Leetcode 1366: Rank Teams by Votes
In a special ranking system, each voter assigns a rank to all participating teams in a competition. The rankings are based on the most first-place votes, then second-place votes, and so on. If there is still a tie, teams are ranked alphabetically.
Problem
Approach
Steps
Complexity
Input: The input consists of an array of strings where each string represents the rankings given by a voter.
Example: votes = ["XYZ", "YZX", "ZXY", "XYZ", "YZX"]
Constraints:
• 1 <= votes.length <= 1000
• 1 <= votes[i].length <= 26
• votes[i].length == votes[j].length for all 0 <= i, j < votes.length
• Each character in a vote string is unique.
Output: The output is a string containing all the teams ranked from highest to lowest based on the voting system.
Example: "XYZ"
Constraints:
• The output string will contain all teams in the order of their ranks.
Goal: The goal is to rank teams based on their positions in the votes, resolving ties using subsequent positions, and sorting alphabetically if necessary.
Steps:
• Count the number of first-place, second-place, third-place, etc., votes for each team.
• Sort teams primarily by the number of first-place votes, then second-place votes, and so on.
• If teams are still tied after considering all positions, sort them alphabetically.
Goal: The input list of votes is constrained by the number of voters and the length of the vote strings.
Steps:
• The number of voters is between 1 and 1000.
• Each vote string contains a number of characters between 1 and 26, representing different teams.
Assumptions:
• The input contains valid rankings where each character is a unique team in every vote string.
• Input: votes = ["XYZ", "YZX", "ZXY", "XYZ", "YZX"]
• Explanation: In this case, team X received the most first-place votes, so it ranks first. Team Y and team Z tie for second place but team Y has the most second-place votes, so it ranks second.
• Input: votes = ["ABCDE", "DCBAE", "EABCD"]
• Explanation: Team A received the most first-place votes across all voters, so it ranks first, followed by team B.
Approach: The approach involves counting the number of votes each team receives for each position and sorting them according to the ranking system.
Observations:
• The problem requires ranking teams based on votes for each position in a list of rankings.
• Sorting the teams by first-place votes, then second-place votes, and so on, will allow us to handle ties effectively.
Steps:
• Create an array to count the number of votes each team receives for each position.
• Sort the teams based on the counts, considering ties at each position.
• If a tie still exists after all positions are considered, sort the teams alphabetically.
Empty Inputs:
• If there is only one voter, their ranking is the final result.
Large Inputs:
• Ensure the solution handles large inputs efficiently, up to 1000 votes and 26 teams.
Special Values:
• If all teams are tied at each position, sort them alphabetically.
Constraints:
• Make sure to handle cases where there are multiple ties at various positions.
string rankTeams(vector<string>& votes) {
vector<vector<int>> count(26, vector<int> (27));
for(char &c: votes[0])
count[c - 'A'][26] = c;
for(string& vote: votes)
for(int i =0; i < vote.size(); i++)
--count[vote[i] - 'A'][i];
sort(count.begin(), count.end());
string res;
for(int i = 0; i < votes[0].length(); i++)
res += count[i][26];
return res;
}
1 : Function Declaration
string rankTeams(vector<string>& votes) {
This function takes a vector of strings 'votes', where each string represents a ranking of teams, and returns a string that represents the ranked order of the teams.
2 : Variable Initialization
vector<vector<int>> count(26, vector<int> (27));
A 2D vector 'count' is initialized to store the ranking positions of 26 teams (A-Z), with each row representing a team and each column representing the position in the votes.
3 : First Vote Processing
for(char &c: votes[0])
The first string in the 'votes' vector is iterated, and for each character (team), its corresponding position in the 'count' array is updated.
4 : Store Team Character
count[c - 'A'][26] = c;
Stores the character representing the team in the last column (index 26) of the 'count' array. This keeps track of the team character in the count table.
5 : Iterate Over Votes
for(string& vote: votes)
The function iterates over all the votes to process each vote string.
6 : Iterate Over Each Vote
for(int i = 0; i < vote.size(); i++)
For each vote string, this loop iterates over each position in the vote to update the corresponding team's score.
7 : Update Team Score
--count[vote[i] - 'A'][i];
For each team in the vote string, the corresponding team's score is decremented based on its position in the vote (lower positions have higher priority).
8 : Sort Teams
sort(count.begin(), count.end());
The 'count' array is sorted based on the scores of the teams, placing the teams with the highest score first.
9 : Initialize Result String
string res;
A string 'res' is initialized to store the result of the ranked teams.
10 : Construct Result String
for(int i = 0; i < votes[0].length(); i++)
The function iterates over the length of the vote strings to construct the result string by adding teams in ranked order.
11 : Add Ranked Team to Result
res += count[i][26];
For each ranked team, its character (team) is appended to the result string 'res'.
12 : Return Result
return res;
The final ranked order of teams is returned as a string.
Best Case: O(n log n)
Average Case: O(n log n)
Worst Case: O(n log n)
Description: The time complexity is dominated by the sorting step, where n is the number of teams.
Best Case: O(n)
Worst Case: O(n)
Description: The space complexity is linear in the number of teams, as we need to store counts for each team.
| LeetCode Solutions Library / DSA Sheets / Course Catalog |
|---|
comments powered by Disqus