Maximum possible cost of buildings with pool view

Consider a series of buildings in a 2-D square plot with different heights and only those buildings are chargeable that have a clear and straight view of the pool from their building. The pool also spans a 2-D square plot with certain coordinates. Given the Coordinates of a square pool, the number of buildings, the charge being paid by customers, and the height of the building at each coordinate, the task is to find the maximum cost possible considering the fee is distributed and charged equally across all heights.

Examples:

Input: pool_coordinates = {{0, 0}, {2, 0}, {0, 2}, {2, 2}}, buildings = 5, charge =10
buildings coordinates = {{0, 0, 50}, {4, 6, 30}, {5, 5, 20}, {0, 1, 10}, {0, 2, 10}}
Output: 500
Explanation: The first three rows are chargeable 
with effective pool view heights as 40, 0 and 10.

5 Buildings with Charge 10

Input: pool_coordinates = {{1, 0}, {3, 0}, {1, 1}, {3, 1}}, buildings = 3, charge = 5 
buildings coordinates = {{0, 0, 50}, {0, 1, 10}, {0, 2, 10}}
Output: 0
Explanation: Second, third and fourth rows are chargeable 
but they don’t have any building at all.

3 Buildings with Charge 5

Approach: To solve the problem follow the below idea:

The problem can be solved using the greedy approach by finding maximum heights from right to left since buildings with bigger heights will block the view of the smaller buildings. The remaining part of these buildings will then be charged equally.

Follow the given steps to solve the problem:

• Find the maximum coordinate where any building is located since that will give us the entire dimension of the grid of the buildings.
• Find the maximum and minimum coordinates of the pool as max_ij and min_ij.
• Find maximum of n, max_ij & min_ij and store it in n.
• Create a grid of size (n + 1) * (n + 1) with all values 0 initially.
• Initialize the heights of the buildings at their corresponding coordinates.
• Traverse from min_ij to max_ij for the rows to find the respective chargeable heights.
• Nested Traverse from n – 1 to 0 for columns.
  • If grid[i][j] comes to be greater than max_cr_val (peak element from right to left along a row) then subtract max_cr_val from grid[i][j] else 
    make grid[i[j] = 0.
  • Update max_cr_val by maximum of max_cr_val and temp (initial grid[i][j] value).
• Traverse again from min_ij to max_ij for rows and from 0 to n for columns.
• Calculate all the overall cost = cost + grid[i][j] (remaining chargeable height) * charge.
• Return cost as the final answer.

Below is the implementation for the above approach:

C++14

// C++ program for above approach

#include <bits/stdc++.h>
using namespace std;

#define ll long long

// Function to find the maximum cost possible
ll Total_Max_Cost(vector<vector<int> >& pool_cr,
                  int& buildings, int& charge,
                  vector<vector<int> >& buildings_cr)
{
    int max_ij = INT_MIN, max_cr_val = INT_MIN, n = INT_MIN;
    int min_ij = INT_MAX, temp, i, j;
    ll cost = 0;

    // Traversing to find the maximum coordinate
    for (i = 0; i < buildings; i++)
        n = max(
            { n, buildings_cr[i][0], buildings_cr[i][1] });

    // Finding minimum and maximum coordinates
    // of the pool
    for (i = 0; i < 4; i++) {

        max_ij = max(max_ij, pool_cr[i][0]);
        min_ij = min(min_ij, pool_cr[i][0]);
    }

    n = max({ n, max_ij, min_ij });

    int grid[n + 1][n + 1];
    memset(grid, 0, sizeof(grid));

    for (i = 0; i < buildings; i++)
        grid[buildings_cr[i][0]][buildings_cr[i][1]]
            = buildings_cr[i][2];

    // Calculate the visible part of every
    // building in front of the pool
    for (i = min_ij; i <= max_ij; i++) {
        max_cr_val = grid[i][n];

        for (j = n - 1; j >= 0; j--) {
            temp = grid[i][j];

            if (grid[i][j] > max_cr_val)
                grid[i][j] -= max_cr_val;
            else
                grid[i][j] = 0;

            max_cr_val = max(max_cr_val, temp);
        }
    }

    for (i = min_ij; i <= max_ij; i++) {
        for (j = 0; j < n + 1; j++)
            cost += charge * grid[i][j];
    }

    return cost;
}

// Driver code
int main()
{
    vector<vector<int> > pool_cr
        = { { 0, 0 }, { 2, 0 }, { 0, 2 }, { 2, 2 } };
    int buildings = 5, charge = 10;
    vector<vector<int> > buildings_cr = { { 0, 0, 50 },
                                          { 4, 6, 30 },
                                          { 5, 5, 20 },
                                          { 0, 1, 10 },
                                          { 0, 2, 10 } };

