8086 machine code, 82 80

Contents of the x.com file:

B7 3D 89 DF B1 80 F3 AA 0D 0A 24 B4 01 CD 21 42 38 D8 74 F7 38 17 77 02 88 17 88 C3 31 D2 3C 0D 75 E9 BF 21 3D B1 5E 31 C0 F3 AE E3 EE 4F BB 04 01 8A 05 D4 0A 86 E0 0D 30 30 89 47 02 3C 30 77 04 88 67 03 43 89 3F 89 DA B4 09 CD 21 47 EB D7 

It only supports repetitions of up to 99 characters.

Source code (served as input for the debug.com assembler), with comments!

a mov bh, 3d ; storage of 128 bytes at address 3d00 mov di, bx mov cl, 80 rep stosb ; zero the array db 0d 0a 24 ; 10b mov ah, 1 int 21 ; input a char inc dx ; calculate the run length cmp al, bl ; is it a repeated character? je 10b cmp [bx], dl ; is the new run length greater than previous? ja 11a mov [bx], dl ; store the new run length ; 11a mov bl, al ; remember current repeating character xor dx, dx ; initialize run length to 0 cmp al, d ; end of input? jne 10b ; no - repeat mov di, 3d21 ; start printing run lengths with char 21 mov cl, 5e ; num of iterations = num of printable characters ; 127 xor ax, ax repe scasb ; look for a nonzero run length jcxz 11b ; no nonzero length - exit dec di mov bx, 104 ; address of output string mov al, [di] ; read the run length aam ; convert to decimal xchg al, ah or ax, 3030 mov [bx+2], ax cmp al, 30 ; was it less than 10? ja 145 mov [bx+3], ah ; output only one digit inc bx ; adjust for shorter string ; 145 mov [bx], di ; store "x=" into output string mov dx, bx ; print it mov ah, 9 int 21 inc di jmp 127 ; repeat ; 150 rcx 50 n my.com w q 

Here are some golfing techniques used here that I think were fun:

• array's address is 3d00, where 3d is the ascii-code for =. This way, the address for array's entry for character x is 3d78. When interpreted as a 2-character string, it's x=.
• Output buffer is at address 104; it overwrites initialization code that is no longer needed. End-of-line sequence 0D 0A 24 is executed as harmless code.
• The aam instruction here doesn't provide any golfing, though it could...
• Writing the number twice, first assuming it's greater than 10, and then correcting if it's smaller.
• Exit instruction is at an obscure address 11b, which contains the needed machine code C3 by luck.

CJam, 27 26 25 bytes

l:S_&{'=L{2$+_S\#)}g,(N}/ 

Try it online.

Example

$ cjam maxseq.cjam <<< "acbaabbbaaaaacc" a=5 c=2 b=3 

How it works

l:S " Read one line from STDIN and store the result in “S”. "; _& " Intersect the string with itself to remove duplicate characters. "; { " For each unique character “C” in “S”: "; '=L " Push '=' and ''. "; { " "; 2$+_ " Append “C” and duplicate. "; S\#) " Get the index of the modified string in “S” and increment it. "; }g " If the result is positive, there is a match; repeat the loop. "; , " Retrieve the length of the string. "; ( " Decrement to obtain the highest value that did result in a match. "; N " Push a linefeed. "; }/ " "; 

J - 52 bytes

Well, a simple approach again.

f=:([,'=',m=:":@<:@#@[`(]m~[,{.@[)@.(+./@E.))"0 1~~. 

Explanation:

f=:([,'=',m=:":@<:@#@[`(]m~[,{.@[)@.(+./@E.))"0 1~~. ~~. Create a set of the input and apply it as the left argument to the following. ([,'=',m=:":@<:@#@[`(]m~[,{.@[)@.(+./@E.))"0 1 The function that does the work "0 1 Apply every element from the left argument (letters) with the whole right argument (text). @.(+./@E.) Check if the left string is in right string. (]m~[,{.@[) If yes, add one letter to the left string and recurse. ":@<:@#@[ If not, return (length of the left string - 1), stringified. [,'=', Append it to the letter + '=' 

Example:

 f 'acbaabbbaaaaacc' a=5 c=2 b=3 f 'aaaabaa' a=4 b=1 

If free-form output is allowed (as in many other answers), I have a 45 bytes version too. These boxes represent a list of boxes (yes, they're printed like that, although SE's line-height breaks them).

 f=:([;m=:<:@#@[`(]m~[,{.@[)@.(+./@E.))"0 1~~. f 'acbaabbbaaaaacc' ┌─┬─┐ │a│5│ ├─┼─┤ │c│2│ ├─┼─┤ │b│3│ └─┴─┘ f 'aaaabaabba' ┌─┬─┐ │a│4│ ├─┼─┤ │b│2│ └─┴─┘ 

Ruby, 72

(a=$*[0]).chars.uniq.map{|b|puts [b,a.scan(/#{b}+/).map(&:size).max]*?=} 

This takes input from command line arguments and outputs to stdout.

