Introduction

In this blog post, we will explore how to solve the LeetCode problem "1366" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

In a special ranking system, each voter gives a rank from highest to lowest to all teams participating in the competition. The ordering of teams is decided by who received the most position-one votes. If two or more teams tie in the first position, we consider the second position to resolve the conflict, if they tie again, we continue this process until the ties are resolved. If two or more teams are still tied after considering all positions, we rank them alphabetically based on their team letter. You are given an array of strings votes which is the votes of all voters in the ranking systems. Sort all teams according to the ranking system described above. Return a string of all teams sorted by the ranking system. Example 1: Input: votes = ["ABC","ACB","ABC","ACB","ACB"] Output: "ACB" Explanation: Team A was ranked first place by 5 voters. No other team was voted as first place, so team A is the first team. Team B was ranked second by 2 voters and ranked third by 3 voters. Team C was ranked second by 3 voters and ranked third by 2 voters. As most of the voters ranked C second, team C is the second team, and team B is the third. Example 2: Input: votes = ["WXYZ","XYZW"] Output: "XWYZ" Explanation: X is the winner due to the tie-breaking rule. X has the same votes as W for the first position, but X has one vote in the second position, while W does not have any votes in the second position. Example 3: Input: votes = ["ZMNAGUEDSJYLBOPHRQICWFXTVK"] Output: "ZMNAGUEDSJYLBOPHRQICWFXTVK" Explanation: Only one voter, so their votes are used for the ranking. Constraints: 1 <= votes.length <= 1000 1 <= votes[i].length <= 26 votes[i].length == votes[j].length for 0 <= i, j < votes.length. votes[i][j] is an English uppercase letter. All characters of votes[i] are unique. All the characters that occur in votes[0] also occur in votes[j] where 1 <= j < votes.length.

Explanation

Here's a breakdown of the solution:

• Analyze Votes: Create a data structure (dictionary/hash map) to store the count of each team at each position.
• Custom Sort: Sort the teams based on their counts at each position. Tie-breaking is handled sequentially by considering subsequent positions and, finally, alphabetically.

• Return Result: Concatenate the sorted teams into a string.

• Runtime Complexity: O(M N log(N)), where M is the number of votes and N is the number of teams.

• Storage Complexity: O(N * N), to store the counts of each team at each position.

Code

    def rankTeams(votes: list[str]) -> str:
    """
    Ranks teams based on the given voting system.

    Args:
        votes: A list of strings representing the votes.

    Returns:
        A string representing the teams sorted according to the ranking system.
    """

    if not votes:
        return ""

    num_teams = len(votes[0])
    team_rankings = {}

    for team in votes[0]:
        team_rankings[team] = [0] * num_teams

    for vote in votes:
        for i, team in enumerate(vote):
            team_rankings[team][i] += 1

    teams = list(team_rankings.keys())

    teams.sort(key=lambda team: team_rankings[team], reverse=True)
    teams.sort(key=lambda team: team, reverse=False)

    teams.sort(key=lambda team: team_rankings[team], reverse=True)


    return "".join(teams)