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:
int countRedBeads(n) {
if(n<=1) return 0;
return 2*(n-1);
}
Option 2:
int countRedBeads(n) {
if (n < 2) {
return 0;
} else {
return n + (n-2);
}
}
Option 3:
int countRedBeads(n)
{
return n<2 ? 0 : 2*n-2;
}
Test cases to validate our solution
#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);
}
