Problem
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.
Discussion
This problem can be solved using 2 data structures- a hashmap and a heap. We use the map to store the votes by each judge and the heap to sort the teams. We create a comparator to sort the teams and initialize the heap using it.
We then traverse the votes array to calculate the votes and create the map.Once this is done we add the keys of the map which are the teams to the heap and pull the elements off one by one to get the final string.
Solution
Java
class Solution {
public String rankTeams(String[] votes) {
HashMap<Character,int[]>map=new HashMap<>();
String answer="";
for(int i=0;i<votes.length;i++){
String vote=votes[i];
for(int j=0;j<vote.length();j++){
char team=vote.charAt(j);
if(!map.containsKey(team)){
int [] counts= new int[vote.length()];
counts[j]++;
map.put(team,counts);
}else{
int [] counts =map.get(team);
counts[j]++;
map.replace(team,counts);
}
}
}
PriorityQueue<Character> pq = new PriorityQueue<>((a, b) -> {
int[] va = map.get(a);
int[] vb = map.get(b);
for (int i = 0; i < va.length; i++) {
if (va[i] != vb[i]) {
return Integer.compare(vb[i], va[i]);
}
}
return Character.compare(a, b);
});
for (char team : map.keySet()) pq.offer(team);
while(!pq.isEmpty()){
char ch=pq.poll();
answer+=ch;
}
return answer;
}
}
