C++ Input Diamond Array - c++

Question

i've had a problem with this code, it skips to 3 asterisks, and my desired program follows the asterisks in order. such as 1 to 2 to 3 asterisks and so on. I want the program to only accept numbers and not special characters.
*
**
***
****
*****
****
***
**
*
Example of desired output ^
Code:
#include <iostream>
using namespace std;
int main() {
cout << "Enter a number: ";
int n;
cin >> n;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n - i; j++) {
cout << " ";
}
for (int j = 1; j <= 2 * i - 1; j++) {
cout << "*";
}
cout << endl;
}
for (int i = n - 1; i >= 1; i--) {
for (int j = 1; j <= n - i; j++) {
cout << " ";
}
for (int j = 1; j <= 2 * i - 1; j++) {
cout << "*";
}
cout << endl;
}
return 0;
}
I input 5 for example, and the asterisks don't follow, it just skips to 3, for example, 1, 3, and so on.

Answer 1

Let me give you some explanation how to find a solution for this problem.
Let us break a big problem into smaller ones.
If we look at the number of asterisks, so, just at the characters, not at the identation or leading spaces, then we see for the input 5:
Row Asterics Number of asterics
0 0
1 * 1
2 ** 2
3 *** 3
4 **** 4
5 ***** 5
6 **** 4
7 *** 3
8 ** 2
9 * 1
10 0
So, we need a function, that will transform the row number to the number of asterics. And, we see a pattern, that looks like a pyramid or triangle.
And this will lead us to the triangular function. Please read here about it. And in this kind of functions the "absolute value" plays a role. In C++ we can use std::abs for this.
Additionally, we can see, that the number of leading spaces to create a diamond is simply the given width minus the number of asterics, divided by 2.
With this know how we can now do the following simple calculation:
Number of stars leading blanks
Row width-abs(row-width) abs(row-width)/2
0 0 5/2 = 2
1 1 4/2 = 2
2 2 3/2 = 1
3 3 2/2 = 1
4 4 1/2 = 0
5 5 0/2 = 0
6 4 1/2 = 0
7 3 2/2 = 0
8 2 3/2 = 1
9 1 4/2 = 2
10 0 5/2 = 2
Additionally we can see the the number of rows is 2 times the width.
With all that, we can come up with a simple program that uses only one for loop.
#include <iostream>
#include <cmath>
#include <algorithm>
#include <string>
int main() {
// Get the max width of the pattern and perform a short input validation
int maxWidth{};
if ((std::cin >> maxWidth) and (maxWidth > 0)) {
// The number of rows for the pattern is dependent on the width. It is a simple relation
const int numberOfRows = 2 * maxWidth;
// Show all rows
for (int row = 1; row < numberOfRows; ++row) {
// Use triangle formular to create star pattern
const std::string starPattern(maxWidth - std::abs(row - maxWidth), '*');
// Create leading spaces
const std::string leadingSpaces(std::abs(row - maxWidth)/2, ' ');
// Show output
std::cout << leadingSpaces << starPattern << '\n';
}
}
else std::cout << "\n*** Error: Invalid input\n\n";
}
Please note. For creating the strings, we use the its constructor number 2. Here you can specify the std::strings length and the filling character.
And if you do not want to use strings, then you need to add again 2 for loops.
Please see:
#include <iostream>
#include <cmath>
int main() {
// Get the max width of the pattern and perform a short input validation
int maxWidth{};
if ((std::cin >> maxWidth) and (maxWidth > 0)) {
// The number of rows for the pattern is dependent on the width. It is a simple relation
const int numberOfRows = 2 * maxWidth;
// Show all rows
for (int row = 1; row < numberOfRows; ++row) {
// Calculate number of characters using the trinagular formular idea
const int numberOfStars = maxWidth - std::abs(row - maxWidth);
const int numberOfSpaces = std::abs(row - maxWidth) / 2;
// print leading spaces
for (int i = 0; i < numberOfSpaces; ++i)
std::cout << ' ';
// print stars
for (int i = 0; i < numberOfStars; ++i)
std::cout << '*';
// Create new line
std::cout << '\n';
}
}
else std::cout << "\n*** Error: Invalid input\n\n";
}
And, if you have even no std::abs the you can create such a function by yourself.
Then the code will be
#include <iostream>
int absValue(int v) {
if (v < 0) v = -v;
return v;
}
int main() {
// Get the max width of the pattern and perform a short input validation
int maxWidth{};
if ((std::cin >> maxWidth) and (maxWidth > 0)) {
// The number of rows for the pattern is dependent on the width. It is a simple relation
const int numberOfRows = 2 * maxWidth;
// Show all rows
for (int row = 1; row < numberOfRows; ++row) {
// Calculate number of characters using the trinagular formular idea
const int numberOfStars = maxWidth - absValue(row - maxWidth);
const int numberOfSpaces = absValue(row - maxWidth) / 2;
// print leading spaces
for (int i = 0; i < numberOfSpaces; ++i)
std::cout << ' ';
// print stars
for (int i = 0; i < numberOfStars; ++i)
std::cout << '*';
// Create new line
std::cout << '\n';
}
}
else std::cout << "\n*** Error: Invalid input\n\n";
}
The output is the same as you have specified it. Please look for the last example with width 5:
And here with width 20: