Given a positive integer **n** print matrix containing an X as in the example below

**Example:**

Input:n= 5Output:x---x -x-x- --x-- -x-x- x---xExplanation:Each line contains exactly n = 5 characters and two 'x's. Each diagonal contains 'x'

Notice that on each line there will be exactly two

`x`

and the other characters are all `-`

. The only exception is the middle line which will have only one `x`

.
How can we use this info?
We can print the matrix line by line. For each line we need to know the two columns where we should print

`x`

.
How can we know these two columns for every line?
We'll store the columns' indices in two variables

`firstX`

and `secondX`

, initially equal to `1`

and `n`

respectively for the first line. How should these variables change from line to line?

`firstX`

should get incremented and `secondX`

should get decremented. Now for each line, you now the 'special' columns. How will you print that line?

We'll use a

`for`

loop to iterate through every column index from `1`

to `n`

. What should we do inside the loop?

We should use an

`if - else`

statement and check if the current index is `firstX`

or `secondX`

and if so, print `x`

. Otherwise, print `-`

.
The task is to print a matrix of size `n x n`

where the diagonals contain the character 'x' and all other positions contain the character '-'.

A single integer `n`

(1 ≤ n ≤ 1000).

A matrix of size `n x n`

with 'x' on the diagonals and '-' elsewhere.

Input:n = 5Output:x---x -x-x- --x-- -x-x- x---x

The core challenge is to correctly place the 'x' characters on both diagonals of the matrix. This problem is significant in understanding how to manipulate and traverse matrices, which is a common task in computer science.

Potential pitfalls include incorrectly indexing the matrix or not properly handling the middle row when `n`

is odd.

To solve this problem, we can use a simple nested loop to construct each row of the matrix. For each row, we determine the positions of the 'x' characters and fill the rest with '-'.

A naive solution would involve manually constructing each row, but this is not scalable or efficient. Instead, we can use a systematic approach to determine the positions of 'x' for each row.

We can use two variables, `firstX`

and `secondX`

, to track the positions of 'x' in each row. Initially, `firstX`

is 0 and `secondX`

is `n-1`

. For each subsequent row, we increment `firstX`

and decrement `secondX`

.

1. Initialize `firstX`

to 0 and `secondX`

to `n-1`

.

2. For each row from 0 to `n-1`

:

- Initialize an empty string for the row.
- For each column from 0 to
`n-1`

: - If the column index is
`firstX`

or`secondX`

, append 'x' to the row string. - Otherwise, append '-'.
- Print the row string.
- Increment
`firstX`

and decrement`secondX`

.

```
public class PrintX {
public static void main(String[] args) {
int n = 5; // Example input
printXMatrix(n);
}
public static void printXMatrix(int n) {
// Initialize the positions of 'x' in the first row
int firstX = 0;
int secondX = n - 1;
// Loop through each row
for (int i = 0; i < n; i++) {
StringBuilder row = new StringBuilder();
// Loop through each column in the current row
for (int j = 0; j < n; j++) {
if (j == firstX || j == secondX) {
row.append('x');
} else {
row.append('-');
}
}
// Print the constructed row
System.out.println(row.toString());
// Update the positions of 'x' for the next row
firstX++;
secondX--;
}
}
}
```

The time complexity of this solution is `O(n^2)`

because we have a nested loop where both the outer and inner loops run `n`

times. The space complexity is `O(n)`

for storing the row string.

Potential edge cases include:

`n = 1`

: The output should be a single 'x'.`n = 2`

: The output should be:xx xx

These cases are handled naturally by the algorithm without any special conditions.

To test the solution, we can use a variety of test cases:

- Simple cases like
`n = 1`

and`n = 2`

. - Odd and even values of
`n`

. - Large values of
`n`

to ensure performance.

When approaching such problems, it's helpful to:

- Break down the problem into smaller parts.
- Use systematic approaches to handle repetitive tasks.
- Think about edge cases and how to handle them.

Practicing similar problems and studying algorithms can improve problem-solving skills.

In this blog post, we discussed how to print a matrix containing an 'X' pattern given a positive integer `n`

. We explored the problem definition, approach, algorithm, and provided a detailed Java implementation. Understanding and solving such problems is crucial for developing strong problem-solving skills in computer science.

For further reading and practice, consider the following resources:

- Matrix Manipulation in Java
- LeetCode for more coding challenges.
- HackerRank for algorithm tutorials and practice problems.