Home » C++ Fizz Buzz alternative version [closed]

C++ Fizz Buzz alternative version [closed]

Solutons:


You should be able to do this equally well with either for-loop or while-loop. However, that is not why your program is failing. As mentioned earlier by @EvilTeach, your program is basically accepting input from stdin(via cin) outside the braces({}) in main() function.

A Function signature in C/C++ is defined as:-

return-type funcName(ArgType1 arg1, ... , ArgTypeN argN) 
{
    // cin goes here
    // Your loop goes here
}

For starters you need to move the cin inside the braces {}. That should get your past your compilation error. It is generally a good idea to print a prompt as well when you are asking for input. As a general rule, if you ask for help, always include any compilation errors you get. That helps zero in on the issue. Welcome to stack overflow.

#include <iostream>
using namespace std;
int main ()
{
       cout << "Enter your number ";      // Prompt for input
       cin >> n;                          // Get the input.
       for (int i = 1; i <= n; i++) 
       {
               if ((i % 15) == 0)
                       cout << "FizzBuzzn";
               else if ((i % 3) == 0)
                       cout << "Fizzn";
               else if ((i % 5) == 0)
                       cout << "Buzzn";
               else
                       cout << i << "n";
       }
       return 0;
}

Naive Solution

for(int i = 1; i <= n; i++)
{
    if(i % 15 == 0)
        cout << "FizzBuzzn";
    else if(i % 3 == 0)  
        cout << "Fizzn";
    else if(i % 5 == 0)
        cout << "Buzzn";
    else
        cout << i << "n";       
}

Problem with Naive solution: % 15 is equivalent to % 3 && % 5.

Hence, there is no point in checking the same condition again.

Efficient Approach

for(int i = 1; i <= n; i++)
{
    string d ="";
    if(i % 3 == 0) d += "Fizz";
    if(i % 5 == 0) d += "Buzz";
    if(d == "") cout << i << "n";
    else cout<< d << "n"; 
}

Problem with the above Efficient solution: ‘%’ is a costly operator.

The complexity of ‘%’ operator is O(n ^ 2)

More Efficient Approach

int c3 = 0;
int c5 = 0;
for(int i = 1; i <= n; i++)
{
    c3++;
    c5++;
    string d = "";
     if(c3 == 3) 
     {
         d += "Fizz"; 
         c3 = 0;
     }
     if(c5 == 5)
     {
         d += "Buzz";
         c5 = 0;
     }   
     if(d == "") cout << i << "n";
     else cout << d << "n";
}

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