Cs50 Tideman Solution Now

printf("The winner is: %d\n", winner);

return 0; } The implementation includes test cases to verify its correctness. For example, consider the following input:

// Allocate memory for voters and candidates *voters_prefs = malloc(*voters * sizeof(voter_t)); candidate_t *candidates_list = malloc(*candidates * sizeof(candidate_t)); Cs50 Tideman Solution

int winner = check_for_winner(candidates_list, candidates); while (winner == -1) { // Eliminate candidate with fewest votes int eliminated = -1; int min_votes = voters + 1; for (int i = 0; i < candidates; i++) { if (candidates_list[i].votes < min_votes) { min_votes = candidates_list[i].votes; eliminated = candidates_list[i].id; } }

// Structure to represent a voter typedef struct voter { int *preferences; } voter_t; printf("The winner is: %d\n", winner); return 0; }

The winner is: 1 This indicates that candidate 1 wins the election.

// Count first-place votes for (int i = 0; i < voters; i++) { for (int j = 0; j < candidates; j++) { if (j == 0) { candidates_list[voters_prefs[i].preferences[j] - 1].votes++; } } } } printf("The winner is: %d\n"

int main() { int voters, candidates; voter_t *voters_prefs; read_input(&voters, &candidates, &voters_prefs);

count_first_place_votes(voters_prefs, voters, candidates_list, candidates);

count_first_place_votes(voters_prefs, voters, candidates_list, candidates);