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Rank Teams by Votes

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.

Solution Explanation: Ranking Teams by Votes

This problem requires sorting teams based on a voting system where each vote ranks teams from highest to lowest. The sorting criteria prioritize teams with the most first-place votes. In case of ties, subsequent positions are considered until the tie is broken. If a tie persists after considering all positions, teams are sorted alphabetically.

The most efficient approach involves using a counting mechanism combined with a custom sorting function. Here's a breakdown:

1. Counting Votes:

We create a data structure to store the vote count for each team at each position. A 2D array or a hashmap is suitable. For each vote in the input votes array, we increment the count for the corresponding team at its respective position. For example, if a vote is "ABC", then:

  • Team 'A' gets 1 vote in position 0.
  • Team 'B' gets 1 vote in position 1.
  • Team 'C' gets 1 vote in position 2.

2. Custom Sorting:

We use a custom comparison function within a sorting algorithm (like sort in Python or Arrays.sort in Java). This function compares two teams based on their vote counts. The comparison follows these steps:

  • Compare Vote Counts: The function iterates through the vote counts for each position (starting from position 0). If the counts for a position differ between the two teams, the team with the higher count is considered "greater."
  • Resolve Ties Alphabetically: If the vote counts are identical for all positions, the comparison falls back to alphabetical order.

Time and Space Complexity:

  • Time Complexity: The dominant operation is the counting of votes and the sorting process. Counting takes O(nm) time, where n is the number of votes and m is the number of teams. The sorting algorithm, depending on the implementation, typically has a time complexity of O(mlog m) or O(m^2) in the worst case (for example, a simple bubble sort). Thus the overall time complexity is O(nm + m^2 * log m) or O(nm + m^2), depending on the sorting method. In practice, the m^2 term is less significant than n*m unless 'm' is exceptionally large.

  • Space Complexity: The space used is primarily for storing the vote counts and potentially an auxiliary array/list for sorting. This leads to a space complexity of O(m^2) (because we store counts for each team at each position).

Code Examples (Python, Java, C++, Go, TypeScript, Rust):

The provided code examples in various languages demonstrate this approach. They all follow the two-step process: counting votes and then sorting using a custom comparator function that handles tie-breaking scenarios efficiently. The choice of data structure (2D array vs hashmap) might slightly affect the constants in the complexity analysis but doesn't change the overall asymptotic behavior.