Largest area in a grid unbounded by towers

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: 12
Explanation: 
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: 4
Explanation:
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]th Row and Y[i]th Column by 1, to denote being guarded by the tower at (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(N3)
Auxiliary Space: O(N2)

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[] = {X1, X2 – X1, …., XN – X(N-1), (L+1) – XN }. Similarly, calculate the spaces between y co-ordinates, dY[] = {Y1, Y2 – Y1, …., YN – Y(N-1), (W + 1) – YN }.
  • 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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output: 

12




 

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

 

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