Cs50 Tideman Solution Apr 2026

3 3 1 2 3 1 3 2 2 1 3 This input represents an election with 3 voters and 3 candidates. The output of the program should be:

// Read in voter preferences for (int i = 0; i < *voters; i++) { (*voters_prefs)[i].preferences = malloc(*candidates * sizeof(int)); for (int j = 0; j < *candidates; j++) { scanf("%d", &(*voters_prefs)[i].preferences[j]); } } } Cs50 Tideman Solution

return 0; } The implementation includes test cases to verify its correctness. For example, consider the following input: 3 3 1 2 3 1 3 2

The winner is: 1 This indicates that candidate 1 wins the election. for (int j = 0

// Function to eliminate candidate void eliminate_candidate(candidate_t *candidates_list, int candidates, int eliminated) { // Decrement vote counts for eliminated candidate for (int i = 0; i < candidates; i++) { if (candidates_list[i].id == eliminated) { candidates_list[i].votes = 0; } } }

// 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++; } } } }

// Structure to represent a voter typedef struct voter { int *preferences; } voter_t;