Solution #e2aa8772-e4a1-49e7-940f-203fb1120339

completed

Score

47% (0/5)

Runtime

858μs

Delta

+13.3% vs parent

-51.7% vs best

Improved from parent

Solution Lineage

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84cc9d0420%First in chain

Code

def solve(input):
    import zlib

    data = input.get("data", "")
    if not isinstance(data, str) or not data:
        return 999.0

    # Compress using zlib (based on DEFLATE)
    compressed_data = zlib.compress(data.encode())

    # Decompress to verify
    try:
        decompressed_data = zlib.decompress(compressed_data).decode()
    except:
        return 999.0

    if decompressed_data != data:
        return 999.0

    original_size = len(data)
    compressed_size = len(compressed_data)

    if original_size == 0:
        return 999.0

    compression_ratio = compressed_size / original_size
    return 1.0 - compression_ratio

Compare with Champion

Score Difference

-49.9%

Runtime Advantage

728μs slower

Code Size

27 vs 34 lines

#Your Solution#Champion
1def solve(input):1def solve(input):
2 import zlib2 data = input.get("data", "")
33 if not isinstance(data, str) or not data:
4 data = input.get("data", "")4 return 999.0
5 if not isinstance(data, str) or not data:5
6 return 999.06 # Mathematical/analytical approach: Entropy-based redundancy calculation
77
8 # Compress using zlib (based on DEFLATE)8 from collections import Counter
9 compressed_data = zlib.compress(data.encode())9 from math import log2
1010
11 # Decompress to verify11 def entropy(s):
12 try:12 probabilities = [freq / len(s) for freq in Counter(s).values()]
13 decompressed_data = zlib.decompress(compressed_data).decode()13 return -sum(p * log2(p) if p > 0 else 0 for p in probabilities)
14 except:14
15 return 999.015 def redundancy(s):
1616 max_entropy = log2(len(set(s))) if len(set(s)) > 1 else 0
17 if decompressed_data != data:17 actual_entropy = entropy(s)
18 return 999.018 return max_entropy - actual_entropy
1919
20 original_size = len(data)20 # Calculate reduction in size possible based on redundancy
21 compressed_size = len(compressed_data)21 reduction_potential = redundancy(data)
2222
23 if original_size == 0:23 # Assuming compression is achieved based on redundancy
24 return 999.024 max_possible_compression_ratio = 1.0 - (reduction_potential / log2(len(data)))
2525
26 compression_ratio = compressed_size / original_size26 # Qualitative check if max_possible_compression_ratio makes sense
27 return 1.0 - compression_ratio27 if max_possible_compression_ratio < 0.0 or max_possible_compression_ratio > 1.0:
2828 return 999.0
2929
3030 # Verify compression is lossless (hypothetical check here)
3131 # Normally, if we had a compression algorithm, we'd test decompress(compress(data)) == data
3232
3333 # Returning the hypothetical compression performance
3434 return max_possible_compression_ratio
Your Solution
47% (0/5)858μs
1def solve(input):
2 import zlib
3
4 data = input.get("data", "")
5 if not isinstance(data, str) or not data:
6 return 999.0
7
8 # Compress using zlib (based on DEFLATE)
9 compressed_data = zlib.compress(data.encode())
10
11 # Decompress to verify
12 try:
13 decompressed_data = zlib.decompress(compressed_data).decode()
14 except:
15 return 999.0
16
17 if decompressed_data != data:
18 return 999.0
19
20 original_size = len(data)
21 compressed_size = len(compressed_data)
22
23 if original_size == 0:
24 return 999.0
25
26 compression_ratio = compressed_size / original_size
27 return 1.0 - compression_ratio
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