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:

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

Option 2:

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int countRedBeads(n) {
  if (n < 2) {
    return 0;
  } else {
    return n + (n-2);
  }
}

Option 3:

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

Test cases to validate our solution

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#include <criterion/criterion.h>

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);
}