Problem solution · Python

Rank Teams by Votes

Rank Teams by Votes: a Python solution using sorting and greedy selection. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sorting and greedy selection
Source
walkccc LeetCode Solutions
Length
25 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Sorting and greedy selection

For Rank Teams by Votes, the implementation first exposes a useful order, then scans that order while making locally justified choices.

  1. Choose the key that reveals the greedy or grouping structure.
  2. Sort the relevant records by that key.
  3. Scan in order, maintaining the invariant that makes each local choice safe.

Code notes

  • 25 lines of Python from the credited upstream file 1366.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Sorting is typically the dominant term unless the subsequent scan uses a more expensive nested operation.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Full codeRank Teams by Votes · PythonPython
Use this to learn the idea, then write your own version.
from dataclasses import dataclass  @dataclassclass Team:  name: str  rank: list[int]   def __init__(self, name: str, teamSize: int):    self.name = name    self.rank = [0] * teamSize  class Solution:  def rankTeams(self, votes: list[str]) -> str:    teamSize = len(votes[0])    teams = [Team(chr(ord('A') + i), teamSize) for i in range(26)]     for vote in votes:      for i in range(teamSize):        teams[ord(vote[i]) - ord('A')].rank[i] += 1     teams.sort(key=lambda x: (x.rank, -ord(x.name)), reverse=True)    return ''.join(team.name for team in teams[:teamSize]) 

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