## The challenge

Two red beads are placed between every two blue beads. There are N blue beads. After looking at the arrangement below work out the number of red beads.

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Implement `count_red_beads(n)` (`countRedBeads(n)`) so that it returns the number of red beads.
If there are less than 2 blue beads return 0.

## The solution in C

Option 1:

 ``````1 2 3 4 `````` ``````int countRedBeads(n) { if(n<=1) return 0; return 2*(n-1); } ``````

Option 2:

 ``````1 2 3 4 5 6 7 `````` ``````int countRedBeads(n) { if (n < 2) { return 0; } else { return n + (n-2); } } ``````

Option 3:

 ``````1 2 3 4 `````` ``````int countRedBeads(n) { return n<2 ? 0 : 2*n-2; } ``````

## Test cases to validate our solution

 `````` 1 2 3 4 5 6 7 8 9 10 11 `````` ``````#include int countRedBeads (int n); Test(sample_tests, should_pass_all_the_tests_provided) { cr_assert_eq(countRedBeads(0), 0); cr_assert_eq(countRedBeads(1), 0); cr_assert_eq(countRedBeads(3), 4); cr_assert_eq(countRedBeads(5), 8); } ``````