Cs50 Tideman Solution Jun 2026

for (int j = i + 1; j < candidate_count; j++)

Once the arrows were locked, the answer was revealed. The winner was the —the one candidate who had arrows pointing at others but no arrows pointing at themselves. Cs50 Tideman Solution

for (int i = 0; i < candidate_count; i++) for (int j = i + 1; j

// Global arrays int preferences[MAX][MAX]; bool locked[MAX][MAX]; j++) Once the arrows were locked

Happy coding, and may your locks always be cycle-free.

Its goal is to determine a winner in an election using a ranked-choice system that satisfies the Condorcet criterion

Finding the is often considered the "rite of passage" for students in Harvard's Intro to Computer Science course. This problem set is infamously difficult—so much so that Harvard even sells "I completed Tideman" t-shirts.