    // Function call
    cout << Total_Max_Cost(pool_cr, buildings, charge,
                           buildings_cr);

    return 0;
}

Java

// Java program for above approach
import java.util.*;
import java.util.Arrays;

class GFG
{

// Function to find the maximum cost possible
static int Total_Max_Cost(int[][] pool_cr,
                  int buildings, int charge,
                  int[][] buildings_cr)
{
    int max_ij = Integer.MIN_VALUE, max_cr_val = Integer.MIN_VALUE, n = Integer.MIN_VALUE;
    int min_ij = Integer.MAX_VALUE, temp, i, j;
    int cost = 0;

    // Traversing to find the maximum coordinate
    for (i = 0; i < buildings; i++)
        n = Math.max(Math.max(n, buildings_cr[i][0]), buildings_cr[i][1]);

    // Finding minimum and maximum coordinates
    // of the pool
    for (i = 0; i < 4; i++) {

        max_ij = Math.max(max_ij, pool_cr[i][0]);
        min_ij = Math.min(min_ij, pool_cr[i][0]);
    }

    n = Math.max(Math.max(n, max_ij), min_ij );

    int grid[][] = new int[n + 1][n + 1];
    for( i = 0; i <n + 1; i++)
    {
        for( j = 0; j <n + 1; j++)
        {
            grid[i][j] = 0;
        }
    }

    for (i = 0; i < buildings; i++)
        grid[buildings_cr[i][0]][buildings_cr[i][1]]
            = buildings_cr[i][2];

    // Calculate the visible part of every
    // building in front of the pool
    for (i = min_ij; i <= max_ij; i++) {
        max_cr_val = grid[i][n];

        for (j = n - 1; j >= 0; j--) {
            temp = grid[i][j];

            if (grid[i][j] > max_cr_val)
                grid[i][j] -= max_cr_val;
            else
                grid[i][j] = 0;

            max_cr_val = Math.max(max_cr_val, temp);
        }
    }

    for (i = min_ij; i <= max_ij; i++) {
        for (j = 0; j < n + 1; j++)
            cost += charge * grid[i][j];
    }

    return cost;
}


// Driver code
public static void main(String[] args)
{
    int[][] pool_cr
        = { { 0, 0 }, { 2, 0 }, { 0, 2 }, { 2, 2 } };
    int buildings = 5, charge = 10;
    int[][] buildings_cr = { { 0, 0, 50 },
                                          { 4, 6, 30 },
                                          { 5, 5, 20 },
                                          { 0, 1, 10 },
                                          { 0, 2, 10 } };

    // Function call
    System.out.print( Total_Max_Cost(pool_cr, buildings, charge,
                           buildings_cr));


}
}

// This code is contributed by sanjoy_62.

Python3

# Python program for above approach

# Function to find the maximum cost possible
import sys
def Total_Max_Cost(pool_cr, buildings, charge, buildings_cr):
  max_ij = -sys.maxsize-1
  max_cr_val = -sys.maxsize-1
  n = -sys.maxsize-1
  min_ij = sys.maxsize
  cost = 0

  #Traversing to find the maximum coordinate
  for i in range(0, buildings):
    n = max(n, buildings_cr[i][0], buildings_cr[i][1])

  # Finding minimum and maximum coordinates of the pool
  for i in range(0, 4):
    max_ij = max(max_ij, pool_cr[i][0])
    min_ij = min(min_ij, pool_cr[i][0])
  n = max(n, max_ij, min_ij)
  grid = [[0 for i in range(n+1)] for j in range(n+1)]

  for i in range(0, buildings):
    grid[buildings_cr[i][0]][buildings_cr[i][1]] = buildings_cr[i][2]

  # Calculate the visible part of every building in front of the pool
  for i in range(min_ij, max_ij + 1):
    max_cr_val = grid[i][n]
    for j in range(n-1,-1,-1):
      temp = grid[i][j]
      if (grid[i][j] > max_cr_val):
        grid[i][j] -= max_cr_val
      else:
        grid[i][j] = 0
      max_cr_val = max(max_cr_val, temp)
  for i in range(min_ij, max_ij + 1):
    for j in range(0, n + 1):
      cost += charge * grid[i][j];
  return cost

# Driver code
if __name__ == "__main__":
  pool_cr = [[0, 0], [2, 0], [0, 2], [2, 2]]
  buildings = 5
  charge = 10
  buildings_cr = [[0, 0, 50], [4, 6, 30], [5, 5, 20], [0, 1, 10],[0, 2, 10]]

  # Function call
  print(Total_Max_Cost(pool_cr,buildings,charge,buildings_cr))

  # This code is contributed by Atul_kumar_Shrivastava

C#

// C# code for the above approach
using System;
class GFG {

// Function to find the maximum cost possible
static int Total_Max_Cost(int[,] pool_cr,
                  int buildings, int charge,
                  int[,] buildings_cr)
{
    int max_ij = Int32.MinValue, max_cr_val = Int32.MinValue, n = Int32.MinValue;
    int min_ij = Int32.MaxValue, temp, i, j;
    int cost = 0;

