Vim, 27, 26, 25 bytes

DJqqYp:s/.\zs[^<C-r>-]<CR>@qq@qD 

Try it online!

Input comes in this format:

e codegolf.stackexchange.com 

My naive first approach is three bytes longer:

i:s/.\zs[^<Right>]<Esc>"addqqYp@a@qq@qdd 

I'm also happy with this answer because it starts with my name.

DJqq:t$|s/.\zs[^<C-r>"]<CR>@qq@qD DJMcMayhem 

See the similarity? Eh?

Less successful approaches:

i:s/.\zs[^<Right>]<Esc>"addqqYp@a@qq@qdd i:s/.\zs[^<Right>]<CR>@q<Esc>"adkqqYp@aq@qdd DJ:s/.\zs[^<C-r>"]<CR>uqqYp@:@qq@qdd DJqq:t.|$s/.\zs[^<C-r>"]<CR>@qq@qdd 

Explanation:

D " Delete everything on this first line J " Remove this empty line qq " Start recording into register 'q' Y " Yank this line p " And paste it :s/ " Run a substitute command on the last line in the buffer. Remove: . " Any character \zs " Move the selection forward (so we don't delete the first one) [^<C-r>-] " Followed by anything *except* for the character we deleted <CR> " Run the command @q " Call register 'q' q " Stop recording @q " Run the recursive macro D " Delete our extra line 

MATL, 20 16 bytes

y-f1X-"t[]@X@q-( 

Try it online! Or verify test cases: 1, 2, 3, 4, 5.

Bonus

Modified code to see the string being gradually shrunk (offline compiler):

y-f1X-"tt.3Y.XxD[]@X@q-(].3Y.XxDvx 

Or try it at MATL Online!

Explanation

y % Implicitly input string and char. Duplicate string onto top - % Subtract. Gives nonzero for chars in the input string that are % different from the input char f % Array of indices of nonzero values 1X- % Remove 1 from that array. This gives an array of the indices of % chars to be removed from the string " % For each t % Duplicate previous string [] % Push empty array @ % Push index of char to be removed. But this index needs to be % corrected to account for the fact that previous chars have % already been removed... X@q- % ... So we correct by subtracting the 0-based iteration index ( % Assign empty array to that position, to remove that char % Implicitly end for each % Implicitly display 

Haskell, 50 bytes

w@(a:x)%c|(d,_:y)<-span(==c)x=w:(a:d++y)%c|0<1=[w] 

Defines a function (%) returning a list of strings.

Explanation

(%) is called as w%c, with w being the input string, and c the character to keep. In short, this definition works by separating w into the first character (a) and the remainder (x), splitting x at the first occurrence of a character other than c, and recursively calling itself with that one character dropped.

w@(a:x)%c -- define (%) with w separated into a and x. |(d,_:y)<-span(==c)x -- split x; bind the string of `c` to d, and the rest -- to _:y, dropping first character and calling the rest y. =w:(a:d++y)%c -- if there was a character to drop, cons w onto the -- sequence gained by recursively calling (%) with that -- character removed (i.e. call with a:d++y). |0<1=[w] -- if nothing needed to be dropped, the result sequence is -- simply the one-element list [w] 

Retina, 28 27 bytes

Byte count assumes ISO 8859-1 encoding.

;{G*1` R1r`(?!^|.*¶?\1$)(.) 

Try it online!

Explanation

;{G*1` 

There's a lot of configuration here. The stage itself is really just G1`, which keeps only the first line, discarding the input character. * turns it into a dry run, which means that the result (i.e. the first line of the string) is printed without actually changing the string. { tells Retina to run both stages in a loop until the string stops changing and ; prevents output at the end of the program.

R1r`(?!^|.*¶?\1$)(.) 

This discards the first character which a) isn't at the beginning of the input, b) isn't equal to the separate input character.

Pip, 22 26 24 22 bytes

Lv+#Paa@oQb?++oPaRA:ox 

Takes string as first command-line argument, character as second. Try it online!

