Bir dize Sve bir dizin listesi verildiğinde X, bu sonucu yeni değeri olarak kullanırken Sher dizindeki öğeyi kaldırarak değiştirin .SS

Örneğin, verilen S = 'codegolf've X = [1, 4, 4, 0, 2],

0 1 2 3 4 5 6 7  |
c o d e g o l f  |  Remove 1
c d e g o l f    |  Remove 4
c d e g l f      |  Remove 4
c d e g f        |  Remove 0
d e g f          |  Remove 2
d e f

Göreviniz bu işlemi gerçekleştirmek, Sher işlemden sonra değerlerini toplamak ve her birini yeni bir satırda sırayla görüntülemek. Son cevap

S = 'codegolf'
X = [1, 4, 4, 0, 2]

Answer:

codegolf
cdegolf
cdeglf
cdegf
degf
def

• Bu kod golf bu yüzden kodunuzu mümkün olduğunca kısa yapın.
• İçindeki değerlerin Xher zaman için geçerli endeksler olduğunu varsayabilir Sve 0 tabanlı veya 1 tabanlı dizinleme kullanabilirsiniz.
• Dize yalnızca [A-Za-z0-9]
• Ya Sya xgöre boş olabilir. Eğer Sboş, öyle izler xde boş olması gerekir.
• Ayrıca S, dize yerine karakterlerin bir listesini de alabilirsiniz .
• Çıktıyı yazdırabilir veya bir dize listesi döndürebilirsiniz. Öncü ve sondaki boşluk kabul edilebilir. Kolayca okunabildiği sürece herhangi bir çıktı şekli iyidir.

Test Durumları

S = 'abc', x = [0]
'abc'
'bc'

S = 'abc', x = []
'abc'

S = 'abc', x = [2, 0, 0]
'abc'
'ab'
'b'
''

S = '', x = []
''

S = 'codegolfing', x = [10, 9, 8, 3, 2, 1, 0]
'codegolfing'
'codegolfin'
'codegolfi'
'codegolf'
'codgolf'
'cogolf'
'cgolf'
'golf'

Haskell, 38 33 bayt

s#i=take i s++drop(i+1)s
scanl(#)

Düz ileri: elemanları indeks i'den önce ve sonra tekrar tekrar alın, tekrar birleştirin ve sonuçları toplayın.

Çevrimiçi deneyin!

Düzenleme: @Lynn 5 bayt kaydetti. Teşekkürler!

JavaScript (ES6), 57 50 48 45 42 bayt

Dizeyi tek tek karakter dizisi olarak alır, orijinalin virgülle ayrılmış dizesini ve ardından her adım için virgülle ayrılmış dizelerin alt dizisini içeren bir dizi çıkarır.

s=>a=>[s+"",a.map(x=>s.splice(x,1)&&s+"")]

• Arnauld sayesinde 3 bayt tasarruf etti , gevşek çıkış özelliğini zaten olduğumdan daha fazla kötüye kullandığımı düşündürdü, bu da beni 3 baytlık bir tasarruf için daha da kötüye kullanmamı sağladı.

Dene

o.innerText=JSON.stringify((f=

s=>a=>[s+"",a.map(x=>s.splice(x,1)&&s+"")]

)([...i.value="codegolf"])(j.value=[1,4,4,0,2]));oninput=_=>o.innerText=JSON.stringify(f([...i.value])(j.value.split`,`))

label,input{font-family:sans-serif;font-size:14px;height:20px;line-height:20px;vertical-align:middle}input{margin:0 5px 0 0;width:100px;}

<label for=i>String: </label><input id=i><label for=j>Indices: </label><input id=j><pre id=o>

açıklama

İki girdiyi curry sözdiziminde parametreler s(dize dizisi) ve a(tamsayı dizisi) aracılığıyla alırız, yani işlev ile çağırırız f(s)(a).

Yeni bir dizi oluşturuyoruz ve orijinal ile başlıyoruz s. Bununla birlikte, splicedaha sonra kullanacağımız yöntem bir diziyi değiştirdiğinden, bir dizesini dönüştürerek bir dizeye dönüştürmemiz gerekir (sadece boş bir dize ekleyin).

Alt mapdiziyi oluşturmak için , tamsayı dizisi üzerinde a( xgeçerli tamsayı nerede ) ve her öğe için, dizinden başlayarak splice1 öğeden sbaşlıyoruz x. Değiştirilmiş solanı döndürüyoruz , yine bir dizeye dönüştürerek bir kopyasını yapıyoruz.

Japt , 6 bayt

åjV uV

Çevrimiçi test edin!

açıklama

UåjV uV   Implicit: U = array of integers, V = string
Uå        Cumulatively reduce U by
  j         removing the item at that index in the previous value,
   V        with an initial value of V.
     uV   Push V to the beginning of the result.

Alternatif:

uQ åjV

UuQ       Push a quotation mark to the beginning of U.
    å     Cumulatively reduce by
     j      removing the item at that index in the previous value,
      V     with an initial value of V.