    // Traversing to find the maximum coordinate
    for (i = 0; i < buildings; i++)
        n = Math.Max(Math.Max(n, buildings_cr[i,0]), buildings_cr[i,1]);

    // Finding minimum and maximum coordinates
    // of the pool
    for (i = 0; i < 4; i++) {

        max_ij = Math.Max(max_ij, pool_cr[i,0]);
        min_ij = Math.Min(min_ij, pool_cr[i,0]);
    }

    n = Math.Max(Math.Max(n, max_ij), min_ij );

    int[,] grid = new int[n + 1, n + 1];
    for( i = 0; i <n + 1; i++)
    {
        for( j = 0; j <n + 1; j++)
        {
            grid[i,j] = 0;
        }
    }

    for (i = 0; i < buildings; i++)
        grid[buildings_cr[i,0], buildings_cr[i,1]]
            = buildings_cr[i, 2];

    // Calculate the visible part of every
    // building in front of the pool
    for (i = min_ij; i <= max_ij; i++) {
        max_cr_val = grid[i, n];

        for (j = n - 1; j >= 0; j--) {
            temp = grid[i,j];

            if (grid[i,j] > max_cr_val)
                grid[i,j] -= max_cr_val;
            else
                grid[i,j] = 0;

            max_cr_val = Math.Max(max_cr_val, temp);
        }
    }

    for (i = min_ij; i <= max_ij; i++) {
        for (j = 0; j < n + 1; j++)
            cost += charge * grid[i,j];
    }

    return cost;
}

// Driver code
public static int Main()
{
    int[,] pool_cr
        = { { 0, 0 }, { 2, 0 }, { 0, 2 }, { 2, 2 } };
    int buildings = 5, charge = 10;
    int[,] buildings_cr = { { 0, 0, 50 },
                                          { 4, 6, 30 },
                                          { 5, 5, 20 },
                                          { 0, 1, 10 },
                                          { 0, 2, 10 } };

    // Function call
    Console.WriteLine( Total_Max_Cost(pool_cr, buildings, charge,
                           buildings_cr));
    return 0;
}
}

// This code is contributed by code_hunt.

Javascript

// javascript program for above approach
const INT_MAX = 2147483647;
const INT_MIN = -2147483648;

// Function to find the maximum cost possible
function Total_Max_Cost(pool_cr, buildings, charge, buildings_cr)
{
    let max_ij = INT_MIN, max_cr_val = INT_MIN, n = INT_MIN;
    let min_ij = INT_MAX, temp, i, j;
    let cost = 0;

    // Traversing to find the maximum coordinate
    for (i = 0; i < buildings; i++)
        n = Math.max(n, Math.max(buildings_cr[i][0], buildings_cr[i][1] ));

    // Finding minimum and maximum coordinates
    // of the pool
    for (i = 0; i < 4; i++) {

        max_ij = Math.max(max_ij, pool_cr[i][0]);
        min_ij = Math.min(min_ij, pool_cr[i][0]);
    }

    n = Math.max(n, Math.max(max_ij, min_ij));

    let grid = new Array(n+1);
    for(let i = 0; i < n+1; i++){
        grid[i] = new Array(n+1).fill(0);
    }

    for (i = 0; i < buildings; i++)
        grid[buildings_cr[i][0]][buildings_cr[i][1]] = buildings_cr[i][2];

    // Calculate the visible part of every
    // building in front of the pool
    for (i = min_ij; i <= max_ij; i++) {
        max_cr_val = grid[i][n];

        for (j = n - 1; j >= 0; j--) {
            temp = grid[i][j];

            if (grid[i][j] > max_cr_val)
                grid[i][j] -= max_cr_val;
            else
                grid[i][j] = 0;

            max_cr_val = Math.max(max_cr_val, temp);
        }
    }

    for (i = min_ij; i <= max_ij; i++) {
        for (j = 0; j < n + 1; j++)
            cost += charge * grid[i][j];
    }

    return cost;
}

// Driver code
let pool_cr = [ [ 0, 0 ], [ 2, 0 ], [ 0, 2 ], [ 2, 2 ] ];
let buildings = 5, charge = 10;
let buildings_cr = [  [ 0, 0, 50 ],
                      [ 4, 6, 30 ],
                      [ 5, 5, 20 ],
                      [ 0, 1, 10 ],
                      [ 0, 2, 10 ] ];

// Function call
console.log(Total_Max_Cost(pool_cr, buildings, charge, buildings_cr));

// The code is contributed by Nidhi goel.

Output

500

Time Complexity: O(N²) // since two nested loops are used so the time taken by the algorithm is n*n
Auxiliary Space: O(N²) // since a 2D matrix is used total space is N*N