Explanation

Loops over characters of input; if the character equals the special character, move on to the next one; if not, delete it and print the string.

An ungolfed version (a, b get cmdline args; o starts with a value of 1, x is ""):

P a Print a L #a-1 Loop len(a)-1 times: I a@o Q b If a[o] string-eQuals b: ++o Increment o E { Else: a RA: o x In-place in a, Replace char At index o with x (i.e. delete it) P a Print a } 

Golfing tricks:

• The loop header for L is only evaluated once, so we can sneak the initial print in there. #Pa-1 won't work because P is low-precedence (it would parse as #P(a-1)), but we can rearrange it to v+#Pa, using the v variable preinitialized to -1.
• The RA: operator returns the new value of a, so we can print that expression instead of having a separate Pa statement.
• Now both of the branches of the if statement are single expressions, so we can use the ternary operator ? instead.

Perl 5, 29 bytes

I got 35 bytes using Strawberry Perl: 31 bytes, plus 1 for -nE instead of -e, plus 3 for space + -i (used for the single-letter input; the longer string is from STDIN).

chomp;say;s/(.)[^$^I]/$1/&&redo 

However, I've no doubt this is doable without chomp; using <<<, which is 29 bytes, even though I can't test it myself using Strawberry.

say;s/(.)[^$^I]/$1/&&redo 

Thus:

perl -im -nE'say;s/(.)[^$^I]/$1/&&redo' <<< "example" 

Perl 6,  47 40  38 bytes

->\a,\b{a,{S/^(."{b}"*:)./$0/}...^{$^a eq $^b}} 

->\a,\b{a,{S/^(."{b}"*:)./$0/}...^&[eq]} 

{$^b;$^a,{S/^(."$b"*:)./$0/}...^&[eq]} 

Expanded:

{ # lambda with two placeholder parameters ｢$a｣ and ｢$b｣ $^b; # declare second parameter $^a, # declare first parameter, and use it to seed the sequence { # bare block lambda with implicit parameter ｢$_｣ S/ # string replace and return ^ # beginning of string ( # capture into ｢$0｣ . # the first character "$b"* # as many ｢$b｣ as possible : # don't allow backtracking ) . # any character ( the one to be removed ) /$0/ # put the captured values back into place } ...^ # repeat that until: ( and throw away the last value ) &[eq] # the infix string equivalence operator/subroutine } 

The reason ...^ was used instead of ... is that &[eq] wouldn't return True until the last value was repeated.

JavaScript (ES6), 74 bytes

(s,c)=>[...t=s.slice(1)].map(d=>c!=d?s+=` `+s[0]+(t=t.replace(d,``)):s)&&s 

Python 3, 72 bytes

def e(i,k): for r in i: if r!=k:i=i[0]+i[1:].replace(r,'',1);print(i) 

Try it online!

e('😋🥕🍎🥓🥑🥓🥑🥒🍆🥔🍆🥓🍅🥜🥓🥔🍅🍄🍆','🥓') 

Go on a diet:

😋🥕🍎🥓🥑🥓🥑🥒🍆🥔🍆🥓🍅🥜🥓🥔🍅🍄🍆 😋🍎🥓🥑🥓🥑🥒🍆🥔🍆🥓🍅🥜🥓🥔🍅🍄🍆 😋🥓🥑🥓🥑🥒🍆🥔🍆🥓🍅🥜🥓🥔🍅🍄🍆 😋🥓🥓🥑🥒🍆🥔🍆🥓🍅🥜🥓🥔🍅🍄🍆 😋🥓🥓🥒🍆🥔🍆🥓🍅🥜🥓🥔🍅🍄🍆 😋🥓🥓🍆🥔🍆🥓🍅🥜🥓🥔🍅🍄🍆 😋🥓🥓🥔🍆🥓🍅🥜🥓🥔🍅🍄🍆 😋🥓🥓🍆🥓🍅🥜🥓🥔🍅🍄🍆 😋🥓🥓🥓🍅🥜🥓🥔🍅🍄🍆 😋🥓🥓🥓🥜🥓🥔🍅🍄🍆 😋🥓🥓🥓🥓🥔🍅🍄🍆 😋🥓🥓🥓🥓🍅🍄🍆 😋🥓🥓🥓🥓🍄🍆 😋🥓🥓🥓🥓🍆 😋🥓🥓🥓🥓 

