Create a matrix with alternating rectangles of O and X | GeeksforGeeks

Create a matrix with alternating rectangles of O and X | GeeksforGeeks
Write a code which inputs two numbers m and n and creates a matrix of size m x n (m rows and n columns) in which every elements is either X or 0. The Xs and 0s must be filled alternatively, the matrix should have outermost rectangle of Xs, then a rectangle of 0s, then a rectangle of Xs, and so on.

Input: m = 3, n = 3
Output: Following matrix 
X X X
X 0 X
X X X

Input: m = 4, n = 5
Output: Following matrix
X X X X X
X 0 0 0 X
X 0 0 0 X
X X X X X

Input:  m = 5, n = 5
Output: Following matrix
X X X X X
X 0 0 0 X
X 0 X 0 X
X 0 0 0 X
X X X X X

Use the code for Printing Matrix in Spiral form.

void fill0X(int m, int n)

{

    /*  k - starting row index

        m - ending row index

        l - starting column index

        n - ending column index

        i - iterator    */

    int i, k = 0, l = 0;

    // Store given number of rows and columns for later use

    int r = m, c = n;

    // A 2D array to store the output to be printed

    char a[m][n];

    char x = 'X'; // Iniitialize the character to be stoed in a[][]

    // Fill characters in a[][] in spiral form. Every iteration fills

    // one rectangle of either Xs or Os

    while (k < m && l < n)

    {

        /* Fill the first row from the remaining rows */

        for (i = l; i < n; ++i)

            a[k][i] = x;

        k++;

        /* Fill the last column from the remaining columns */

        for (i = k; i < m; ++i)

            a[i][n-1] = x;

        n--;

        /* Fill the last row from the remaining rows */

        if (k < m)

        {

            for (i = n-1; i >= l; --i)

                a[m-1][i] = x;

            m--;

        }

        /* Print the first column from the remaining columns */

        if (l < n)

        {

            for (i = m-1; i >= k; --i)

                a[i][l] = x;

            l++;

        }

        // Flip character for next iteration

        x = (x == '0')? 'X': '0';

    }

    // Print the filled matrix

    for (i = 0; i < r; i++)

    {

        for (int j = 0; j < c; j++)

            printf("%c ", a[i][j]);

        printf("\n");

    }

}

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