Bu, dizindeki öğeyi kaldırmanın "hiçbir şey yapmaması ve bu nedenle orijinal dizeyi döndürmesi nedeniyle çalışır .

Kabuk , 7 bayt

G§+oh↑↓

Önce dizeyi alır, sonra (1 tabanlı) endeksler. Çevrimiçi deneyin!

açıklama

G§+oh↑↓
G        Scan from left by function:
           Arguments are string, say s = "abcde", and index, say i = 3.
      ↓    Drop i elements: "de"
     ↑     Take i elements
   oh      and drop the last one: "ab"
 §+        Concatenate: "abde"
         Implicitly print list of strings on separate lines.

Python 2 , 43 bayt

s,i=input()
for i in i+[0]:print s;s.pop(i)

Çevrimiçi deneyin!

Kolayca okunabildiği sürece herhangi bir çıktı şekli iyidir.

Bu, karakter listeleri olarak yazdırılır.

Python 2 , 47 bayt

This could be shortened to 43 bytes, as @LuisMendo pointed out, but that's already @ErktheOutgolfer's solution.

a,b=input();print a
for i in b:a.pop(i);print a

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Java 8, 78 bytes

This is a curried lambda, from int[] to a consumer of StringBuilder or StringBuffer. Output is printed to standard out.

l->s->{System.out.println(s);for(int i:l)System.out.println(s.delete(i,i+1));}

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05AB1E, 11 bytes

v=ā0m0yǝÏ},

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v           # For each index:
 =          #   Print without popping
  ā         #   Push range(1, len(a) + 1)
   0m       #   Raise each to the power of 0. 
            #   This gives a list of equal length containing all 1s
     0yǝ    #   Put a 0 at the location that we want to remove
        Ï   #   Keep only the characters that correspond to a 1 in the new list
         }, # Print the last step

Mathematica, 70 bytes

(s=#;For[t=1,t<=Length@#2,Print@s;s=StringDrop[s,{#2[[t]]+1}];t++];s)&

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R, 46 32 bytes

function(S,X)Reduce(`[`,-X,S,,T)

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Takes input as a list of characters and X is 1-based. Reduce is the R equivalent of fold, the function in this case is [ which is subset. Iterates over -X because negative indexing in R removes the element, and init is set to S, with accum=TRUE so we accumulate the intermediate results.

R, 80 bytes

function(S,X,g=substring)Reduce(function(s,i)paste0(g(s,0,i-1),g(s,i+1)),X,S,,T)

2-argument function, takes X 1-indexed. Takes S as a string.

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Haskell, 33 bytes

scanl(\s i->take i s++drop(i+1)s)

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PowerShell, 94 84 bytes

param($s,$x)$a=[Collections.Generic.list[char]]$s;$x|%{-join$a;,$a|% r*t $_};-join$a

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Takes input $s as a string and $x as an explicit array. We then create $a based on $s as a list.

Arrays in PowerShell are fixed size (for our purposes here), so we need to use the lengthy [System.Collections.Generic.list] type in order to get access to the .removeAt() function, which does exactly what it says on the tin.

I sacrificed 10 bytes to include two -join statements to make the output pretty. OP has stated that outputting a list of chars is fine, so I could output just $a rather than -join$a, but that's really ugly in my opinion.

Saved 10 bytes thanks to briantist.

Python 2, 50 bytes

• Takes in string and list of indices

s,i=input()
for j in i+[0]:print s;s=s[:j]+s[j+1:]

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05AB1E, 9 7 bytes

=svõyǝ=

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=s        # Print original string, swap with indices.
  v       # Loop through indices...
   õyǝ    # Replace current index with empty string.

-2 thanks to idea from @ETHProductions.

Retina, 58 bytes

¶\d+
¶¶1$&$*
+1`(?=.*¶¶.(.)*)(((?<-1>.)*).(.*)¶)¶.*
$2$3$4

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¶\d+

Match the indices (which are never on the first line, so are always preceded by a newline).

¶¶1$&$*

Double-space the indices, convert to unary, and add 1 (because zeros are hard in Retina).

+1`

Repeatedly change the first match, which is always the current value of the string.

   (?=.*¶¶.(.)*)

Retrieve the next index in $#1.

                (           .    ¶)

Capture the string, including the $#1th character and one newline.

                 ((?<-1>.)*) (.*)

Separately capture the prefix and suffix of the $#1th character of the string.

                                   ¶.*

Match the index.

$2$3$4

Replace the string with itself and the index with the prefix and suffix of the $#1th character.

Pyth, 8 bytes

.u.DNYEz

Demonstration

Reduce, starting with the string and iterating over the list of indices, on the deletion function.

PowerShell, 54 58 bytes

param($s,$x),-1+$x|%{$z=$_;$i=0;-join($s=$s|?{$z-ne$i++})}

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Explanation

Takes input as a char array ([char[]]).