05AB1E, 26 25 bytes

¬ˆ[¦Ðg_#¬²k0Qi²ˆë¯J?,]¯J? ¬ˆ Put the first character into an array [ While true ¦ Remove the first character Ð Triplicate g_# if the string is empty, break ¬²k0Qi if the first character is equal to the one specified in the input ²ˆ Add it to the array ë Else ¯J? Display the array , Display the remaining string ] End while ¯J? Display the last string 

Try it online!

Please note that ¬²k0Q could be rewritten ¬²Q, but for some reason it doesn't work when the current character is a quote mark: Q returns the actual string instead of a boolean and it causes an infinite loop.

This code can be golfed further since ¯J? is duplicated. Moving this part in the loop would remove the duplication and would also allow to drop the closing square bracket.

Python 2, 71 66 bytes:

f,m=input();k=f[0] while f:a=f[0]==m;k+=f[0]*a;f=f[1+a:];print k+f 

A full program. Takes 2 inputs through STDIN in the format '<String>','<Char>'.

Also, here is a recursive solution currently at 140 bytes:

Q=lambda c,p,k='',j=1,l=[]:c and Q(c[1:],p,k+c[0]*(j<2)+c[0]*(c[0]==p),j+1,l+[k+c])or'\n'.join(sorted({*l},key=l.index))+('\n'+k)*(k not in l) 

This one should be called in the format print(Q('<String>','<Char>')).

Ruby, 148 139 97 90 83 77 62 bytes

a,c=$*;p s=""+a;i=1;while d=s[i];(d!=c)?(s[i]="";p s):i+=1;end 

Not sure if amateur code is accepted on this exchange but I'm interested in learning to code golf although I'm terrible at it, any help on how I'd get this program looking as small as the others here?

EDIT:

Replaced puts with p

Removed a tonne of whitespace and counted bytes correctly thanks to Wheat Wizard

Thanks to challenger5 went from s=gets.chop;c=gets.chop; to s,c=gets.chop,gets.chop;

replaced then with ; and gets.chop with gets[0] thanks Mhutter!

Taking input as command line variables now, eg. prog.rb helloworld l

Thanks to numerous improvements by jeroenvisser101 replacing a=s.dup with s=""+a and the previous if statement if s[i]!=c;s[i]="";p s;else i+=1;end with (d!=c)?(s[i]="";p s):i+=1; huge improvement!

c90, 129 125 bytes

with whitespace:

main(q, x) { for (char **v = x, *a = v[1], *b = a, *c;*b++;) for (c = a; c == a | *c == *v[2] && *b != *v[2] && putchar(*c), ++c != b || *b != *v[2] && !puts(b);); } 

without whitespace:

main(q,x){for(char**v=x,*a=v[1],*b=a,*c;*b++;)for(c=a;c==a|*c==*v[2]&&*b!=*v[2]&&putchar(*c),++c!=b||*b!=*v[2]&&!puts(b););} 

ungolfed:

#include <stdio.h> int main(int argc, char **argv) { char *a = argv[1]; for (char *b = a + 1; *b; b++) { if (*b == *argv[2]) { continue; } putchar(*a); for (char *c = a + 1; c != b; c++) { if (*c == *argv[2]) { putchar(*c); } } puts(b); } } 

This takes a pointer to the start of the string, and loops, iterating this pointer until it reaches the end of the string. Within the loop, it prints the first character, then any instances of the second argument it finds between the start of the string and the pointer. After this, it calls puts on the pointer, printing out the rest of the string.

