Solution #0d6c3416-be65-451b-b1fd-d46853707c7c

completed

Score

19% (0/5)

Runtime

189μs

Delta

-50.5% vs parent

-80.1% vs best

Regression from parent

Solution Lineage

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f22b171153%Same as parent
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b96fbcb340%Improved from parent
84cc9d0420%First in chain

Code

def solve(input):
    data = input.get("data", "")
    if not isinstance(data, str) or len(data) == 0:
        return 999.0
    
    # Use Run-Length Encoding (RLE) for compression
    def rle_compress(data):
        if not data:
            return ""
        compressed = []
        prev_char = data[0]
        count = 1
        for char in data[1:]:
            if char == prev_char:
                count += 1
            else:
                compressed.append((prev_char, count))
                prev_char = char
                count = 1
        compressed.append((prev_char, count))
        return compressed

    def rle_decompress(compressed_data):
        decompressed = ''.join(char * count for char, count in compressed_data)
        return decompressed

    compressed_data = rle_compress(data)
    decompressed_data = rle_decompress(compressed_data)

    if decompressed_data != data:
        return 999.0

    original_size = len(data)
    compressed_size = sum(len(char) + len(str(count)) for char, count in compressed_data)

    return 1.0 - (compressed_size / float(original_size))

Compare with Champion

Score Difference

-77.4%

Runtime Advantage

59μs slower

Code Size

36 vs 34 lines

#Your Solution#Champion
1def solve(input):1def solve(input):
2 data = input.get("data", "")2 data = input.get("data", "")
3 if not isinstance(data, str) or len(data) == 0:3 if not isinstance(data, str) or not data:
4 return 999.04 return 999.0
5 5
6 # Use Run-Length Encoding (RLE) for compression6 # Mathematical/analytical approach: Entropy-based redundancy calculation
7 def rle_compress(data):7
8 if not data:8 from collections import Counter
9 return ""9 from math import log2
10 compressed = []10
11 prev_char = data[0]11 def entropy(s):
12 count = 112 probabilities = [freq / len(s) for freq in Counter(s).values()]
13 for char in data[1:]:13 return -sum(p * log2(p) if p > 0 else 0 for p in probabilities)
14 if char == prev_char:14
15 count += 115 def redundancy(s):
16 else:16 max_entropy = log2(len(set(s))) if len(set(s)) > 1 else 0
17 compressed.append((prev_char, count))17 actual_entropy = entropy(s)
18 prev_char = char18 return max_entropy - actual_entropy
19 count = 119
20 compressed.append((prev_char, count))20 # Calculate reduction in size possible based on redundancy
21 return compressed21 reduction_potential = redundancy(data)
2222
23 def rle_decompress(compressed_data):23 # Assuming compression is achieved based on redundancy
24 decompressed = ''.join(char * count for char, count in compressed_data)24 max_possible_compression_ratio = 1.0 - (reduction_potential / log2(len(data)))
25 return decompressed25
2626 # Qualitative check if max_possible_compression_ratio makes sense
27 compressed_data = rle_compress(data)27 if max_possible_compression_ratio < 0.0 or max_possible_compression_ratio > 1.0:
28 decompressed_data = rle_decompress(compressed_data)28 return 999.0
2929
30 if decompressed_data != data:30 # Verify compression is lossless (hypothetical check here)
31 return 999.031 # Normally, if we had a compression algorithm, we'd test decompress(compress(data)) == data
3232
33 original_size = len(data)33 # Returning the hypothetical compression performance
34 compressed_size = sum(len(char) + len(str(count)) for char, count in compressed_data)34 return max_possible_compression_ratio
3535
36 return 1.0 - (compressed_size / float(original_size))36
Your Solution
19% (0/5)189μs
1def solve(input):
2 data = input.get("data", "")
3 if not isinstance(data, str) or len(data) == 0:
4 return 999.0
5
6 # Use Run-Length Encoding (RLE) for compression
7 def rle_compress(data):
8 if not data:
9 return ""
10 compressed = []
11 prev_char = data[0]
12 count = 1
13 for char in data[1:]:
14 if char == prev_char:
15 count += 1
16 else:
17 compressed.append((prev_char, count))
18 prev_char = char
19 count = 1
20 compressed.append((prev_char, count))
21 return compressed
22
23 def rle_decompress(compressed_data):
24 decompressed = ''.join(char * count for char, count in compressed_data)
25 return decompressed
26
27 compressed_data = rle_compress(data)
28 decompressed_data = rle_decompress(compressed_data)
29
30 if decompressed_data != data:
31 return 999.0
32
33 original_size = len(data)
34 compressed_size = sum(len(char) + len(str(count)) for char, count in compressed_data)
35
36 return 1.0 - (compressed_size / float(original_size))
Champion
97% (3/5)130μs
1def solve(input):
2 data = input.get("data", "")
3 if not isinstance(data, str) or not data:
4 return 999.0
5
6 # Mathematical/analytical approach: Entropy-based redundancy calculation
7
8 from collections import Counter
9 from math import log2
10
11 def entropy(s):
12 probabilities = [freq / len(s) for freq in Counter(s).values()]
13 return -sum(p * log2(p) if p > 0 else 0 for p in probabilities)
14
15 def redundancy(s):
16 max_entropy = log2(len(set(s))) if len(set(s)) > 1 else 0
17 actual_entropy = entropy(s)
18 return max_entropy - actual_entropy
19
20 # Calculate reduction in size possible based on redundancy
21 reduction_potential = redundancy(data)
22
23 # Assuming compression is achieved based on redundancy
24 max_possible_compression_ratio = 1.0 - (reduction_potential / log2(len(data)))
25
26 # Qualitative check if max_possible_compression_ratio makes sense
27 if max_possible_compression_ratio < 0.0 or max_possible_compression_ratio > 1.0:
28 return 999.0
29
30 # Verify compression is lossless (hypothetical check here)
31 # Normally, if we had a compression algorithm, we'd test decompress(compress(data)) == data
32
33 # Returning the hypothetical compression performance
34 return max_possible_compression_ratio