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How to Solve Simple Beads Count in C

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 ; return 2*(n-1); }
Code language: C++ (cpp)

Option 2:

int countRedBeads(n) { if (n < 2) { return ; } else { return n + (n-2); } }
Code language: C++ (cpp)

Option 3:

int countRedBeads(n) { return n<2 ? : 2*n-2; }
Code language: C++ (cpp)

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(), ); cr_assert_eq(countRedBeads(1), ); cr_assert_eq(countRedBeads(3), 4); cr_assert_eq(countRedBeads(5), 8); }
Code language: C++ (cpp)

See also  How to Check for All Inclusive in C
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