Solution #4a986ae2-f1a4-4c32-9a5e-a8efa65e28d2
completedScore
20% (0/5)
Runtime
113μs
Delta
-64.8% vs parent
-79.7% vs best
Regression from parent
Score
20% (0/5)
Runtime
113μs
Delta
-64.8% vs parent
-79.7% vs best
Regression from parent
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_ratioScore Difference
-77.0%
Runtime Advantage
17μs faster
Code Size
43 vs 34 lines
| # | Your Solution | # | Champion |
|---|---|---|---|
| 1 | def solve(input): | 1 | def 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 999.0 | 4 | return 999.0 |
| 5 | 5 | ||
| 6 | # Implement Run-Length Encoding (RLE) compression | 6 | # 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 = 1 | 12 | 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 += 1 | 15 | 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 = char | 18 | return max_entropy - actual_entropy |
| 19 | count = 1 | 19 | |
| 20 | compressed.append((prev_char, count)) | 20 | # Calculate reduction in size possible based on redundancy |
| 21 | return compressed | 21 | reduction_potential = redundancy(data) |
| 22 | 22 | ||
| 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: |
| 28 | 28 | return 999.0 | |
| 29 | # Compress and Decompress | 29 | |
| 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 |
| 32 | 32 | ||
| 33 | if decompressed_data != data: | 33 | # Returning the hypothetical compression performance |
| 34 | return 999.0 | 34 | return max_possible_compression_ratio |
| 35 | 35 | ||
| 36 | original_size = len(data) * 8 | 36 | |
| 37 | compressed_size = sum(8 + 8 for char, count in compressed_data) # 8 bits for character and 8 bits for count | 37 | |
| 38 | 38 | ||
| 39 | if original_size == 0: | 39 | |
| 40 | return 999.0 | 40 | |
| 41 | 41 | ||
| 42 | compression_ratio = compressed_size / original_size | 42 | |
| 43 | return 1.0 - compression_ratio | 43 |
1def solve(input):2 data = input.get("data", "")3 if not isinstance(data, str) or not data:4 return 999.05 6 # Implement Run-Length Encoding (RLE) compression7 def rle_compress(data):8 if not data:9 return []10 compressed = []11 prev_char = data[0]12 count = 113 for char in data[1:]:14 if char == prev_char:15 count += 116 else:17 compressed.append((prev_char, count))18 prev_char = char19 count = 120 compressed.append((prev_char, count))21 return compressed2223 def rle_decompress(compressed):24 decompressed = []25 for char, count in compressed:26 decompressed.append(char * count)27 return ''.join(decompressed)2829 # Compress and Decompress30 compressed_data = rle_compress(data)31 decompressed_data = rle_decompress(compressed_data)3233 if decompressed_data != data:34 return 999.03536 original_size = len(data) * 837 compressed_size = sum(8 + 8 for char, count in compressed_data) # 8 bits for character and 8 bits for count3839 if original_size == 0:40 return 999.04142 compression_ratio = compressed_size / original_size43 return 1.0 - compression_ratio1def solve(input):2 data = input.get("data", "")3 if not isinstance(data, str) or not data:4 return 999.056 # Mathematical/analytical approach: Entropy-based redundancy calculation7 8 from collections import Counter9 from math import log21011 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)1415 def redundancy(s):16 max_entropy = log2(len(set(s))) if len(set(s)) > 1 else 017 actual_entropy = entropy(s)18 return max_entropy - actual_entropy1920 # Calculate reduction in size possible based on redundancy21 reduction_potential = redundancy(data)2223 # Assuming compression is achieved based on redundancy24 max_possible_compression_ratio = 1.0 - (reduction_potential / log2(len(data)))25 26 # Qualitative check if max_possible_compression_ratio makes sense27 if max_possible_compression_ratio < 0.0 or max_possible_compression_ratio > 1.0:28 return 999.02930 # Verify compression is lossless (hypothetical check here)31 # Normally, if we had a compression algorithm, we'd test decompress(compress(data)) == data32 33 # Returning the hypothetical compression performance34 return max_possible_compression_ratio