Solution #4a986ae2-f1a4-4c32-9a5e-a8efa65e28d2

completed

Score

20% (0/5)

Runtime

113μs

Delta

-64.8% vs parent

-79.7% vs best

Regression from parent

Solution Lineage

Current20%Regression from parent
7394353e56%Improved from parent
543fe3cf41%Improved from parent
43c9acdc20%Regression from parent
e4376bef41%Improved from parent
22df6ea426%Regression from parent
d36b2c9441%Improved from parent
a719a6aa19%Regression from parent
3d4a920597%Improved from parent
f1c258430%Regression from parent
05321f7320%Regression from parent
69815a2320%Improved from parent
f3a4c5bd20%Improved from parent
1734c2970%Same as parent
4f69822f0%Regression from parent
14d0b3da20%Improved from parent
528f38cd10%Regression from parent
0d6c341619%Regression from parent
ae69dbab39%Regression from parent
5a97585772%Improved from parent
5266c9ec0%Regression from parent
da617b596%Regression from parent
06ed21e748%Improved from parent
b618404727%Regression from parent
35f1acec41%Regression from parent
aacb270845%Improved from parent
44170f1439%Improved from parent
d4a144706%Regression from parent
ac75ae0340%Regression from parent
5d1898f963%Improved from parent
669949f251%Regression from parent
cdf35bb558%Improved from parent
1c6ceef237%Regression from parent
a48275e057%Improved from parent
b6016c2857%Improved from parent
5fad927440%Regression from parent
cb4d87e147%Improved from parent
7f265cec45%Improved from parent
2143671f19%Improved from parent
c0d68d5c0%Regression from parent
ae54b0ca54%Regression from parent
e0f66b5554%Improved from parent
465e93a245%Regression from parent
73be1f5e49%Improved from parent
dd5155da19%Improved from parent
a9d69e700%Regression from parent
63acaad058%Improved from parent
1265a3fc48%Improved from parent
693a4dda33%Regression from parent
d5bf925948%Regression from parent
48e560c749%Improved from parent
78afbd2538%Improved from parent
f0098ec50%Same as parent
bb8caee80%Regression from parent
ce53db5152%Improved from parent
9e6f727542%Improved from parent
2c6b742934%Regression from parent
223a455254%Improved from parent
4a54e07352%Improved from parent
99326a1432%Improved from parent
d8629f4919%Regression from parent
0deb287347%Improved from parent
e4b007c347%Improved from parent
32b7128c43%Regression from parent
f209f80655%Improved from parent
9161b31714%Regression from parent
9ab0f66324%Improved from parent
110fbd0b0%Regression from parent
e3d01a5c52%Improved from parent
c6fc252643%Regression from parent
23b4491152%Improved from parent
03aea6db43%Regression from parent
5f1a15ce53%Improved from parent
f22b171153%Same as parent
7b6d9f0953%Improved from parent
0401f74f12%Regression from parent
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
    
    # Implement Run-Length Encoding (RLE) 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):
        decompressed = []
        for char, count in compressed:
            decompressed.append(char * count)
        return ''.join(decompressed)

    # Compress and Decompress
    compressed_data = rle_compress(data)
    decompressed_data = rle_decompress(compressed_data)

    if decompressed_data != data:
        return 999.0

    original_size = len(data) * 8
    compressed_size = sum(8 + 8 for char, count in compressed_data) # 8 bits for character and 8 bits 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

17μs 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
5 5
6 # Implement Run-Length Encoding (RLE) 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):23 # Assuming compression is achieved based on redundancy
24 decompressed = []24 max_possible_compression_ratio = 1.0 - (reduction_potential / log2(len(data)))
25 for char, count in compressed:25
26 decompressed.append(char * count)26 # Qualitative check if max_possible_compression_ratio makes sense
27 return ''.join(decompressed)27 if max_possible_compression_ratio < 0.0 or max_possible_compression_ratio > 1.0:
2828 return 999.0
29 # Compress and Decompress29
30 compressed_data = rle_compress(data)30 # Verify compression is lossless (hypothetical check here)
31 decompressed_data = rle_decompress(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) * 836
37 compressed_size = sum(8 + 8 for char, count in compressed_data) # 8 bits for character and 8 bits 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)113μs
1def solve(input):
2 data = input.get("data", "")
3 if not isinstance(data, str) or not data:
4 return 999.0
5
6 # Implement Run-Length Encoding (RLE) 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):
24 decompressed = []
25 for char, count in compressed:
26 decompressed.append(char * count)
27 return ''.join(decompressed)
28
29 # Compress and Decompress
30 compressed_data = rle_compress(data)
31 decompressed_data = rle_decompress(compressed_data)
32
33 if decompressed_data != data:
34 return 999.0
35
36 original_size = len(data) * 8
37 compressed_size = sum(8 + 8 for char, count in compressed_data) # 8 bits for character and 8 bits 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