This must be compiled on a system where sizeof(int) == sizeof(char*). +3 bytes otherwise.

This is the first time I've tried code golfing here, so I'm sure there are some optimizations to be made.

Java 10, 155 140 139 124 bytes

c->s->{var r=s+"\n";for(int i=0;++i<s.length();)if(s.charAt(i)!=c)r+=(s=s.substring(0,i)+s.substring(i--+1))+"\n";return r;} 

Try it online.

Explanation:

c->s->{ // Method with character and String parameters and String return-type var r=s+"\n"; // Result-String, starting at the input-String with trailing new-line for(int i=0;++i<s.length();) // Loop over the characters of the String, skipping the first if(s.charAt(i)!=c) // If the current character and the input-character are equal r+=(s=s.substring(0,i)+s.substring(i--+1)) // Remove this character from the String `s` +"\n"; // And append the new `s` with trailing new-line to the result-String return r;} // Return the result-String 

Old 139 bytes recursive answer:

void c(String s,int c){System.out.println(s);for(int i=1;i<s.length();)if(s.charAt(i++)!=c){c(s.substring(0,i-1)+s.substring(i),c);break;}} 

-1 bytes thanks to @Eugene. (Next time make a comment instead of editing someone else's post, please.)

Try it online.

Explanation:

void c(String s,int c){ // Method with String and integer parameters and no return-type System.out.println(s); // Print the input-String with trailing new-line for(int i=1;i<s.length();)// Loop over the characters of the String, skipping the first if(s.charAt(i++)!=c){ // If the current character and the input-character are equal c(s.substring(0,i-1)+s.substring(i),c); // Remove this character, and do a recursive call break;}} // And stop the loop 

JavaScript (ES6), 63 64 69

Thanks to @Shaggy for shaving 1 byte!

Returning a single string with newlines

s=>c=>[...s].map((x,i,z)=>i&&x!=c&&(s+=` `+z.join(z[i]='')))&&s 

F= s=>c=>[...s].map((x,i,z)=>i&&x!=c&&(s+=` `+z.join(z[i]='')))&&s function update() { var s=S.value,c=C.value[0] O.textContent=F(s)(c) } update()

<input id=S value='Hello world!' oninput='update()'> <input id=C value='!' oninput='update()'> <pre id=O></pre>

Dyalog APL, 27 bytes

{×i←⊃1+⍸⍺≠1↓⎕←⍵:⍺∇⍵/⍨i≠⍳≢⍵}

⍺ is the excluded character, ⍵ is the initial string

print argument; find index i of the first non-⍺ after the first char; if found, call recursively with i removed

Mathematica, 64 bytes

Most@FixedPointList[StringReplace[#,b_~~Except@a:>b,1]&,a=#2;#]& 

Anonymous function. Takes two strings as input, and returns a list of strings as output. Works by repeatedly removing the first non-instance of the character.

PHP, 88 84 86 85 82 81 78 bytes

1 byte saved thanks to @IsmaelMiguel, 3 bytes thanks to @user59178, 3 bytes inspired by @user59178

while($b=substr("$argv[1]\n",$i++))$a>""&$b[0]!=$argv[2]?print$a.$b:$a.=$b[0]; 

takes input from command line arguments; run with php -r <code> '<string>' <character>

• appends a newline to the input for an implicit final print.
  That adds 5 4 bytes of code, but saves on the output and an additional echo$a;.

05AB1E, 26 24 23 bytes

Thanks @Kade for 2 bytes!
Thanks @Emigna for 1 byte!

¬UDvy²k0Êiy¡¬s¦yý«Xs«=¦ 

Uses the CP-1252 encoding. Try it online!

y²k0Ê could be y²Ê but the " messes it up.

This probably could be golfed more because « is repeated twice. Please leave a comment if you have any suggestions or ways to golf it down more.

