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This approach involves sorting the scores in descending order, then iteratively assigning ranks to each score while handling ties appropriately. This can be efficiently achieved using a sorting algorithm followed by a traversal to assign ranks. By maintaining an index while sorting, we can assign ranks directly to the original scores.
Time Complexity: O(n log n) due to the sorting operation.
Space Complexity: O(1), only a fixed amount of additional memory is used for sorting.
1import java.util.ArrayList;
2import java.util.Collections;
3import java.util.Comparator;
4
5class ScoreEntry {
6 int id;
7 double score;
8 ScoreEntry(int id, double score) {
9 this.id = id;
10 this.score = score;
11 }
12}
13
14public class RankScores {
15 public static void assignRanks(ArrayList<ScoreEntry> scores) {
16 Collections.sort(scores, new Comparator<ScoreEntry>() {
17 public int compare(ScoreEntry a, ScoreEntry b) {
18 return Double.compare(b.score, a.score);
19 }
20 });
21 int rank = 1;
22 for (int i = 0; i < scores.size(); i++) {
23 if (i == 0 || scores.get(i).score != scores.get(i - 1).score) {
24 rank = i + 1;
25 }
26 System.out.println("Score: " + scores.get(i).score + ", Rank: " + rank);
27 }
28 }
29 public static void main(String[] args) {
30 ArrayList<ScoreEntry> scores = new ArrayList<>();
31 scores.add(new ScoreEntry(1, 3.50));
32 scores.add(new ScoreEntry(2, 3.65));
33 scores.add(new ScoreEntry(3, 4.00));
34 scores.add(new ScoreEntry(4, 3.85));
35 scores.add(new ScoreEntry(5, 4.00));
36 scores.add(new ScoreEntry(6, 3.65));
37
38 assignRanks(scores);
39 }
40}This Java implementation stores scores in an `ArrayList` and sorts them using a comparator for descending order. The method iterates over the sorted list to print each score along with its corresponding rank, managing ties as described.
This approach uses a bucket sort strategy, particularly efficient when scores have a limited range of decimal places. This reduces the complexity substantially in scenarios where HTTP (High Throughput Processing) is required. Counting occurrences of each score allows direct assignment of ranks in descending order of scores efficiently.
Time Complexity: O(n + k), where n is the number of scores, and k is the number of buckets (constant in this case).
Space Complexity: O(k), which is fixed and depends on BUCKET_COUNT.
1
In JavaScript, a Map structure is used to count instances of scores in buckets calculated by integer conversion. Sorted in descending order, each bucket's contents are examined, and ranks are adapted accordingly. This allows fast iteration and logging for real-time HTTP applications.