**L * W**with**0s.** - Traverse
**X[]**and**Y[],**and for every**(X[i], Y[i])**, mark all the cells in the**X[i]**Row and^{th}**Y[i]**Column by 1, to denote being guarded by the tower at^{th}**(X[i], Y[i])**. - Then traverse the matrix and for each cell, and find the largest sub-matrix which is left unguarded, i.e. largest submatrix consisting of 0s. Print the corresponding area.

**Time Complexity: **O(N^{3})**Auxiliary Space:** O(N^{2})

**Efficient Approach: **The above approach can be optimized using the following Greedy technique:

- Sort both the lists of co-ordinates
**X[]**and**Y[]**. - Calculate the empty spaces between the x co-ordinates, i.e.
**dX[] = {X**. Similarly, calculate the spaces between y co-ordinates,_{1}, X_{2 }– X_{1}, …., X_{N }– X_{(N-1)}, (L+1) – X_{N}}**dY[] = {Y**._{1}, Y_{2 }– Y_{1}, …., Y_{N }– Y_{(N-1)}, (W + 1) – Y_{N}} - Traverse
**dX[]**and**dY[]**and calculate their respective maximums. - Calculate the product of their maximums and print it as the required longest unbounded area.

Below is the implementation of the above approach:

## C++

`// C++ Program for the above approach` `#include <algorithm>` `#include <iostream>` `using` `namespace` `std;` `// Function to calculate the largest` `// area unguarded by towers` `void` `maxArea(` `int` `point_x[], ` `int` `point_y[], ` `int` `n,` ` ` `int` `length, ` `int` `width)` `{` ` ` `// Sort the x-coordinates of the list` ` ` `sort(point_x, point_x + n);` ` ` `// Sort the y-coordinates of the list` ` ` `sort(point_y, point_y + n);` ` ` `// dx --> maximum uncovered` ` ` `// tiles in x coordinates` ` ` `int` `dx = point_x[0];` ` ` `// dy --> maximum uncovered` ` ` `// tiles in y coordinates` ` ` `int` `dy = point_y[0];` ` ` `// Calculate the maximum uncovered` ` ` `// distances for both x and y coordinates` ` ` `for` `(` `int` `i = 1; i < n; i++) {` ` ` `dx = max(dx, point_x[i]` ` ` `- point_x[i - 1]);` ` ` `dy = max(dy, point_y[i]` ` ` `- point_y[i - 1]);` ` ` `}` ` ` `dx = max(dx, (length + 1)` ` ` `- point_x[n - 1]);` ` ` `dy = max(dy, (width + 1)` ` ` `- point_y[n - 1]);` ` ` `// Largest unguarded area is` ` ` `// max(dx)-1 * max(dy)-1` ` ` `cout << (dx - 1) * (dy - 1);` ` ` `cout << endl;` `}` `// Driver Code` `int` `main()` `{` ` ` `// Length and width of the grid` ` ` `int` `length = 15, width = 8;` ` ` `// No of guard towers` ` ` `int` `n = 3;` ` ` `// Array to store the x and` ` ` `// y coordinates` ` ` `int` `point_x[] = { 3, 11, 8 };` ` ` `int` `point_y[] = { 8, 2, 6 };` ` ` `// Function call` ` ` `maxArea(point_x, point_y,` ` ` `n, length, width);` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `import` `java.io.*;` `import` `java.util.*;` `class` `GFG{` `// Function to calculate the largest` `// area unguarded by towers` `static` `void` `maxArea(` `int` `[] point_x, ` `int` `[] point_y,` ` ` `int` `n, ` `int` `length, ` `int` `width)` `{` ` ` ` ` `// Sort the x-coordinates of the list` ` ` `Arrays.sort(point_x);` ` ` `// Sort the y-coordinates of the list` ` ` `Arrays.sort(point_y);` ` ` `// dx --> maximum uncovered` ` ` `// tiles in x coordinates` ` ` `int` `dx = point_x[` `0` `];` ` ` `// dy --> maximum uncovered` ` ` `// tiles in y coordinates` ` ` `int` `dy = point_y[` `0` `];` ` ` `// Calculate the maximum uncovered` ` ` `// distances for both x and y coordinates` ` ` `for` `(` `int` `i = ` `1` `; i < n; i++)` ` ` `{` ` ` `dx = Math.max(dx, point_x[i] -` ` ` `point_x[i - ` `1` `]);` ` ` `dy = Math.max(dy, point_y[i] -` ` ` `point_y[i - ` `1` `]);` ` ` `}` ` ` `dx = Math.max(dx, (length + ` `1` `) -` ` ` `point_x[n - ` `1` `]);` ` ` ` ` `dy = Math.max(dy, (width + ` `1` `) -` ` ` `point_y[n - ` `1` `]);` ` ` `// Largest unguarded area is` ` ` `// max(dx)-1 * max(dy)-1` ` ` `System.out.println((dx - ` `1` `) * (dy - ` `1` `));` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` ` ` `// Length and width of the grid` ` ` `int` `length = ` `15` `, width = ` `8` `;` ` ` `// No of guard towers` ` ` `int` `n = ` `3` `;` ` ` `// Array to store the x and` ` ` `// y coordinates` ` ` `int` `point_x[] = { ` `3` `, ` `11` `, ` `8` `};` ` ` `int` `point_y[] = { ` `8` `, ` `2` `, ` `6` `};` ` ` `// Function call` ` ` `maxArea(point_x, point_y, n,` ` ` `length, width);` `}` `}` `// This code is contributed by akhilsaini` |

