Make a "Ceeeeeeee" program

Question

Once I wrote a JavaScript program that would take as input a string and a character and would remove all characters except for the first one and the character given as input, one by one.

For example, computing this with inputs codegolf.stackexchange.com and e for the character yields:

codegolf.stackexchange.com
cdegolf.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
ceechange.com
ceehange.com
ceeange.com
ceenge.com
ceege.com
ceee.com
ceeecom
ceeeom
ceeem
ceee

It keeps the first character and all es. All other characters are removed one by one.

Your task is to write a program (or function) that takes two inputs and outputs (or returns) a string that accomplishes this effect.

Specifications

• You can assume that the string will not contain any newlines.
• The second input will always be one character.
• If the answer is in the form of a function, you may return an array of strings containing each line in the output.
• The output can contain a trailing newline.

Test Cases

Test Cases, s:

Test Cases
Tst Cases
Ts Cases
TsCases
Tsases
Tsses
Tsss

Make a "Ceeeeeeee" program, e:

Make a "Ceeeeeeee" program
Mke a "Ceeeeeeee" program
Me a "Ceeeeeeee" program
Mea "Ceeeeeeee" program
Me "Ceeeeeeee" program
Me"Ceeeeeeee" program
MeCeeeeeeee" program
Meeeeeeeee" program
Meeeeeeeee program
Meeeeeeeeeprogram
Meeeeeeeeerogram
Meeeeeeeeeogram
Meeeeeeeeegram
Meeeeeeeeeram
Meeeeeeeeeam
Meeeeeeeeem
Meeeeeeeee

Hello World!, !:

Hello World!
Hllo World!
Hlo World!
Ho World!
H World!
HWorld!
Horld!
Hrld!
Hld!
Hd!
H!

Hello World!, z:

Hello World!
Hllo World!
Hlo World!
Ho World!
H World!
HWorld!
Horld!
Hrld!
Hld!
Hd!
H!
H

alphabet, a:

alphabet
aphabet
ahabet
aabet
aaet
aat
aa

upperCASE, e:

upperCASE
uperCASE
uerCASE
ueCASE
ueASE
ueSE
ueE
ue

This is code-golf, so the shortest code (in bytes) wins.

Answer 1

V, 12 bytes

òYpó.“„a]òd

Try it online!

Hexdump:

00000000: f259 70f3 2e93 8412 615d f264            .Yp.....a].d

I have tested this with the latest version of V available before the challenge, and everything runs correctly, making this answer competing.

Explanation:

ò         ò    " Recursively:
 Yp            "   Duplicate this line
   ó           "   Remove:
    .a]      "     A compressed regex
            d  " Delete our extra line

The compressed regex translates to

.\zs[^e]

Which means

.           " Any character
 \zs        " Leave the previous character out of the selection
    [^e]    " Any character except for 'e' (Or whatever is given for input)

Non-competing version (11 bytes)

This version uses a shortcut for Yp that wasn't available when this challenge was posted.

Answer 2

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

Answer 3

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

Answer 4

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]

Answer 5

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.

Answer 6

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"

Answer 7

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.

Answer 8

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.

Answer 9

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>')).

Answer 10

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.

Answer 11

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:

Answer 12

JavaScript (ES6), 74 bytes

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

Answer 13

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!

Answer 14

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.

Answer 15

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;.

Answer 16

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

Answer 17

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.

Answer 18

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.

Answer 19

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

Answer 20

JavaScript (ES6), 64 69

Returning a single string with newlines

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

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

Answer 21

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.

Answer 22

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.

Answer 23

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:]

Answer 24

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

Answer 25

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]

Answer 26

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
""

Answer 27

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))
        }
    }
}

Answer 28

Mathematica, 78 bytes

Damn you Martin Ender, I was almost first :p

(i=2;a={c=#};While[i<=Length@c,If[c[[i]]==#2,i++,c=c~Drop~{i};a=a~Append~c]];a)&

Unnamed function; straightforward implementation with a While loop and a few temporary variables.

Answer 29

JavaScript ES6, 89 bytes

I thought this would be an easy challenge, but I'm pretty sure I'm missing something here..

Uses recursion and returns an array of strings

(c,i=1,r)=>f=a=>a[i]?a[i++]==c?f(a):f(g=a.slice(0,i-1)+a.slice(i--),(r=r||[a]).push(g)):r

F=
  (c,i=1,r)=>f=a=>a[i]?a[i++]==c?f(a):f(g=a.slice(0,i-1)+a.slice(i--),(r=r||[a]).push(g)):r

G=_=>A.value.length>1 && (O.innerHTML = F(E.value)(A.value).join`
`)

G()

<input id=A oninput='G()' value='alphabet'>
<input id=E oninput='G()' value='a'>
<pre id=O>

Answer 30

Python 3, 67 63 142 Bytes

a,b,l1=input(),input(),0
for n in range(len(a)):
    l2=l1
    l1=''.join([a[:len(a)-n]]+[x for x in a[len(a)-n:] if x==b])
    if l2!=l1: print(l1)

It takes a as the string and b as the char, then goes through each letter in the string and adds it to the list if it's the first one or the character. Then it joins and prints.

