I encountered a tricky bonus question on our Loops In C quiz that I couldn't figure out. The question involved generating a specific pattern based on an input integer. For example, the outputs for inputs 2, 3, and 4 look like this:
- For input 4:
```
4444444
4333334
4322234
4321234
4322234
4333334
4444444
```
- For input 2:
```
222
212
222
```
- For input 3:
```
33333
32223
32123
32223
33333
```
I answered all the other quiz questions correctly, but this one, worth 20 bonus points, stumped me. Below is my attempt at the solution:
```c
#include
int main() {
int n;
printf("Input: ");
scanf("%d", &n);
printf("nOutput:n");
for(int i = 0; i <= ((2 * n) - 1); i++)
{
for(int j = 0; j <= ((2 * n) - 1); j++)
{
if(i == 0 || i == (2 * n) - 1 || j == 0 || j == (2 * n) - 1)
{
printf("%d", n);
}
else
{
printf(" ");
}
}
printf("n");
}
return 0;
}
```
This logic prints a box with the outer layer being the input number, but I'm struggling with printing the inner layers. I'd appreciate any guidance on tackling this problem and suggestions for where to find similar practice problems!
6 Answers
To solve this pattern, focus on the maximum distance from the center in terms of coordinates. You don't need to build layers explicitly. Just calculate the value for each cell based on its distance from the center of the grid, using the formula for the printed number: `max(n - i, n - j, i - n + 1, j - n + 1)` for whatever edge you're considering. This allows for a more straightforward approach without too much complexity in looping.
You're on the right track with your existing code! For the inner loops, you might simplify it by determining the maximum value from these calculations: `n - i`, `i - n + 2`, `n - j`, and `j - n + 2`. Using the max value from those gives you the correct number to print for each position.
I'd recommend breaking down the problem by going row by row instead of treating it as a fully layered box. Notice how the middle row is the most crucial—it's something like `4321234`. From there, changing numbers in subsequent rows helps you build the pattern in a structured way without overcomplicating the problem.
If you're looking to add a diagonal cross to the pattern, consider using a condition like `if (i == j || i + j == 2 * (n - 1))` to handle those elements. It helps in creating the crossing effect in the box. Keep experimenting with different conditions to see what fits your desired output!
To understand the pattern, note that you're creating a grid of size `(2n-1) x (2n-1)`, which means your center is at `(n-1, n-1)`. The value at any coordinates `(i, j)` corresponds to the shortest distance to the edge, which needs to incorporate both row and column movements. After establishing the width with `2n - 1`, try implementing your logic in segments and gradually build from those center lines. That's often a more intuitive way of coding this out!
I think this resembles those puzzle problems you find on coding practice sites like LeetCode or HackerRank. If you're keen on tackling similar challenges, check out those platforms. They usually have a wide range of problems. For this specific one, keep in mind that the general rule is based on distance—focusing on specific lines and building the solution from the mid-line outwards might also help!
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