How to Invite More Women in C

The challenge

Task

King Arthur and his knights are having a New Years party. Last year Lancelot was jealous of Arthur, because Arthur had a date and Lancelot did not, and they started a duel.

To prevent this from happening again, Arthur wants to make sure that there are at least as many women as men at this year’s party. He gave you a list of integers of all the party goers.

Arthur needs you to return true if he needs to invite more women or false if he is all set.

Input/Output

• [input] integer array L ($a in PHP)

An array (guaranteed non-associative in PHP) representing the genders of the attendees, where -1 represents women and 1 represents men.

2 <= L.length <= 50

• [output] a boolean valuetrue if Arthur need to invite more women, false otherwise.

The solution in C

Option 1:

#include <stddef.h>

int invite_more_women(int *arr, size_t count) {
    int sum = 0;
    
    for (int i = 0; i < count; ++i) sum += arr[i];
    
    return (sum>0)? 1 : 0;
}

Option 2:

#include <stddef.h>

int invite_more_women(int *arr, size_t count) {
    int balance = 0;
    while (count--)
        balance += *arr++;
    return balance > 0;
}

Option 3:

#include <stdbool.h>
#include <stddef.h>

bool invite_more_women(const int attendee_genders[], size_t count) {
  ptrdiff_t sum = 0;
  for (size_t i = 0; i < count; i++)
    sum += attendee_genders[i];
  return sum > 0;
}

Test cases to validate our solution

#include <stdbool.h>
#include <stddef.h>
#include <criterion/criterion.h>

extern void do_test (size_t count, const int array[count], bool expected);

#define ARR_LEN(array) (sizeof(array) / sizeof *(array))

#define sample_test(array, expected) do_test(ARR_LEN(array), array, expected)

Test(tests_suite, sample_tests)
{
	sample_test(((int[]){1, -1, 1}), true);
	sample_test(((int[]){-1, -1, -1}), false);
	sample_test(((int[]){1, -1}), false);
	sample_test(((int[]){1, 1, 1}), true);
	do_test(0, NULL, false);
}