Maximum area that can be saved in the given Matrix

Consider a matrix of N rows and M columns. Initially, there is a cell at co-ordinate (X, Y), which is blue. Then the task is to place the Yellow color in an optimal cell such that the area with yellow color after ending below simulation is the maximum possible. The simulation takes place as follows (after putting a yellow color in a cell):

• The cells, that are adjacent to blue cells and not colored yellow becomes blue.
• The cells, that are adjacent to yellow cells and not colored blue becomes yellow.

Explanation of simulation

Examples:

Input: N = 3, M = 2, X = 1, Y = 2

Output: 4

Explanation: The blue colored cell is placed at (1, 2) initially. It will be optimal to place yellow cell at (2, 2). By doing so we will get yellow colored area 4 sq. units, which is maximum possible.

Explanation of test case 1

Input: N = 3, M = 4, X = 2, Y = 2

Output: 6

Explanation: It can be verified that placing yellow color at optimal cell will give us the maximum yellow colored area as 6 sq. units.

Approach: Implement the idea below to solve the problem

This problem is an observation based. Let us the observations:

• The optimal way to choose the starting cell of yellow color is the adjacent cell of the first blue cell. We can place the first yellow color in a cell just above, just below, just left or just right of the initial blue cell.
• Out of the above four choices, wherever you place the first yellow color cell, a part of the matrix will be blue.
  • Say, If you place the yellow cell just below the blue cell, all the rows above the blue will be colored with blue with the simulation for sure.
  • If we place the yellow cell above the blue cell, the lower part will get blue with the simulation.
  • Similarly for right and left positions.

Steps were taken to solve the problem:

• Create a variable let say Total to store the total possible area of matrix and initialize it with N*M
• Create six variables let say, A, B, C, D, C1, C2 and C3
• Set all the above variables as follows:
  • A = X*M
  • B = (N-X+1)*M
  • C = Y*N
  • D = (M-Y+1)*N
  • C1 = min (A, B)
  • C2 = min (C, D)
  • C3 = min (C1, C2)
• Output (Total – C3)

Code to implement the approach:

C++

//code by flutterfly
#include <iostream>

// Function declaration
void Max_Area(int N, int M, int X, int Y);

// Driver Function
int main()
{
    // Inputs
    int N = 3, M = 2, X = 1, Y = 1;

    // Function call
    Max_Area(N, M, X, Y);

    return 0;
}

void Max_Area(int N, int M, int X, int Y)
{

    // Total area of the matrix
    int total = N * M;

    // a,b,c,d are no.of blue cells according
    // to the position of the yellow cell
    int A, B, C, D, C1, C2, C3;

    A = X * M;                 // The above grids are blue
    B = (N - X + 1) * M;        // The below grids are blue
    C = Y * N;                 // The left grids are blue
    D = (M - Y + 1) * N;        // The right grids are blue
    C1 = std::min(A, B);
    C2 = std::min(C, D);
    C3 = std::min(C1, C2);

    // c3 has minimum of a,b,c,d
    // total-c3 gives you the number
    // of yellow grids
    std::cout << total - C3 << std::endl;
}

Java

// Java code to implement the approach

// Driver Class
public class Main {

    // Driver Function
    public static void main(String[] args)
    {
        // Inputs
        int N = 3, M = 2, X = 1, Y = 1;

        // Function call
        Max_Area(N, M, X, Y);
    }
    public static void Max_Area(int N, int M, int X, int Y)
    {

        // Total area of the matrix
        int total = N * M;

        // a,b,c,d are no.of blue cells according
        // to the position of the yellow cell
        int A, B, C, D, C1, C2, C3;

        A = X * M; // The above grids are blue
        B = (N - X + 1) * M; // The below grids are blue
        C = Y * N; // The left grids are blue
        D = (M - Y + 1) * N; // The right grids are blue
        C1 = Math.min(A, B);
        C2 = Math.min(C, D);
        C3 = Math.min(C1, C2);

        // c3 has minimum of a,b,c,d
        // total-c3 gives you number
        // of yellow grids
        System.out.println(total - C3);
    }
}

Python

# code by flutterfly
def Max_Area(N, M, X, Y):
    # Total area of the matrix
    total = N * M

    # a,b,c,d are no. of blue cells according
    # to the position of the yellow cell
    A = X * M          # The above grids are blue
    B = (N - X + 1) * M  # The below grids are blue
    C = Y * N          # The left grids are blue
    D = (M - Y + 1) * N  # The right grids are blue
    C1 = min(A, B)
    C2 = min(C, D)
    C3 = min(C1, C2)

    # C3 has the minimum of a, b, c, d
    # total - C3 gives you the number
    # of yellow grids
    print(total - C3)

# Inputs
N, M, X, Y = 3, 2, 1, 1

# Function call
Max_Area(N, M, X, Y)

C#

//code by flutterfly
using System;

public class MainClass
{
    // Function declaration
    public static void Max_Area(int N, int M, int X, int Y)
    {
        // Total area of the matrix
        int total = N * M;

        // a,b,c,d are no. of blue cells according
        // to the position of the yellow cell
        int A = X * M;          // The above grids are blue
        int B = (N - X + 1) * M;  // The below grids are blue
        int C = Y * N;          // The left grids are blue
        int D = (M - Y + 1) * N;  // The right grids are blue
        int C1 = Math.Min(A, B);
        int C2 = Math.Min(C, D);
        int C3 = Math.Min(C1, C2);

        // C3 has the minimum of a, b, c, d
        // total - C3 gives you the number
        // of yellow grids
        Console.WriteLine(total - C3);
    }

    // Driver Function
    public static void Main(string[] args)
    {
        // Inputs
        int N = 3, M = 2, X = 1, Y = 1;

        // Function call
        Max_Area(N, M, X, Y);
    }
}

Javascript

// Function declaration
function Max_Area(N, M, X, Y) {
    // Total area of the matrix
    const total = N * M;

    // a, b, c, d are no. of blue cells according
    // to the position of the yellow cell
    const A = X * M;         // The above grids are blue
    const B = (N - X + 1) * M;  // The below grids are blue
    const C = Y * N;         // The left grids are blue
    const D = (M - Y + 1) * N;  // The right grids are blue
    const C1 = Math.min(A, B);
    const C2 = Math.min(C, D);
    const C3 = Math.min(C1, C2);

    // C3 has the minimum of a, b, c, d
    // total - C3 gives you the number
    // of yellow grids
    console.log(total - C3);
}

// Inputs
const N = 3, M = 2, X = 1, Y = 1;

// Function call
Max_Area(N, M, X, Y);

Output

4

Time Complexity: O(1)
Auxiliary Space: O(1)