C#, 122 117 112 bytes

IEnumerable F(string s,char c){for(int i=0;i<s.Length;++i)if(i<1||s[i]!=c)yield return i>0?s=s.Remove(i--,1):s;} 

Ungolfed :

public IEnumerable F(string s, char c) { for (int i = 0; i < s.Length; ++i) { if (i < 1 || s[i] != c) yield return i > 0 ? s = s.Remove(i--, 1) : s; } } 

Returns a collection of strings.

TSQL, 127 bytes(Excluding variable definitions)

DECLARE @1 VARCHAR(100)='codegolf.stackexchange.com' DECLARE @2 CHAR(1) = 'o' DECLARE @ char=LEFT(@1,1)WHILE patindex('%[^'+@2+']%',@1)>0BEGIN SET @1=STUFF(@1,patindex('%[^'+@2+']%',@1),1,'')PRINT @+@1 END 

Formatted:

DECLARE @1 VARCHAR(100) = 'codegolf.stackexchange.com' DECLARE @2 CHAR(1) = 'o' DECLARE @ CHAR = LEFT(@1, 1) WHILE patindex('%[^' + @2 + ']%', @1) > 0 BEGIN SET @1 = STUFF(@1, patindex('%[^' + @2 + ']%', @1), 1, '') PRINT @ + @1 END 

C#, 135 138 :( 137 bytes

Golfed:

IEnumerable<string>F(string s,char c){int i=1,l;for(;;){yield return s;l=s.Length;while(i<l&&s[i]==c)i++;if(i==l)break;s=s.Remove(i,1);}} 

Ungolfed:

 IEnumerable<string> F(string s, char c) { int i = 1, l; for (;;) { yield return s; l = s.Length; while (i < l && s[i] == c) i++; if (i == l) break; s = s.Remove(i, 1); } } 

Function returns collection of strings.

EDIT1: @psycho noticed that algorithm was not implemented properly.

EDIT2: Created variable for s.Length. One byte saved thanks to @TheLethalCoder.

Python 2 - 65 73 Bytes

lambda s,c:[s[0]+c*s[1:i].count(c)+s[i+1:]for i in range(len(s))] 

And a 76 Bytes recursive solution, because despite being longer than the firt one, I kinda like it better:

f=lambda s,c,i=1:s[i:]and[[s]+f(s[:i]+s[i+1:],c,i),f(s,c,i+1)][s[i]==c]or[s] 

Racket 194 bytes

(let p((s s)(n 1)(t substring)(a string-append))(displayln s)(cond[(>= n(string-length s))""] [(equal? c(string-ref s n))(p(a(t s 0 n)(t s n))(+ 1 n)t a)][else(p(a(t s 0 n)(t s(+ 1 n)))n t a)])) 

Ungolfed:

(define (f s c) (let loop ((s s) (n 1)) (displayln s) (cond [(>= n (string-length s))""] [(equal? c (string-ref s n)) (loop (string-append (substring s 0 n) (substring s n)) (add1 n))] [else (loop (string-append (substring s 0 n) (substring s (add1 n))) n)]))) 

Testing:

(f "Test cases" #\s) (f "codegolf.stackexchange.com" #\e) 

Output:

Test cases Tst cases Tst cases Ts cases Tscases Tsases Tsses Tsses Tsss Tsss "" codegolf.stackexchange.com cdegolf.stackexchange.com cegolf.stackexchange.com cegolf.stackexchange.com ceolf.stackexchange.com celf.stackexchange.com cef.stackexchange.com ce.stackexchange.com cestackexchange.com cetackexchange.com ceackexchange.com ceckexchange.com cekexchange.com ceexchange.com ceexchange.com ceechange.com ceehange.com ceeange.com ceenge.com ceege.com ceee.com ceee.com ceeecom ceeeom ceeem ceee "" 

Swift 3 - 151 147 bytes

Swift isn't the ideal language for golfing, particularly when it relates to string indexing. This is the best I could do:

func c(a:String,b:String){print(a);var q=Array(a.characters),i=1;while i<q.count{if "\(q[i])" != b{q.remove(at:i);c(a:String(q),b:b);break};i=i+1}} 

Unfortunately, Swift needs spaces around != (but not for ==), and Swift 3 dropped the ++ operator. The trick for both of these is to convert to a character array which allows integer indexing, and using string interpolation of a character to convert back to a String ("\(c)").