EDIT: Fixed the problem mentioned by @R. Kap where it included copies of the first character as well, and shortened it to 63 bytes in the meantime.

EDIT 2: Now I realize you have to print out each string along the way. Brilliant. Well, this code is no longer golfed, but rather driven to the hole on a moped.

Answer 31

Groovy, 34 bytes

{a,b->a.inject{i,r->i+=r==b?b:""}}

Answer 32

JavaScript, 91 bytes

function(p,k){for(var i=1;i<p.length;)p[i]!=k?console.log(p=p.slice(0,i)+p.slice(i+1)):i++}

^{Huge thanks to ETHproductions for the help!}

Answer 33

Java 8, 104 bytes

Golfed:

(x,y)->{String q="(.["+y+"]*)",s=x;while(!x.matches(q))s+="\n"+(x=x.replaceFirst(q+".","$1"));return s;}

Ungolfed

(x, y) -> {
        String q = "(.[" + y + "]*)", s = x; // Store pattern for matching, and start our 'list' of strings.
        while (!x.matches(q)) {              // While our string doesn't fit a valid 'Ceeeeee' pattern...
            s += "\n" + (                    // We append our list of strings with a newline and...
            x = x.replaceFirst(q + ".", "$1")// A version of the string where the next character that isn't our given character is removed through regex replacement.
            );
        }
        return s;                            // We then return the newline-delimited string.
    }

Output:

Hello world!
Hllo world!
Hll world!
Hllworld!
Hllorld!
Hllrld!
Hllld!
Hlll!
Hlll

See it here!

Function is a lambda (BiFunction used in test code). Takes a String and a Character, and returns a String.

Answer 34

SmileBASIC, 65 bytes

INPUT S$,C$?S$@L
IF I&&S$[I]!+C$THEN S$[I]=""?S$ELSE I=I+1
GOTO@L

Answer 35

Excel, 52 Bytes

=LEFT(A1)&REPT(A2,LEN(A1)-LEN(SUBSTITUTE(A1,A2,"")))

where A1 and A2 are the string and the substitute letter respectively

I realise that may not actually answer the question, the wording is functions may return arrays but I'm guessing I need all the intermediate steps. So I've added

VBA, 142 137 (-4) bytes

Sub x(a, b)
For i = 2 To Len(a)
c = Mid(a, i, 1)
If c = "" Then End 'or d = 1 / Len(c) but then we quit on an error
If b<>c Then a = Replace(a, c, "", , 1): i = 1: Debug.?a
Next
End Sub

Which I think is the right byte count but I'm not sure how extra lines are treated. called in the same way as the Excel version. Returns all the steps in the Immediate Window

Answer 36

Python 3, 135 bytes

(lambda a:print(a[0])or[a.__setitem__(0,a[0][0]+a[0][1:].replace(x,'',1))or print(a[0]) for x in a[0][1:] if x!=a[1]])([input(),input()])

Answer 37

Python 3: 108 99 bytes

def g(s,o):
 r,p=s[0],print
 for i in range(1,len(s)):
  if s[i]==o:r+=s[i]
  else:p(r+s[i:])
 p(r)

Then invoke the function with:

g('alphabet', 'a')

Output:

alphabet
aphabet
ahabet
aabet
aaet
aat
aa

Try it online!

Answer 38

Python 3, 87 83 bytes

s=open(0).read()
p,c,s=s[0],s[-2],s[:-3]
for h in s:p+=s[:h==c];s=s[1:];print(p+s)

Explanation: I chose python 3 instead of python 2, because it allows to use file descriptors in open (and 0 is the descriptor of the standard input). In each loop iteration p keeps track of stored prefix. s[:h==c] works as follows: boolean expression h==c is casted into integer (0 or 1), and then the corresponding amount of characters from the beginning of s are taken.

