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Fastest-algorithm competitions are won by the answer with the smallest asymptotic time complexity. For challenges based on actual runtime, use [fastest-code] instead.
4
votes
(Cops) Fast and Golfiest Season 1: The Lord of the Strings
Concatenated Words, Python, 177 bytes, \$O(n)\$ time assuming input is a set where \$n\$ is the number of words, Cracked by Cursor Coercer, 7 points
This is \$O(2^L)\$ where \$L\$ is the length of eac …
4
votes
(Cops) Fast and Golfiest Season 1: The Lord of the Strings
Generate Parenthesis, ><> (Fish), 178 bytes, \$O(n!)\$, Cracked by C--
Time: 2023-07-21 12:35:24Z
Hover over any symbol to see what it does. If this appears glitched unformulated version …
2
votes
(Cops) Fast and Golfiest Season 1: The Lord of the Strings
Reverse words in string, Python, 47 bytes, \$O(n)\$ complexity, Cracked by The Thonnu, 6 point
lambda g:print(*map(str.strip,g.split()[::-1]))
Given an input string s, reverse the order of the words …
1
vote
(Cops) Fast and Golfiest Season 1: The Lord of the Strings
Generate Parenthesis, Rust, 193 bytes, \$O(n!)\$ Cracked by C--, 1 point
C-- generously helped me save 1 point before cracking it in an actually significant way
fn f(z:i32)->Vec<String>{if z<1{vec![ …
1
vote
(Robbers) Fast and Golfiest Season 1: The Lord of the Strings
C (GCC), 66 bytes (7.04% reduction), cracks Sum of Scores for buit strings
f(s,l,i)char*s;{for(i=0;s[i]&&s[l+i]==s[i];++i);s=l?i+f(s,l-1):i;}
Attempt This Online!
Based on This C tip from G B