## Python3

`# Python3 program for the above approach` `# Function to calculate the largest` `# area unguarded by towers` `def` `maxArea(point_x, point_y, n,` ` ` `length, width):` ` ` ` ` `# Sort the x-coordinates of the list` ` ` `point_x.sort()` ` ` `# Sort the y-coordinates of the list` ` ` `point_y.sort()` ` ` `# dx --> maximum uncovered` ` ` `# tiles in x coordinates` ` ` `dx ` `=` `point_x[` `0` `]` ` ` `# dy --> maximum uncovered` ` ` `# tiles in y coordinates` ` ` `dy ` `=` `point_y[` `0` `]` ` ` `# Calculate the maximum uncovered` ` ` `# distances for both x and y coordinates` ` ` `for` `i ` `in` `range` `(` `1` `, n):` ` ` `dx ` `=` `max` `(dx, point_x[i] ` `-` ` ` `point_x[i ` `-` `1` `])` ` ` `dy ` `=` `max` `(dy, point_y[i] ` `-` ` ` `point_y[i ` `-` `1` `])` ` ` ` ` `dx ` `=` `max` `(dx, (length ` `+` `1` `) ` `-` ` ` `point_x[n ` `-` `1` `])` ` ` ` ` `dy ` `=` `max` `(dy, (width ` `+` `1` `) ` `-` ` ` `point_y[n ` `-` `1` `])` ` ` `# Largest unguarded area is` ` ` `# max(dx)-1 * max(dy)-1` ` ` `print` `((dx ` `-` `1` `) ` `*` `(dy ` `-` `1` `))` `# Driver Code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` ` ` `# Length and width of the grid` ` ` `length ` `=` `15` ` ` `width ` `=` `8` ` ` `# No of guard towers` ` ` `n ` `=` `3` ` ` `# Array to store the x and` ` ` `# y coordinates` ` ` `point_x ` `=` `[ ` `3` `, ` `11` `, ` `8` `]` ` ` `point_y ` `=` `[ ` `8` `, ` `2` `, ` `6` `]` ` ` `# Function call` ` ` `maxArea(point_x, point_y, n,` ` ` `length, width)` `# This code is contributed by akhilsaini` |

## C#

`// C# Program for the above approach` `using` `System;` `class` `GFG{` `// Function to calculate the largest` `// area unguarded by towers` `static` `void` `maxArea(` `int` `[] point_x, ` `int` `[] point_y,` ` ` `int` `n, ` `int` `length, ` `int` `width)` `{` ` ` ` ` `// Sort the x-coordinates of the list` ` ` `Array.Sort(point_x);` ` ` `// Sort the y-coordinates of the list` ` ` `Array.Sort(point_y);` ` ` `// dx --> maximum uncovered` ` ` `// tiles in x coordinates` ` ` `int` `dx = point_x[0];` ` ` `// dy --> maximum uncovered` ` ` `// tiles in y coordinates` ` ` `int` `dy = point_y[0];` ` ` `// Calculate the maximum uncovered` ` ` `// distances for both x and y coordinates` ` ` `for` `(` `int` `i = 1; i < n; i++)` ` ` `{` ` ` `dx = Math.Max(dx, point_x[i] -` ` ` `point_x[i - 1]);` ` ` `dy = Math.Max(dy, point_y[i] -` ` ` `point_y[i - 1]);` ` ` `}` ` ` `dx = Math.Max(dx, (length + 1) -` ` ` `point_x[n - 1]);` ` ` `dy = Math.Max(dy, (width + 1) -` ` ` `point_y[n - 1]);` ` ` `// Largest unguarded area is` ` ` `// max(dx)-1 * max(dy)-1` ` ` `Console.WriteLine((dx - 1) * (dy - 1));` `}` `// Driver Code` `static` `public` `void` `Main()` `{` ` ` ` ` `// Length and width of the grid` ` ` `int` `length = 15, width = 8;` ` ` `// No of guard towers` ` ` `int` `n = 3;` ` ` `// Array to store the x and` ` ` `// y coordinates` ` ` `int` `[] point_x = { 3, 11, 8 };` ` ` `int` `[] point_y = { 8, 2, 6 };` ` ` `// Function call` ` ` `maxArea(point_x, point_y, n,` ` ` `length, width);` `}` `}` `// This code is contributed by akhilsaini` |

**Output:**

12

**Time Complexity:** (N * log N)**Auxiliary Space:** O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.