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CS50 Problem Set 3
2026-03-18

CS50 Problem Set 3#

Here’s my answer for the CS50 Problem Set 3. Hope that will help you a bit.

Problem 1: Sort#

answers.txt
sort1 uses: bubble sort
How do you know?: When sorting the sorted numbers, bubble sort's time complexity is omega(n); while sorting the random numbers, bubble sort's time complexity is O(n^2).
sort2 uses: merge sort
How do you know?: the time sort2 uses to sort every kind of txt is the least between the three sorts.
sort3 uses: selection sort
How do you know?: the time it uses to sort three kinds of txt is almost the same, but it cost more time than sort2.

Problem 2: Plurality#

plurality.c
#include <cs50.h>
#include <stdio.h>
#include <string.h>
// Max number of candidates
#define MAX 9
// Candidates have name and vote count
typedef struct
{
string name;
int votes;
} candidate;
// Array of candidates
candidate candidates[MAX];
// Number of candidates
int candidate_count;
// Function prototypes
bool vote(string name);
void print_winner(void);
int main(int argc, string argv[])
{
// Check for invalid usage
if (argc < 2)
{
printf("Usage: plurality [candidate ...]\n");
return 1;
}
// Populate array of candidates
candidate_count = argc - 1;
if (candidate_count > MAX)
{
printf("Maximum number of candidates is %i\n", MAX);
return 2;
}
for (int i = 0; i < candidate_count; i++)
{
candidates[i].name = argv[i + 1];
candidates[i].votes = 0;
}
int voter_count = get_int("Number of voters: ");
// Loop over all voters
for (int i = 0; i < voter_count; i++)
{
string name = get_string("Vote: ");
// Check for invalid vote
if (!vote(name))
{
printf("Invalid vote.\n");
}
}
// Display winner of election
print_winner();
}
// Update vote totals given a new vote
bool vote(string name)
{
for (int i = 0; i < candidate_count; i++)
{
if (strcmp(candidates[i].name, name) == 0)
{
candidates[i].votes += 1;
return true;
}
}
return false;
}
// Print the winner (or winners) of the election
void print_winner(void)
{
int highest = 0;
string name;
for (int i = 0; i < candidate_count; i++)
{
if (candidates[i].votes > highest)
{
highest = candidates[i].votes;
}
}
for (int i = 0; i < candidate_count; i++)
{
if (candidates[i].votes == highest)
{
printf("%s\n", candidates[i].name);
}
}
return;
}

Problem 3: Runoff#

runoff.c
#include <cs50.h>
#include <stdio.h>
#include <string.h>
// Max voters and candidates
#define MAX_VOTERS 100
#define MAX_CANDIDATES 9
// preferences[i][j] is jth preference for voter i
int preferences[MAX_VOTERS][MAX_CANDIDATES];
// Candidates have name, vote count, eliminated status
typedef struct
{
string name;
int votes;
bool eliminated;
} candidate;
// Array of candidates
candidate candidates[MAX_CANDIDATES];
// Numbers of voters and candidates
int voter_count;
int candidate_count;
// Function prototypes
bool vote(int voter, int rank, string name);
void tabulate(void);
bool print_winner(void);
int find_min(void);
bool is_tie(int min);
void eliminate(int min);
int main(int argc, string argv[])
{
// Check for invalid usage
if (argc < 2)
{
printf("Usage: runoff [candidate ...]\n");
return 1;
}
// Populate array of candidates
candidate_count = argc - 1;
if (candidate_count > MAX_CANDIDATES)
{
printf("Maximum number of candidates is %i\n", MAX_CANDIDATES);
return 2;
}
for (int i = 0; i < candidate_count; i++)
{
candidates[i].name = argv[i + 1];
candidates[i].votes = 0;
candidates[i].eliminated = false;
}
voter_count = get_int("Number of voters: ");
if (voter_count > MAX_VOTERS)
{
printf("Maximum number of voters is %i\n", MAX_VOTERS);
return 3;
}
// Keep querying for votes
for (int i = 0; i < voter_count; i++)
{
// Query for each rank
for (int j = 0; j < candidate_count; j++)
{
string name = get_string("Rank %i: ", j + 1);
// Record vote, unless it's invalid
if (!vote(i, j, name))
{
printf("Invalid vote.\n");
return 4;
}
}
printf("\n");
}
// Keep holding runoffs until winner exists
while (true)
{
// Calculate votes given remaining candidates
tabulate();
// Check if election has been won
bool won = print_winner();
if (won)
{
break;
}
// Eliminate last-place candidates
int min = find_min();
bool tie = is_tie(min);
// If tie, everyone wins
if (tie)
{
for (int i = 0; i < candidate_count; i++)
{
if (!candidates[i].eliminated)
{
printf("%s\n", candidates[i].name);
}
}
break;
}
// Eliminate anyone with minimum number of votes
eliminate(min);
// Reset vote counts back to zero
for (int i = 0; i < candidate_count; i++)
{
candidates[i].votes = 0;
}
}
return 0;
}
// Record preference if vote is valid
bool vote(int voter, int rank, string name)
{
for (int i = 0; i < candidate_count; i++)
{
if (strcmp(candidates[i].name, name) == 0)
{
preferences[voter][rank] = i;
return true;
}
}
return false;
}
// Tabulate votes for non-eliminated candidates
void tabulate(void)
{
for (int i = 0; i < voter_count; i++)
{
for (int j = 0; j < candidate_count; j++)
{
if (!candidates[preferences[i][j]].eliminated == true)
{
candidates[preferences[i][j]].votes += 1;
break;
}
}
}
return;
}
// Print the winner of the election, if there is one
bool print_winner(void)
{
int limit = voter_count / 2;
for (int i = 0; i < candidate_count; i++)
{
if (!candidates[i].eliminated && (candidates[i].votes > limit))
{
printf("%s\n", candidates[i].name);
return true;
}
}
return false;
}
// Return the minimum number of votes any remaining candidate has
int find_min(void)
{
int min = 0;
for (int i = 0; i < candidate_count; i++)
{
if (!candidates[i].eliminated)
{
if (i == 0)
{
min = candidates[i].votes;
}
if (candidates[i].votes < min)
{
min = candidates[i].votes;
}
}
}
return min;
}
// Return true if the election is tied between all candidates, false otherwise
bool is_tie(int min)
{
for (int i = 0; i < candidate_count; i++)
{
if (!candidates[i].eliminated && candidates[i].votes != min)
{
return false;
}
}
return true;
}
// Eliminate the candidate (or candidates) in last place
void eliminate(int min)
{
for (int i = 0; i < candidate_count; i++)
{
if (!candidates[i].eliminated && candidates[i].votes == min)
{
candidates[i].eliminated = true;
}
}
return;
}
CS50 Problem Set 3
https://tech.kinghua0629.com/posts/cs50-problem-set-3/
Author
Ziwen Hua
Published at
2026-03-18
License
CC BY-NC-SA 4.0