Ungolfed:

func c(a:String, b:String) { print(a) var q = Array(a.characters) var i = 1 while i < q.count { if "\(q[i])" != b { q.remove(at:i) c(a: String(q), b: b) break } i=i+1 } } 

Previous, non-recursive solution

func c(a:String,b:String){var q=Array(a.characters),e={q.removeFirst()},z="\(e())";print(a);while !q.isEmpty{"\(e())"==b ? z=z+b : print(z+String(q))}} 

func c(a:String, b:String) { var q = Array(a.characters) var z = "\(q.removeFirst())" print(a) while !q.isEmpty { if "\(q.removeFirst())" == b { z = z + b }else{ print(z + String(q)) } } } 

Pyke, 26 19 17 bytes

jljjhF3<Q/Q*jih>s 

Try it here!

 - Q = input j - j = input_2 ljjhF3 - for (i, j, j[0]) for i in range(len(j)) < - j[:i] Q/ - ^.count(Q) Q* - ^*Q s - sum(j[0], ^, V) jih> - j[i+1:] 

Vyxal, 77 bits^v1, 9.625 bytes

ḣ≬‡⁰≠Ǒ⟇↔Ṡ+ 

Try it Online!

Explained

The very first explanation generated using Luminespire, a currently work-in-progress explanation formatting assistant. Now with metadata!

ḣ≬‡⁰≠Ǒ⟇↔Ṡ+­⁡​‎‎⁡⁠⁡‏‏​⁡⁠⁡‌⁢​‎‎⁡⁠⁢‏⁠‎⁡⁠⁢⁤‏‏​⁡⁠⁡‌⁣​‎‎⁡⁠⁣‏⁠‎⁡⁠⁢⁢‏‏​⁡⁠⁡‌⁤​‎‎⁡⁠⁤‏⁠‎⁡⁠⁢⁡‏‏​⁡⁠⁡‌⁢⁡​‎‎⁡⁠⁢⁣‏‏​⁡⁠⁡‌⁢⁢​‎‎⁡⁠⁣⁡‏‏​⁡⁠⁡‌⁢⁣​‎‎⁡⁠⁣⁢‏‏​⁡⁠⁡‌­ ḣ # ‎⁡Push first character of input, then all of input minus first character ≬ ↔ # ‎⁢Repeat and collect results of applying the following on the top of the stack until invariant: ‡ Ǒ # ‎⁣ Find the first index of: ⁰≠ # ‎⁤ a character that isn't the second input ⟇ # ‎⁢⁡ Remove that character from the top of the stack Ṡ # ‎⁢⁢Convert each item to a string via vectorised summation + # ‎⁢⁣Prepend the first letter to each 

C (gcc), 70 bytes

f(x,v,i)char*x,*i;{for(puts(i=x);*++i;)*i-v&&puts(x,strcpy(i--,i+1));} 

Try it online!

C (gcc), 65 bytes(32-bit) by c--

f(x,v){char*i;for(puts(i=x);*++i;)*i-v&&puts(x,strcpy(i--,i+1));} 

Try it online!

R, 91 bytes

\(s,p,`+`=utf8ToInt,w=which(+s!=+p)){for(j in w[1!=w])s=c(s,intToUtf8((+s)[F<-c(F,-j)]));s} 

Attempt This Online!

This solution outputs a vector/array of strings. Although removal of the 2nd character, not equal to the pattern argument from the string can be done with sub, I could not construct a short enough regex that would result in a solution of less than 100 bytes (it was 116 bytes).

Instead, I have applied the classic scheme, which has often proved to give the golfiest solutions: string -> charcodes -> string.