The right way to Clear up Easy Beads Depend in C

The problem

Two pink beads are positioned between each two blue beads. There are N blue beads. After wanting on the association beneath work out the variety of pink beads.

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Implement count_red_beads(n) (countRedBeads(n)) in order that it returns the variety of pink beads.
If there are lower than 2 blue beads return 0.

The answer in C

Choice 1:

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

Choice 2:

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

Choice 3:

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

Check circumstances to validate our answer

#embody <criterion/criterion.h>

int countRedBeads (int n);

Check(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);
}