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.

Examples:

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 Input: m = 6, n = 7 Output: Following matrix X X X X X X X X 0 0 0 0 0 X X 0 X X X 0 X X 0 X X X 0 X X 0 0 0 0 0 X X X X X X X X

**We strongly recommend to minimize the browser and try this yourself first.**

This question was asked in campus recruitment of Shreepartners Gurgaon. I followed the following approach.

**1) ** Use the code for Printing Matrix in Spiral form.

**2) **Instead of printing the array, inserted the element ‘X’ or ‘0’ alternatively in the array.

Following is implementation of the above approach.

## CPP

#include <stdio.h> // Function to print alternating rectangles of 0 and X 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"); } } /* Driver program to test above functions */ int main() { puts("Output for m = 5, n = 6"); fill0X(5, 6); puts("\nOutput for m = 4, n = 4"); fill0X(4, 4); puts("\nOutput for m = 3, n = 4"); fill0X(3, 4); return 0; }

## Java

// Java code to demonstrate the working. import java.io.*; class GFG { // Function to print alternating // rectangles of 0 and X static 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[][] = new char[m][n]; // Iniitialize the character // to be stoed in a[][] char x = 'X'; // 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++) System.out.print(a[i][j] + " "); System.out.println(); } } /* Driver program to test above functions */ public static void main (String[] args) { System.out.println("Output for m = 5, n = 6"); fill0X(5, 6); System.out.println("Output for m = 4, n = 4"); fill0X(4, 4); System.out.println("Output for m = 3, n = 4"); fill0X(3, 4); } } // This code is contributed by vt_m

Output:

Output for m = 5, n = 6 X X X X X X X 0 0 0 0 X X 0 X X 0 X X 0 0 0 0 X X X X X X X Output for m = 4, n = 4 X X X X X 0 0 X X 0 0 X X X X X Output for m = 3, n = 4 X X X X X 0 0 X X X X X

Time Complexity: O(mn)

Auxiliary Space: O(mn)

Please suggest if someone has a better solution which is more efficient in terms of space and time.

This article is contributed by **Deepak Bisht**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above