Printing stars using nested loops

Question

I just got back from my interview where I had to solve the following problem:

Write a C++ program that prints the following pattern using nested for loops.

*
***
*****
***
*

Here is my solution:

int inc = 2;
for (int i = 1; i >= 1; i += inc){
    for (int j = 0; j < i; j++ )
        std::cout<<'*';
    std::cout<<endl;
    if ( i == 5 )
        inc = -2;
}

I'm just curious to know if there are better solutions (any criteria for better) or just different approaches.

Answer 1

This answer is simpler and shorter:

for(;0;)
    for(;0;)
        ;
std::cout << "*\n***\n*****\n***\n*\n";

Perhaps you'd do better to consider how you could generalise your answer by parametrising it e.g. by the depth and the angle of the tree it outputs.

Answer 2

Edit Oh well, not using nested for... Sorry about that

int inc = 2;
for (int i = 1; i >= 1; i += inc){
    std::cout<<std::string(i,'*')<<endl;
    if ( i == 5 )
        inc = -2;
}

Answer 3

This allows for reusability - any number of lines, whatever increment, to get the sideways triangle format:

int NUMBER_LINES = 5;
int INC = 2;
int stars = 1;
for( int i = 1; i <= NUMBER_LINES; i++ )
{
   for( int j = 0; j < stars; j++ )
      std::cout<<'*';
   std::cout<<std::endl;
   if( i < (NUMBER_LINES / 2 + NUMBER_LINES % 2) )
      stars += INC;
   else if( NUMBER_LINES % 2 == 0 && i == NUMBER_LINES / 2 )
      ;
   else
      stars -= INC;
}

Answer 4

@ecatmur hit the nail on the head: unless we know which are the moving parts, the hard-coded answer is the best. Otherwise, we can only guess what’s supposed to be variable:

1. Does the number of lines change?
2. Does the step increment change?
3. Does the printed symbol change? E.g. + instead of *, or maybe even [], necessitating a new data type (string instead of char).
4. …

Every solution will necessarily make some assumptions about these questions so it’s good to clear them up beforehand, or at least to be aware of them. As an interviewer, this is what I’d expect as the answer to this question.

Answer 5

I'm not sure about better, but this qualifies as a "different" solution.

int count;
for (int i = -2; i <= 2; i++){
    count = 1 + 2 * (2 - abs(i));
    for (int j = 0; j < count; j++ )
        std::cout<<'*';
    std::cout<<std::endl;
}
std::cin >> count;
return 0;

Answer 6

A friend of mine missed the part about the for loops and came up with the following recursive version:

const int MAX_STARS = 5;
const int INC = 2; 
const int STARTING_STARS = 1;

void printRowOfStars(int length){
         while( length-- > 0){
             std::cout<<'*';                
         }
         std::cout<<'\n';
     }

void printTriangle(int i = STARTING_STARS){
     assert (MAX_STARS - STARTING_STARS) % INC == 0;
     printRowOfStars(i);
     if (i == MAX_STARS)
        return;
     printTriangle(i+INC);
     printRowOfStars(i);
     }

int main()
{
    printTriangle();
    system("pause");
    return 0;
}

I liked it so I thought why not post it