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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.
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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.
1from collections import Counter
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3def rank_scores(scores, bucket_size):
4 count = Counter(int(score / bucket_size) for score in scores)
5 rank = 1
6 for score_index in sorted(count.keys(), reverse=True):
7 score = score_index * bucket_size
8 for _ in range(count[score_index]):
9 print(f"Score: {score:.2f}, Rank: {rank}")
10 rank += count[score_index]
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12scores = [3.50, 3.65, 4.00, 3.85, 4.00, 3.65]
13bucket_size = 0.01
14rank_scores(scores, bucket_size)Python uses a Counter from the collections module to tally the scores into bucketed indices. It then processes these counts in descending order of buckets to output ranks efficiently. This bucket implementation relies on Python's ability to handle large integers and dynamic indexing smartly.