P.S. The loop in the initial version was:

for h in s:
 if h==c:p+=h
 s=s[1:];print(p+s)

Answer 39

JavaScript (ES2015) - 86 bytes

let f=([a,...s],c,r=[])=>r[0]==a+s?r:f(a+s.replace(eval(`/[^${c}]/`),''),c,[a+s,...r])

Answer 40

C, 83 79 bytes

char x[];i;m(char*v,c){for(x[i]=*v;*v;)*++v-c?printf("%s%s\n",x,v):(x[++i]=c);}

Answer 41

sed, 76 (code) + 3 (-nr flags) = 79 bytes

N;s/(.*)\n(.).*/\2\1/
:L
h;s/.//;p;x
/^(.)(.\1*)$/q
s/^(.)(.\1*)./\1\2/
b L

Commented:

# Read in both lines, and swap the order so we can use
# back references to check for the special character
N;s/(.*)\n(.).*/\2\1/

:L

# Erase the first character (temporarily) and print the line
h;s/.//;p;x

# Terminate once only the second character in the buffer doesn't match the first
/^(.)(.\1*)$/q

# Erase the first non-matching character
s/^(.)(.\1*)./\1\2/

b L

Unfortunately, back references aren't permitted within character classes, so we're forced to repeat ourselves a bit on the last three lines. If it were possible, we could replace b L with t L, and remove /^(.)(.\1*)$/q entirely, thus dropping this down to 65 bytes.

Usage:

$ cat in
Test Cases
s
$ sed -nrf foo.sed < in
Test Cases
Tst Cases
Ts Cases
TsCases
Tsases
Tsses
Tsss

Answer 42

Python 3, 94 93 Bytes

b=list(input())
c=input()
i=1
while i<len(b):
 while b[i]==c:i+=1
 b.pop(i);print(''.join(b))

Thanks to @Challenger5 for pointing out that I had misread the question

Answer 43

Scala, 107 82 bytes

Thanks to corvus_192 for a scala code-golf discussion link (appreciated!)

Golfed

1.to(s.size-1)./:(s take 1)((p,i)=>if(s(i)!=c){println(p+s.drop(i));p}else p+s(i))

Ungolfed

(1 until s.length).foldLeft((s.take(1), List[String]())) { case ((prefix, acc), i) =>
  s.charAt(i) match {
    case char if char != c => (prefix, acc :+ (prefix + s.drop(i)))
    case char => (prefix + char, acc)
  }
}._2.foreach(println)

Using foldLeft means we can 'accumulate' the prefix, and print a new line when the next character does not equal 'c'. Short-hand for foldLeft is /:

Ungolfed, we can also accumulate the results and just print the results at the end (no side effects), plus the match statement makes it a little cleaner

Answer 44

C#, 91 bytes

s=>c=>{var r=s;for(int i=1;i<s.Length;)if(s[i++]!=c)r+="\n"+(s=s.Remove(--i,1));return r;};

Anonymous function which merges each line of output into a string and returns the result, since the function may return an array of strings instead of will return.

Full program with ungolfed method and test cases:

using System;

namespace CeeeeeeeeProgram
{
    class Program
    {
        static void Main(string[] args)
        {
            Func<string, Func<char, string>> f =
            s => c =>
            {
                var r = s;
                for (int i = 1; i < s.Length; )
                    if (s[i++] != c)
                        r += "\n" + (s = s.Remove(--i, 1));
                return r;
            };

            // test cases:
            Console.WriteLine(f("codegolf.stackexchange.com")('e'));
            Console.WriteLine(f("Test Cases")('s'));
            Console.WriteLine(f("Make a \"Ceeeeeeee\" program")('e'));
            Console.WriteLine(f("Hello World!")('!'));
            Console.WriteLine(f("Hello World!")('z'));
            Console.WriteLine(f("alphabet")('a'));
            Console.WriteLine(f("upperCASE")('e'));
        }
    }
}

Alternatively, an anonymous function with no return type which will print the output to stdout line by line will count 105 bytes:

(s,c)=>{Console.WriteLine(s);for(int i=1;i<s.Length;)if(s[i++]!=c)Console.WriteLine(s=s.Remove(--i,1));};

Answer 45

Bash + Unix utilities, 89 88 85 84 82 bytes

z=$1
for((;${#z}-n;)){
echo "$z"
n=${#z}
z=`sed "s/^\(.[$2]*\)[^$2]/\1/"<<<"$z"`
}

Try it online!

Note that you can't write just $2 instead of [$2] in the regex; if you do that, it will work most of the time, but it will fail if the second argument is a special character like * or /.

Edit: Thanks to @ckjbgames for 3 bytes.

Answer 46

Clojure, 123 bytes

(fn[[f & p]l](loop[[n & m :as r]p a f](println(str a(apply str r)))(if m(recur(if(= n l)(rest m)m)(if(= n l)(str a n)a)))))

Definitely not optimal. I tried getting rid of the two ifs in the recur, but the inclusion of a let made it bigger!

Loops over the string, popping the first char. If it equals the supplied character, it's added to the accumulator, and dropped from the remaining string.

