Construct a Matrix with no element exceeding X and sum of two adjacent elements not exceeding Y

Last Updated : 13 Dec, 2022

Given four integers N, M, X and Y, the task is to construct a N * M matrix such that each cell consists of a value in the range [0, X] such that sum of any two adjacent cells should be less than or equal to Y and the total sum of the matrix should be maximum.

Example:

Input: N = 3, M = 3, X = 5, Y = 3
Output:
3 0 3
0 3 0
3 0 3
Explanation:
All the values of the matrix are between 0 to 5.
The Sum of any two adjacent cells is equal to 3.
The total sum of the matrix is 15, which is maximum possible.

Input: N = 3, M = 3, X = 3, Y = 5
Output:
3 2 3
2 3 2
3 2 3

Approach:
Follow the steps below to solve the problem:

In order to satisfy the two necessary conditions, the matrix needs to be filled with only the following two numbers alternately:

First number = minimum(X, Y)
Second number = minimum(2X, Y) - First Number
Illustration:
N = 3, M = 3, X = 5, Y = 3
First Number = Minimum(X, Y) = Minimum(5, 3) = 3
Second Number = Minimum(2X, Y) - 3 = Minimum(10, 3) - 3 = 3 - 3 = 0
Therefore, sum of any two adjacent cells = 3 + 0 = 3(= Y)

Finally, print the two numbers alternately, first number followed by the second number to maximize the sum of the matrix.

Below is the implementation of the above approach.

C++

// C++ implementation of
// the above approach
#include <iostream>
using namespace std;

// Function to print the
// required matrix
void FindMatrix(int n, int m,
                int x, int y)
{
    int a, b, i, j;

    // For 1*1 matrix
    if (n * m == 1) {
        if (x > y) {
            cout << y << "\n";
        }
        else {
            cout << x << "\n";
        }
        return;
    }

    // Greater number
    a = min(x, y);

    // Smaller number
    b = min(2 * x, y) - a;

    // Sets/Resets for alternate
    // filling of the matrix
    bool flag = true;

    // Print the matrix
    for (i = 0; i < n; i++) {
        for (j = 0; j < m; j++) {
            if (flag)
                cout << a << ' ';
            else
                cout << b << ' ';
            flag = !flag;
        }

        // If end of row is reached
        if (((n % 2 != 0 && m % 2 == 0)
             || (n % 2 == 0 && m % 2 == 0)))
            flag = !flag;
        cout << "\n";
    }
}

// Driver Code
int main()
{

    int N, M, X, Y;
    N = 3;
    M = 3;
    X = 5;
    Y = 3;
    FindMatrix(N, M, X, Y);
}

Java

// Java implementation of
// the above approach
import java.io.*;
public class GFG{

// Function to print the
// required matrix
static void FindMatrix(int n, int m,
                       int x, int y)
{
    int a, b, i, j;

    // For 1*1 matrix
    if (n * m == 1)
    {
        if (x > y) 
        {
            System.out.print(y + "\n");
        }
        else 
        {
            System.out.print(x + "\n");
        }
        return;
    }

    // Greater number
    a = Math.min(x, y);

    // Smaller number
    b = Math.min(2 * x, y) - a;

    // Sets/Resets for alternate
    // filling of the matrix
    boolean flag = true;

    // Print the matrix
    for (i = 0; i < n; i++) 
    {
        for (j = 0; j < m; j++)
        {
            if (flag)
                System.out.print(a + " ");
            else
                System.out.print(b + " ");
            flag = !flag;
        }

        // If end of row is reached
        if (((n % 2 != 0 && m % 2 == 0) || 
             (n % 2 == 0 && m % 2 == 0)))
            flag = !flag;
        System.out.print("\n");
    }
}

// Driver Code
public static void main(String[] args)
{
    int N, M, X, Y;
    N = 3;
    M = 3;
    X = 5;
    Y = 3;
    FindMatrix(N, M, X, Y);
}
}

// This code is contributed by gauravrajput1

Python3

# Python3 implementation of
# the above approach

# Function to print the
# required matrix
def FindMatrix(n, m, x, y):

    # For 1*1 matrix
    if (n * m == 1):
        if (x > y):
            print(y)
        else:
            print(x)
            
        return

    # Greater number
    a = min(x, y)

    # Smaller number
    b = min(2 * x, y) - a

    # Sets/Resets for alternate
    # filling of the matrix
    flag = True

    # Print the matrix
    for i in range(n):
        for j in range(m):
            
            if (flag):
                print(a, end = " ")
            else:
                print(b, end = " ")
                
            flag = not flag

        # If end of row is reached
        if (((n % 2 != 0 and m % 2 == 0) or
             (n % 2 == 0 and m % 2 == 0))):
            flag = not flag
            
        print ()

# Driver Code
N = 3
M = 3
X = 5
Y = 3

FindMatrix(N, M, X, Y)

# This code is contributed by chitranayal

// C# implementation of
// the above approach
using System;
class GFG{

// Function to print the
// required matrix
static void FindMatrix(int n, int m,
                       int x, int y)
{
    int a, b, i, j;

    // For 1*1 matrix
    if (n * m == 1)
    {
        if (x > y) 
        {
            Console.Write(y + "\n");
        }
        else 
        {
            Console.Write(x + "\n");
        }
        return;
    }

    // Greater number
    a = Math.Min(x, y);

    // Smaller number
    b = Math.Min(2 * x, y) - a;

    // Sets/Resets for alternate
    // filling of the matrix
    bool flag = true;

    // Print the matrix
    for (i = 0; i < n; i++) 
    {
        for (j = 0; j < m; j++)
        {
            if (flag)
                Console.Write(a + " ");
            else
                Console.Write(b + " ");
            flag = !flag;
        }

        // If end of row is reached
        if (((n % 2 != 0 && m % 2 == 0) || 
             (n % 2 == 0 && m % 2 == 0)))
            flag = !flag;
        Console.Write("\n");
    }
}

// Driver Code
public static void Main(String[] args)
{
    int N, M, X, Y;
    N = 3;
    M = 3;
    X = 5;
    Y = 3;
    FindMatrix(N, M, X, Y);
}
}

// This code is contributed by Amit Katiyar

JavaScript

<script>

// Javascript implementation of
// the above approach

// Function to print the
// required matrix
function FindMatrix(n, m, x, y)
{
    var a, b, i, j;

    // For 1*1 matrix
    if (n * m == 1)
    {
        if (x > y) 
        {
            document.write(y);
        }
        else 
        {
            document.write(x);
        }
        return;
    }

    // Greater number
    a = Math.min(x, y);

    // Smaller number
    b = Math.min(2 * x, y) - a;

    // Sets/Resets for alternate
    // filling of the matrix
    var flag = true;

    // Print the matrix
    for (i = 0; i < n; i++) 
    {
        for (j = 0; j < m; j++)
        {
            if (flag)
                document.write(a + " ");
            else
                document.write(b + " ");
                
            flag = !flag;
        }

        // If end of row is reached
        if (((n % 2 != 0 && m % 2 == 0) || 
             (n % 2 == 0 && m % 2 == 0)))
            flag = !flag;
            
        document.write("<br>");
    }
}
    
// Driver Code
var N, M, X, Y;
N = 3;
M = 3;
X = 5;
Y = 3;

FindMatrix(N, M, X, Y);

// This code is contributed by Khushboogoyal499

</script>

Output:

3 0 3 
0 3 0 
3 0 3

Time complexity: O(N * M)
Space complexity: O(1)

deepanshujindal634

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