Solution #2c8087b0-079f-4f10-a068-d4b09f60d2bc

completed

Score

20% (0/5)

Runtime

127μs

Delta

-2.6% vs parent

-79.7% vs best

Regression from parent

Solution Lineage

Current20%Regression from parent
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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 not data:
        return 999.0

    # Approach: Dynamic Programming with Run-length Encoding

    def compress_rle(s):
        if not s:
            return []
        compressed = []
        current_char = s[0]
        current_count = 1
        for char in s[1:]:
            if char == current_char:
                current_count += 1
            else:
                compressed.append((current_char, current_count))
                current_char = char
                current_count = 1
        compressed.append((current_char, current_count))
        return compressed

    def decompress_rle(compressed):
        decompressed = []
        for char, count in compressed:
            decompressed.append(char * count)
        return ''.join(decompressed)

    compressed_data = compress_rle(data)
    decompressed_data = decompress_rle(compressed_data)

    if decompressed_data != data:
        return 999.0

    original_size = len(data)
    compressed_size = sum(len(char) + 1 for char, _ in compressed_data)  # Assuming 1 byte for count

    if original_size == 0:
        return 999.0

    compression_ratio = compressed_size / original_size
    return 1.0 - compression_ratio

Compare with Champion

Score Difference

-77.0%

Runtime Advantage

3μs faster

Code Size

43 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 not data:3 if not isinstance(data, str) or not data:
4 return 999.04 return 999.0
55
6 # Approach: Dynamic Programming with Run-length Encoding6 # Mathematical/analytical approach: Entropy-based redundancy calculation
77
8 def compress_rle(s):8 from collections import Counter
9 if not s:9 from math import log2
10 return []10
11 compressed = []11 def entropy(s):
12 current_char = s[0]12 probabilities = [freq / len(s) for freq in Counter(s).values()]
13 current_count = 113 return -sum(p * log2(p) if p > 0 else 0 for p in probabilities)
14 for char in s[1:]:14
15 if char == current_char:15 def redundancy(s):
16 current_count += 116 max_entropy = log2(len(set(s))) if len(set(s)) > 1 else 0
17 else:17 actual_entropy = entropy(s)
18 compressed.append((current_char, current_count))18 return max_entropy - actual_entropy
19 current_char = char19
20 current_count = 120 # Calculate reduction in size possible based on redundancy
21 compressed.append((current_char, current_count))21 reduction_potential = redundancy(data)
22 return compressed22
2323 # Assuming compression is achieved based on redundancy
24 def decompress_rle(compressed):24 max_possible_compression_ratio = 1.0 - (reduction_potential / log2(len(data)))
25 decompressed = []25
26 for char, count in compressed:26 # Qualitative check if max_possible_compression_ratio makes sense
27 decompressed.append(char * count)27 if max_possible_compression_ratio < 0.0 or max_possible_compression_ratio > 1.0:
28 return ''.join(decompressed)28 return 999.0
2929
30 compressed_data = compress_rle(data)30 # Verify compression is lossless (hypothetical check here)
31 decompressed_data = decompress_rle(compressed_data)31 # Normally, if we had a compression algorithm, we'd test decompress(compress(data)) == data
3232
33 if decompressed_data != data:33 # Returning the hypothetical compression performance
34 return 999.034 return max_possible_compression_ratio
3535
36 original_size = len(data)36
37 compressed_size = sum(len(char) + 1 for char, _ in compressed_data) # Assuming 1 byte for count37
3838
39 if original_size == 0:39
40 return 999.040
4141
42 compression_ratio = compressed_size / original_size42
43 return 1.0 - compression_ratio43
Your Solution
20% (0/5)127μs
1def solve(input):
2 data = input.get("data", "")
3 if not isinstance(data, str) or not data:
4 return 999.0
5
6 # Approach: Dynamic Programming with Run-length Encoding
7
8 def compress_rle(s):
9 if not s:
10 return []
11 compressed = []
12 current_char = s[0]
13 current_count = 1
14 for char in s[1:]:
15 if char == current_char:
16 current_count += 1
17 else:
18 compressed.append((current_char, current_count))
19 current_char = char
20 current_count = 1
21 compressed.append((current_char, current_count))
22 return compressed
23
24 def decompress_rle(compressed):
25 decompressed = []
26 for char, count in compressed:
27 decompressed.append(char * count)
28 return ''.join(decompressed)
29
30 compressed_data = compress_rle(data)
31 decompressed_data = decompress_rle(compressed_data)
32
33 if decompressed_data != data:
34 return 999.0
35
36 original_size = len(data)
37 compressed_size = sum(len(char) + 1 for char, _ in compressed_data) # Assuming 1 byte for count
38
39 if original_size == 0:
40 return 999.0
41
42 compression_ratio = compressed_size / original_size
43 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