(defn ceee [[first-char & rest-phrase] letter] ; Deconstruct the first letter off right from the parameter list

  (loop [[next-char & rest-remaining :as r] rest-phrase ; Deconstruct the remaining string apart
         acc first-char] ; Keep track of the letters to show at the start

    (println (str acc (apply str r))) ; Print the prefix accumulator, then the remaining phrase

    (if rest-remaining ; Recur if there's part of the phrase left
      (recur (if (= next-char letter)
               (rest rest-remaining)
               rest-remaining)
             (if (= next-char letter)
               (str acc next-char)
               acc)))))

Answer 47

PHP, 84 Bytes

a REGEX based answer

for($a=$argv[$c=1];$c;$a=preg_replace("#^(.+)[^$argv[2]]#","$1",$a,1,$c))echo"$a\n";

You can use the second input as REGEX for chars that are not removed

Examples

Answer 48

Japt -R, 18 16 bytes

¬ËUoÈ¥Vª!YªY>E
â

Try it online!

Unpacked & How it works

Uq mDEF{UoXYZ{X==V||!Y||Y>E
Uâ

Uq mDEF{    Split input into chars and map...
  UoXYZ{      Filter on the original input...
    X==V||      Keeping chars same as 2nd input (V),
    !Y||        1st char,
    Y>E         and indices exceeding outer index
\n          Close all braces implicitly and save the result to U
Uâ          Take unique elements from U

Turns out that trying to use regex is way too verbose here.

The -R flag joins the resulting array of strings with newline, saving three bytes <space>qR at the end. Also, this code exploits the JS feature of Array.map and Array.filter, where the index is passed as the 2nd argument (E and Y in the code above).

Answer 49

Yabasic, 102 bytes

An Anonymous function that takes input as two strings and outputs to the console.

Input""s$,v$
For i=2To Len(s$)If Mid$(s$,i,1)<>v$Then?s$:s$=Mid$(s$,1,i-1)+Mid$(s$,i+1):i=1Fi
Next
?s$

Try it online!

Answer 50

Stax, 10 bytes

Ä▬ä+QÜ├┌9◘

Run and debug it

Unpacked, ungolfed, and commented, it looks like this.

Q       print the input string without popping
1:/     split after the first character, and push both parts separately
        e.g. "H" "ello World!"
c,-     remove instances of the character from the second part
m       for each remaining character, run the rest of the program and print the result
  |-    remove the first instance of the character
  b+    copy both strings on the stack and append

Run this one

Answer 51

PowerShell, 94 89 84 79 bytes

param($s,$c)$s
for(;++$y-lt$s.length){if($s[$y]-cne$c){($s=$s.remove($y--,1))}}

Try it online!

Works by just chipping out a letter, starting at the 2nd, if it doesn't match and prints the resulting string; otherwise just increments i and this goes until it falls off. Had to use the case-sensitive -ne because PowerShell is insensitive by default. It takes a string and a char.

Answer 52

Pascal (FPC), 195 bytes

var s:string;c:char;i,j:word;begin readln(s);read(c);i:=2;while i<length(s)+1do begin if c=s[i]then i:=i+1 else begin writeln(s);for j:=i to length(s)do s[j]:=s[j+1];Dec(s[0])end;end;write(s)end.

Try it online!

String is in first line of input, character is in second line of input.

Answer 53

Assembly (MIPS, SPIM), 271 bytes

.text
.globl main
main:
lw $4 8($5)
lb $10 ($4)
lw $8 4($5) 
move $9 $8
i:
move $4 $8
li $2 4
syscall
k:
move $4 $9
addi $9 1
lb $2 ($9)
beq $2 $10 k
beq $2 $0 e
l:
lb $2 ($4)
sb $2 1($4)
addi $4 -1
bge $4 $8 l
li $4 10 
li $2 11
syscall
addi $8 1
j i
e:
li $2 10
syscall

Try it online!

I've realized when not doing syscalls, the 'a' and 'v' registers are free, which allowed me to save a few bytes. It's also a huge code smell, but this is code golf, so that's irrelevant. Takes the input via command-line arguments (I could do STDIN equivalent, but that costs more), and outputs to system out.

A detailed version that explains what all this does can be found here

Answer 54

JavaScript (ES6), 56 Bytes

F=(s,c)=>s[0]+s.slice(1).replace(eval(`/[^${c}]/g`),'');

Trey it right here:

F=(s,c)=>s[0]+s.slice(1).replace(eval(`/[^${c}]/g`),'');

function Answer() {
  O.textContent=F(S.value,C.value[0])
}
Answer()

<input id=S value='Make a "Ceeeeeeee" program' oninput='Answer()'>
<input id=C value='e' oninput='Answer()'>
<pre id=O></pre>