Iterates through the array of indices ($x) plus an injected first element of -1, then for each one, assigns the current element to $z, initializes $i to 0, then iterates through the array of characters ($s), returning a new array of only the characters whose index ($i) does not equal (-ne) the current index to exclude ($z). This new array is assigned back to $s, while simultaneously being returned (this happens when the assignment is done in parentheses). That returned result is -joined to form a string which is sent out to the pipeline.

Injecting -1 at the beginning ensures that the original string will be printed, since it's the first element and an index will never match -1.

q/kdb+, 27 10 bytes

Solution:

{x _\0N,y}

Examples:

q){x _\0N,y}["codegolf";1 4 4 0 2]
"codegolf"
"cdegolf"
"cdeglf"
"cdegf"
"degf"
"def"
q){x _\0N,y}["abc";0]
"abc"
"bc"
q){x _\0N,y}["abc";()]
"abc"
q){x _\0N,y}["abc";2 0 0]
"abc"
"ab"
,"b"
""    
q){x _\0N,y}["";()]
""

Explanation:

Takes advantage of the converge functionality \ as well as drop _.

{x _\0N,y}
{        } / lambda function, x and y are implicit variables
     0N,y  / join null to the front of list (y), drop null does nothing
   _\      / drop over (until result converges) printing each stage
 x         / the string (need the space as x_ could be a variable name)

Notes:

If we didn't need to print the original result, this would be 2 bytes in q:

q)_\["codegolfing";10 9 8 3 2 1 0]
"codegolfin"
"codegolfi"
"codegolf"
"codgolf"
"cogolf"
"cgolf"
"golf"

Perl 5, 55 bytes (54 + "-l")

sub{print($s=shift);for(@_){substr$s,$_,1,"";print$s}}

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MATL, 8 bytes

ii"t[]@(

Indexing is 1-based.

Try it online! Or verify the test cases.

Explanation

i      % Input string. Input has to be done explicitly so that the string
       % will be displayed even if the row vector of indices is empty
i      % Input row vector of indices
"      % For each
  t    %   Duplicate current string
  []   %   Push empty array
  @    %   Push current index
  (    %   Assignment indexing: write [] to string at specified index
       % End (implicit). Display stack (implicit)

C# (.NET Core), 87 87 74 70 bytes

S=>I=>{for(int i=0;;S=S.Remove(I[i++],1))System.Console.WriteLine(S);}

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Just goes to show that recursion isn't always the best solution. This is actually shorter than my original invalid answer. Still prints to STDOUT rather than returning, which is necessary because it ends with an error.

-4 bytes thanks to TheLethalCoder

C (gcc), 99 bytes

j;f(s,a,i)char*s;int*a;{puts(s);for(j=0;j<i;j++){memmove(s+a[j],s+a[j]+1,strlen(s)-a[j]);puts(s);}}

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Takes the string, array, and the length of the array.

Pyth, 10 bytes

Rule changes saved me 1 byte:

V+E0Q .(QN

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Pyth, 11 bytes

V+E0sQ .(QN

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Gaia, 9 bytes

+⟪Seḥ+⟫⊣ṣ

^{I should really add a "delete at index" function...}

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Explanation

+          Add the string to the list
 ⟪Seḥ+⟫⊣   Cumulatively reduce by this block:
  S         Split around index n
   e        Dump the list
    ḥ       Remove the first char of the second part
     +      Concatenate back together
        ṣ  Join the result with newlines

V, 12 bytes

òdt,GÙ@-|xHx

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This is 1-indexed, input is like:

11,10,9,4,3,2,1,
codegolfing

Explanation

ò              ' <M-r>ecursively until error
 dt,           ' (d)elete (t)o the next , (errors when no more commas)
    G          ' (G)oto the last line
     Ù         ' Duplicate it down
        |      ' Goto column ...
      @-       ' (deleted number from the short register)
         x     ' And delete the character there
          H    ' Go back home
           x   ' And delete the comma that I missed

Swift 3, 80 bytes

func f(l:[String],c:[Int]){var t=l;for i in c{print(t);t.remove(at:i)};print(t)}

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Pyth, 8 bytes

+zm=z.Dz

Test suite!

explanation

+zm=z.DzdQ    # implicit: input and iteration variable
  m      Q    # for each of the elements of the first input (the array of numbers, Q)
     .Dzd     # remove that index from the second input (the string, z)
   =z         # Store that new value in z
+z            # prepend the starting value

Python 2, 54

X,S=input()
for a in X:
    print S
    S=S[:a]+S[a+1:]
print S

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APL, 31 30 28 bytes

{⎕←⍺⋄⍵≡⍬:⍬⋄⍺[(⍳⍴⍺)~⊃⍵]∇1↓⍵}

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C# (Mono), 85 bytes

s=>a=>{for(int i=-2;++i<a.Count;)System.Console.WriteLine(s=i<0?s:s.Remove(a[i],1));}

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