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To meet spec, I turned my answer into a function that takes a string argument. To my surprise, it was two bytes smaller! … ' sorted(map(chr,range(65,91)),key=s.upper().count) # Sort by that function ''.join(sorted(map(chr,range(65,91)),key=s.upper().count)) # Concatenate to string …

brainfuck, 238 207 bytes ,[[<+<+>>-]++++[<-------->-]-[<<-->>-----]<<+++++[----[----[------[------>>]]]]>[>>]>,]<<++++++++[<[++++<<]<++++<[++++<<]>>[>>]>>[>>]<-]<[<<]-[<+>---]>[>.>]<[<<]<<[<<]>[>>]>[ …

. @) from the string. This is okay, because only letters and spaces will ever be present. …

Make a list of all the lines in the input, define a function that tells you how much leading whitespace a string has, and make a sorted list of values that function spits out for each line of input. … l # string with the line f(l) # amount of leading whitespace d.index(f(l)) # where it is in list of whitespace …

pb - 89 bytes v[4]w[Y!-1]{b[35]^}w[B!0]{t[B]vvv>>b[T]^^^<}v>>>w[Y!4]{b[35]v}w[X!0]{b[35]^[Y]b[35]v[4]<} This is the kind of challenge pb was made for! Not that it's competitive for this kind of cha …

brainfuck - 25 23 bytes ,[>,]<-----.<,<[<]>[.>] If your language of choice does not support functions, whole programs are also acceptable. This is a whole program that takes input from STDIN. …

I wrote a custom puzzle for this challenge, which allows me to take input and check output against a known-correct "string". … This is the string I used for my byte count: MOV 99 ANY MOV -46 ACC ADD UP JEZ D ADD ACC ANY JRO -5 D:ADD UP MOV UP ANY MOV UP ANY That's the text of each non-empty node, seperated by newlines (effectively …

pb (NONCOMPETING), 125 bytes ^w[B!0]{>}w[B!45]{<w[B=10]{t[T+1]b[0]}}v[X]vw[T!0]{vb[124]<[X]b[124]>w[B=0]{>}t[T-1]}w[X!1]{<b[95]}<w[B!0]{^}w[Y!-1]{b[47]>^} The version of pbi that you need to run this …

brainfuck - 156 bytes ++++++++++>,[>>+++++[<+++++++>-],]<....[.---<<]>>+++>>+++.---.[.>>]<<.+++.[<]>.>>+++.---.<[.>>]<<<.>>.---[<]>.--->>+++.---[.>>]<<..+++.---[<]>[+++.>>]<<.... This is probably g …

brainfuck, 656 bytes +[[>>>,[>++++[<-------->-]]<]<<[>]<-[+[<+>>+<-]----[>>+<<----]>>+[<[-<]<[>]>>-]-<[[-]>+<]>[[-[->]<<+>]<->>>]<<<[>>>+<<<-]<<-]>>>>>[[<+>>+<-]----[>-<----]>--[----<]<[>]>[----<]<[>] …

Python 2.7 - 45 bytes print':)'*__import__('random').randint(0,255) (Originally this printed :)\n, the newline was removed upon OP's clarification) The (slightly) more idiomatic way to do this is …

Python, 228 198 182 166 146 145 bytes Counter (72 bytes): ior1=0 w=lambda i:i.split()if ior1 else sum(ord(c)in(9,10,32)for c in i) Splitter (73 bytes): ior1=0 w=lambda i:i.split()if i or 1else s …

pb, 80 bytes b[72]>b[101]>b[108]>b[108]>b[111]>b[44]>>b[87]>b[111]>b[114]>b[108]>b[100]>b[33] Super naive. I tried to golf it down by keeping 108 (the character code for "l") in T, either by doing …

pb - 449 bytes ^w[B!0]{w[B=40]{b[39]}t[B]w[B!0]{w[B=T]{^b[1]}w[B=T+1]{^b[1]}w[B=T+2]{^b[2]}^[Y]^>}<[X]^w[B!1]{>}t[1]b[0]w[T!0]{>w[B=1]{t[T+1]b[0]}w[B=2]{t[T-1]b[0]}}b[3]vw[B!0]{>}^w[B!3]{b[0]<}b[0]vb …

pb, 648 bytes ^w[B!0]{t[B]vb[T]^>}vb[-1]w[X!0]{<t[64]w[T!0]{w[B!0]{b[B-1]v}^[Y]t[T-1]}}w[B!-1]{w[B=0]{b[27]}t[26]w[T!0]{w[B!0]{b[B-1]v}^[Y]t[T-1]}>}b[0]w[X!0]{<w[B!0]{b[1]v}^[Y]w[B=0]{b[32]}w[B=1]{b[0 …