GolfScript, 26 bytes

:s.&{61{2$=}s%1,/$-1=,n+}% 

Try it online.

Explanation:

• :s saves the input string in the variable s for later use.
• .& extracts the unique characters in the input, which the rest of the code in the { }% loop then iterates over.
• 61 pushes the number 61 (ASCII code for an equals sign) on top of the current character on the stack, to act as an output delimiter.
• {2$=}s% takes the string s and replaces its characters with a 1 if they equal the current character being iterated over, or 0 if they don't. (It also leaves the current character on the stack for output.)
• 1,/ takes this string of ones and zeros, and splits it at zeros.
• $ sorts the resulting substrings, -1= extracts the last substring (which, since they all consist of repetitions of the same character, is the longest), and , returns the length of this substring.
• n+ stringifies the length and appends a newline to it.

Ps. If the equals signs in the output are optional, the 61 can be omitted (and the 2$ replaced by 1$), for a total length of 24 bytes:

:s.&{{1$=}s%1,/$-1=,n+}% 

Python 3, 69 bytes

s=input() for c in set(s): i=0 while-~i*c in s:i+=1 print(c,'=',i) 

Try it online!

Even golfed Python can be very readable. I think this code is fully idiomatic except for the -~i for i+1 and the single-letter variables. Thanks to pxeger for saving 1 byte.

Example runs:

>>> helloworld e = 1 d = 1 h = 1 l = 2 o = 1 r = 1 w = 1 >>> acbaabbbaaaaacc a = 5 c = 2 b = 3 

CoffeeScript, 109 bytes

I like regex.

f=(s)->a={};a[t[0]]=t.length for t in s.match(/((.)\2*)(?!.*\1)/g).reverse();(k+'='+v for k,v of a).join '\n' 

Here is the compiled JavaScript you can try in your browser's console

f = function(s) { var a, t, _i, _len, _ref; a = {}; _ref = s.match(/((.)\2*)(?!.*\1)/g).reverse(); for (_i = 0, _len = _ref.length; _i < _len; _i++) { t = _ref[_i]; a[t[0]] = t.length; } return a; }; 

Then you can call

f("acbaabbbaaaaacc") 

to get

c=2 a=5 b=3 

Pyth, 24 25 26 (or 29)

=ZwFY{Z=bkW'bZ~bY)p(Yltb 

Test can be done here: link

Outputs in the format:

('a', 5) ('c', 2) ('b', 3) 

Explanation:

=Zw Store one line of stdin in Z FY{Z For Y in set(Z): =bk b='' W'bZ while b in Z: ~bY b+=Y ) end while p(Yltb print (Y, len(b)-1) 

Python:

k="" Z=copy(input()) for Y in set(Z): b=copy(k) while (b in Z): b+=Y print(_tuple(Y,len(tail(b)))) 

For proper (a=5) output, use:

=ZwFY{Z=bkW'bZ~bY)p++Y"="`ltb 

29 characters

C, 126 125 119 bytes

l,n,c[256];main(p){while(~(p=getchar()))n*=p==l,c[l=p]=c[p]>++n?c[p]:n;for(l=256;--l;)c[l]&&printf("%c=%d\n",l,c[l]);} 

Running:

$ gcc seq.c 2>& /dev/null $ echo -n 'acbaabbbaaaaacc' | ./a.out c=2 b=3 a=5 

Mathematica, 74 72 69

Print[#[[1,1]],"=",Max[Tr/@(#^0)]]&/@Split@Characters@#~GatherBy~Max& % @ "acbaabbbaaaaacc" 

a=5 c=2 b=3 

Not very good but strings are not Mathematica's best area. Getting better though. :-)

C# (LinQPad)

146

This is tsavino's answer but shorter. Here, I used Distinct() instead of GroupBy(c=>c). Also the curly braces from the foreach-loop are left out:

void v(string i){foreach(var c in i.Distinct())Console.WriteLine(c+"="+(from Match m in Regex.Matches(i,"["+c+"]+")select m.Value.Length).Max());} 

