Solution #2143671f-2d07-4fe6-add9-613d671f0cb9

completed

Score

19% (0/5)

Runtime

201μs

Delta

New score

-80.1% vs best

Improved from parent

Solution Lineage

Current19%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):
    try:
        data = input["data"] if isinstance(input, dict) and "data" in input else ""
        if not isinstance(data, str):
            data = str(data)
        n = len(data)
        if n == 0:
            return 0.0

        def encode_runs(s):
            chars = [s[i] for i in range(len(s))]
            idx = list(range(1, len(chars)))
            cuts = [0] + [i for i in idx if chars[i] != chars[i - 1]] + [len(chars)]
            runs = [(chars[cuts[i]], cuts[i + 1] - cuts[i]) for i in range(len(cuts) - 1)]
            return runs

        def decode_runs(runs):
            return "".join([ch * cnt for ch, cnt in runs])

        runs = encode_runs(data)
        if decode_runs(runs) != data:
            return 999.0

        compressed_size = sum([1 + len(str(cnt)) for ch, cnt in runs])
        ratio = compressed_size / float(n)
        return float(ratio)
    except:
        return 999.0

Compare with Champion

Score Difference

-77.4%

Runtime Advantage

71μs slower

Code Size

28 vs 34 lines

#Your Solution#Champion
1def solve(input):1def solve(input):
2 try:2 data = input.get("data", "")
3 data = input["data"] if isinstance(input, dict) and "data" in input else ""3 if not isinstance(data, str) or not data:
4 if not isinstance(data, str):4 return 999.0
5 data = str(data)5
6 n = len(data)6 # Mathematical/analytical approach: Entropy-based redundancy calculation
7 if n == 0:7
8 return 0.08 from collections import Counter
99 from math import log2
10 def encode_runs(s):10
11 chars = [s[i] for i in range(len(s))]11 def entropy(s):
12 idx = list(range(1, len(chars)))12 probabilities = [freq / len(s) for freq in Counter(s).values()]
13 cuts = [0] + [i for i in idx if chars[i] != chars[i - 1]] + [len(chars)]13 return -sum(p * log2(p) if p > 0 else 0 for p in probabilities)
14 runs = [(chars[cuts[i]], cuts[i + 1] - cuts[i]) for i in range(len(cuts) - 1)]14
15 return runs15 def redundancy(s):
1616 max_entropy = log2(len(set(s))) if len(set(s)) > 1 else 0
17 def decode_runs(runs):17 actual_entropy = entropy(s)
18 return "".join([ch * cnt for ch, cnt in runs])18 return max_entropy - actual_entropy
1919
20 runs = encode_runs(data)20 # Calculate reduction in size possible based on redundancy
21 if decode_runs(runs) != data:21 reduction_potential = redundancy(data)
22 return 999.022
2323 # Assuming compression is achieved based on redundancy
24 compressed_size = sum([1 + len(str(cnt)) for ch, cnt in runs])24 max_possible_compression_ratio = 1.0 - (reduction_potential / log2(len(data)))
25 ratio = compressed_size / float(n)25
26 return float(ratio)26 # Qualitative check if max_possible_compression_ratio makes sense
27 except:27 if max_possible_compression_ratio < 0.0 or max_possible_compression_ratio > 1.0:
28 return 999.028 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
19% (0/5)201μs
1def solve(input):
2 try:
3 data = input["data"] if isinstance(input, dict) and "data" in input else ""
4 if not isinstance(data, str):
5 data = str(data)
6 n = len(data)
7 if n == 0:
8 return 0.0
9
10 def encode_runs(s):
11 chars = [s[i] for i in range(len(s))]
12 idx = list(range(1, len(chars)))
13 cuts = [0] + [i for i in idx if chars[i] != chars[i - 1]] + [len(chars)]
14 runs = [(chars[cuts[i]], cuts[i + 1] - cuts[i]) for i in range(len(cuts) - 1)]
15 return runs
16
17 def decode_runs(runs):
18 return "".join([ch * cnt for ch, cnt in runs])
19
20 runs = encode_runs(data)
21 if decode_runs(runs) != data:
22 return 999.0
23
24 compressed_size = sum([1 + len(str(cnt)) for ch, cnt in runs])
25 ratio = compressed_size / float(n)
26 return float(ratio)
27 except:
28 return 999.0
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