136

I tried using a lambda expression instead of the normal query syntax but since I needed a Cast<Match> first, the code became 1 character longer... Anyhow, since it can be executed in LinQPad, you can use Dump() instead of Console.WriteLine():

void v(string i){foreach(var c in i.Distinct())(c+"="+(from Match m in Regex.Matches(i,"["+c+"]+")select m.Value.Length).Max()).Dump();} 

Further study of the code got me thinking about the Max(). This function also accepts a Func. This way I could skip the Select part when using the lambda epxression:

void v(string i){foreach(var c in i.Distinct())(c+"="+Regex.Matches(i,"["+c+"]+").Cast<Match>().Max(m=>m.Value.Length)).Dump();} 

Thus, final result:

128

Update:

Thanks to the tip from Dan Puzey, I was able to save another 6 characters:

void v(string i){i.Distinct().Select(c=>c+"="+Regex.Matches(i,"["+c+"]+").Cast<Match>().Max(m=>m‌​.Value.Length)).Dump();} 

Length:

122

R, 73 bytes

paste(names(b<-by((r<-rle(el(strsplit(scan(,""),""))))$l,r$v,max)),"=",b) 

Try it online!

I would like to mention that this solution uses rle function, which was introduced in the R version 3.6.2 (Dec. 2019); therefore my answer and the earlier one from 2014 are not directly comparable.

Ruby, 58

h={} gets.scan(/(.)\1*/){h[$1]=[h[$1]||0,$&.size].max} p h 

Takes input from STDIN, outputs it to STDOUT in the form {"a"=>5, "c"=>2, "b"=>3}

C# in LINQPad - 159 Bytes

Well, at least I beat T-SQL ;P Won't beat anyone else, but I thought I'd share it anyway.

void v(string i){foreach(var c in i.GroupBy(c=>c)){Console.WriteLine(c.Key+"="+(from Match m in Regex.Matches(i,"["+c.Key+"]+")select m.Value.Length).Max());}} 

Usage:

v("acbaabbbaaaaacc"); 

Suggestions are always welcome!

Powershell 80 77 72

$x=$args;[char[]]"$x"|sort -u|%{"$_="+($x-split"[^$_]"|sort)[-1].length} 

You need to run it on console...

Perl - 65 71 76 characters

My first code golf!

For each answer, copy to golf.pl and run as:

echo acbaabbbaaaaacc | perl golf.pl 

My shortest solution prints each character as many times as it appears, since that is not prohibited by the rules.

$_=$i=<>;for(/./g){$l=length((sort$i=~/$_*/g)[-1]);print"$_=$l "} 

My next-shortest solution (85 90 characters) only prints each character once:

<>=~s/((.)\2*)(?{$l=length$1;$h{$2}=$l if$l>$h{$2}})//rg;print"$_=$h{$_} "for keys %h 

F# - 106

let f s= let m=ref(Map.ofList[for c in 'a'..'z'->c,0]) String.iter(fun c->m:=(!m).Add(c,(!m).[c]+1))s;m 

In FSI, calling

f "acbaabbbaaaaacc" 

gives

val it : Map<char,int> ref = {contents = map [('a', 8); ('b', 4); ('c', 3); ('d', 0); ('e', 0); ('f', 0); ('g', 0); ('h', 0); ('i', 0); ...];} 

However, to print it without the extra information, call it like this:

f "acbaabbbaaaaacc" |> (!) |> Map.filter (fun _ n -> n > 0) 

which gives

val it : Map<char,int> = map [('a', 8); ('b', 4); ('c', 3)] 

Javascript, 116 bytes

y=x=prompt();while(y)r=RegExp(y[0]+'+','g'),alert(y[0]+'='+x.match(r).sort().reverse()[0].length),y=y.replace(r,'') 

Sample output:

lollolllollollllollolllooollo l=4 o=3 acbaabbbaaaaacc a=5 c=2 b=3 helloworld h=1 e=1 l=2 o=1 w=1 r=1 d=1 

T-SQL (2012) 189 171

Edit: removed ORDER BY because rules allow any output order.

Takes input from a CHAR variable, @a, and uses a recursive CTE to create a row for each character in the string and figures out sequential occurrences.

After that, it's a simple SELECT and GROUP BY with consideration for the order of the output.

Try it out on SQL Fiddle.

WITH x AS( SELECT @a i,''c,''d,0r,1n UNION ALL SELECT i,SUBSTRING(i,n,1),c,IIF(d=c,r+1,1),n+1 FROM x WHERE n<LEN(i)+2 ) SELECT d+'='+LTRIM(MAX(r)) FROM x WHERE n>2 GROUP BY d 

Assigning the variable:

DECLARE @a CHAR(99) = 'acbaabbbaaaaacc'; 

Sample output:

a=5 c=2 b=3 

Haskell - 113 120 bytes

import Data.List main=interact$show.map(\s@(c:_)->(c,length s)).sort.nubBy(\(a:_)(b:_)->a==b).reverse.sort.group 

Tested with

$ printf "acbaabbbaaaaacc" | ./sl [('a',5),('b',3),('c',2)] 

JavaScript [83 bytes]

prompt().match(/(.)\1*/g).sort().reduce(function(a,b){return a[b[0]]=b.length,a},{}) 

^{Run this code in the browser console.}

For input "acbaabbbaaaaacc" the console should output "Object {a: 5, b: 3, c: 2}".

JavaScript - 91

for(i=0,s=(t=prompt()).match(/(.)\1*/g);c=s[i++];)t.match(c+c[0])||alert(c[0]+'='+c.length) 

EDIT: My first solution obeys the rules, but it prints several times single char occurrences like abab => a=1,b=1,a=1,b=1 so I came out with this (101 chars), for those not satisfied with my first one:

for(i=0,s=(t=prompt()).match(/((.)\2*)(?!.*\1)/g);c=s[i++];)t.match(c+c[0])||alert(c[0]+'='+c.length) 

Husk, 18 bytes

mȯJ'=§eo;←osL►Lk←g 

Try it online!

11 bytes if we ignore the exact input format and display as a list of pairs instead.

Husk, 16 bytes

mS:(:'=s▲`mg¹#)U 

Try it online!

Explanation:

m U # for each of the unique elements of the input S: # prepend the element to (:'= ) # '=' prepended to ▲ # the maximum of # # the number of times it occurs in `mg¹ # each group of identical items in the input, s # converted into a string 

I saw that Razetime had answered this in Husk, so I had to have a go too...

Jelly, 12 bytes

ŒriƇṀj”=ʋⱮQY 

Try it online!

-2 bytes thanks to Unrelated String!

8 bytes to output as a list of pairs

How it works

ŒriƇṀj”=ʋⱮQY - Main link. Takes a string S on the left ʋ - Group the previous 4 links into a dyad f(S, C): Œr - Run-length encode S Ƈ - For each sublist in rle(S), keep it if: i - It contains C Ṁ - Maximum As all elements are in the form [C, n] for integers n, Python considers the maximum as the greatest n j”= - Join with "="; Yields [C, "=", n] Q - Get the unique characters of S Ɱ - For each unique character C in S, yield f(C, S) Y - Join by newlines and output 

Bash / GNU core utils (40)

fold -1|uniq -c|sort -k2 -k1r|uniq -f1 

Try it online!

Output when input is "acbaabbbaaaaacc":

 5 a 3 b 2 c 

Explanation:

• fold -1 adds a newline after each character
• uniq -c counts the number of consecutive identical lines
• sort -k2 -k1r sorts the result so as the maximum count for each character comes first
• uniq -f1 keeps only the first line for each character

05AB1E, 17 bytes

ÙvIγʒyå}€gZy…ÿ=ÿ, 

Try it online!

ÙvIγʒyå}€gZy…ÿ=ÿ, Argument s Ùv For each c in s uniquified Iγ Split s into chunks of consecutive equal elements ʒyå} Filter: Keep elements that contain c €g Map to element length Z Get max y…ÿ=ÿ, Print "c=max" 

Julia, 85

f(s)=(l=0;n=1;a=Dict();[c==l?n+=1:(n>get(a,l,1)&&(a[l]=n);n=1;l=c) for c in s*" "];a) julia> f("acbaabbbaaaaacc") {'a'=>5,'c'=>2,'b'=>3} 

Python3 - 111, 126, 115 114 111 bytes

Executable code that will read 1 line (only use lowercase letters a-z)

d={}.fromkeys(map(chr,range(97,123)),0) for c in input():d[c]+=1 [print("%s=%d"%(p,d[p]))for p in d if d[p]>0] 

Edit: Excluded unnecessary output on request from @Therare

The output looks nice

~/codegolf $ python3 maxseq.py helloworld l=3 o=2 h=1 e=1 d=1 w=1 r=1 

JavaScript - 141 137 125

I don't like regex :)

function g(a){i=o=[],a=a.split('');for(s=1;i<a.length;){l=a[i++];if(b=l==a[i])s++;if(!b|!i){o[l]=o[l]>s?o[l]:s;s=1}}return o} 

Run

console.log(g("acbaabbbaaaaacc")); 

outputs

[ c: 2, a: 5, b: 3 ]