00119-medium-replaceall

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type ReplaceAll<
  S extends string,
  From extends string,
  To extends string
> = From extends ""
  ? S
  : S extends `${infer F}${From}${infer U}`
  ? `${F}${To}${ReplaceAll<U, From, To>}`
  : S;

Solution by LennyLee1998 #35196 

type ReplaceAll<S extends string, From extends string, To extends string, ACC extends string = ''> = From extends '' ? S
  : S extends `${infer Left}${From}${infer Right}` ? ReplaceAll<Right, From, To, `${ACC}${Left}${To}`>
  : `${ACC}${S}`

Solution by eunsukimme #35139 

type ReplaceAll<Str, From extends string, To extends string> = 
Str extends `${infer R1}${From}${infer R2}` ? `${ReplaceAll<R1, From, To>}${To}${ReplaceAll<R2, From, To>}` : Str;

Solution by kyzinatra #35028 

// your answers
type ReplaceAll<S extends string, From extends string, To extends string> = From extends '' ? S : S extends `${infer Start}${From}${infer End}` ? `${Start}${To}${ReplaceAll<End, From, To>}` : S;

Solution by LeeKangHyun #35010

We can solve easily with Replace type and recursive left and right types.

// your answers
type ReplaceAll<S extends string, From extends string, To extends string> = From extends '' ? S : S extends `${infer L}${From}${infer R}` ? `${ReplaceAll<L, From, To>}${To}${ReplaceAll<R, From, To>}` : S;

Solution by dev-jaemin #35005 

type ReplaceAll<S extends string, From extends string, To extends string> =  From extends '' ? S : S extends `${infer First}${From}${infer Rest}` ? `${First}${To}${ReplaceAll<Rest,From,To>}` : S

Solution by devshinthant #34575 

type ReplaceAll<S extends string, From extends string, To extends string> = From extends '' ? S : S extends `${infer prefix}${From}${infer suffix}` ? `${prefix}${To}${ReplaceAll<suffix, From, To>}` : S

Solution by ouzexi #33998 

type ReplaceAll<S extends string, From extends string, To extends string, K extends string = ''> = S extends `${infer L}${From extends '' ? never : From}${infer R}`
? ReplaceAll<R, From, To,`${K}${L}${To}`>
: `${K}${S}`

Solution by IamSmaLLR #33920 

// 你的答案
type ReplaceAll<
  S extends string,
  From extends string,
  To extends string
> = From extends ""
  ? S
  : S extends `${infer V}${From}${infer R}`
  ? `${V}${To}${ReplaceAll<`${R}`, From, To>}`
  : S;

Solution by shawnliu200058 #33917 

type ReplaceAll<S extends string, From extends string, To extends string> = From extends '' 
? S 
: S extends `${infer F}${From}${infer R}`
// 下面这样会递归替换，但是不需要处理已经替换过的字符串，也就是只需要递归处理右侧剩余的
// ? ReplaceAll<`${F}${To}${R}`, From, To> 
? `${F}${To}${ReplaceAll<R, From, To>}`
: S

Solution by ScriptBloom #33896 

type ReplaceAll<S extends string, T extends string, P extends string> =
  T extends ''
    ? S
    : S extends `${infer L}${T}${infer R}`
      ? `${ReplaceAll<L, T, P>}${P}${ReplaceAll<R, T, P>}`
      : S

Solution by Danny101201 #33849

119 - ReplaceAll

by Anthony Fu (@antfu) #medium #template-literal

Question

Implement ReplaceAll<S, From, To> which replace the all the substring From with To in the given string S

For example

type replaced = ReplaceAll<'t y p e s', ' ', ''> // expected to be 'types'

View on GitHub: https://tsch.js.org/119

...

type ReplaceAll<S extends string, From extends string, To extends string> = From extends ``
  ? S
  : S extends `${infer L}${From}${infer R}`
    ? `${L}${To}${ReplaceAll<R, From, To>}`
    : S

Solution by veralex #33782

type ReplaceAll<S extends string, From extends string, To extends string> = S extends ${infer F}${From}${infer E}? ${F}${To}${ReplaceAll<E, From, To>}:S

Solution by rookiewxy #33688 

type ReplaceAll<
  S extends string,
  From extends string,
  To extends string
> = From extends ""
  ? S
  : S extends `${infer Start}${From}${infer End}`
  ? `${Start}${To}${ReplaceAll<End, From, To>}`
  : S;

Solution by fyuanz #33685 

type ReplaceAll<S extends string, From extends string, To extends string> = From extends '' 
? S : S extends `${infer left}${From}${infer right}` 
? `${ReplaceAll<left,From,To>}${To}${ReplaceAll<right,From,To>}` : S

Solution by loevray #33349 

type ReplaceAll<S extends string, From extends string, To extends string> = From extends '' ? S : S extends `${infer F}${From}${infer L}` ? `${F}${To}${ReplaceAll<L, From, To>}` : S

Solution by Tubring25 #33313 

// 你的答案
type ReplaceAll<T extends string, U  , V extends string> = T extends `${infer A}${infer B}`
  ? A extends U
    ? `${V}${ReplaceAll<`${B}`,U,V>}`
    : `${A}${ReplaceAll<`${B}`,U,V>}`
  : T;

Solution by 2531800823 #33251 

// your answers

type ReplaceAll<S extends string, From extends string, To extends string> = 
  From extends ''
    ? S
    : S extends `${infer F}${From}${infer R}`
      ? `${F}${To}${ReplaceAll<R, From, To>}`
      : S

Solution by KeithChou #33196 

type ReplaceAll<S extends string, From extends string, To extends string> = From extends ""
  ? S
  : S extends `${infer L}${From}${infer R}`
    ? `${ReplaceAll<L, From, To>}${To}${ReplaceAll<R, From, To>}`
    : S

Solution by Taneros #33152 

// 你的答案
type ReplaceAll<S extends string, From extends string, To extends string> = From extends ''?S
:S extends `${infer L}${From}${infer R}` ? `${L}${To}${ReplaceAll<R,From,To>}`
:S

Solution by walker-hzx #32938 

type ReplaceAll<S extends string, From extends string, To extends string> = 
  From extends ''
    ? S
    : S extends `${infer H}${From}${infer T}`
        ? T extends ""
            ? `${H}${To}`
            : `${H}${To}${ReplaceAll<T, From, To>}`
        : S```

Solution by keyurparalkar #32411 

type ReplaceAll<S extends string, From extends string, To extends string> = 
  From extends '' 
    ? S
    : S extends `${infer Head}${From}${infer Tail}`
      ? `${Head}${To}${ReplaceAll<Tail, From, To>}`
      : S

Solution by dev-hobin #32376 

type ReplaceAll<S extends string, From extends string, To extends string> =
  S extends `${infer A}${From extends '' ? never : From}${infer B}`
    ? `${A}${To}${ReplaceAll<B, From, To>}`
    : S

Solution by nivenice #32372 

type ReplaceAll<
  S extends string,
  From extends string,
  To extends string,
> = S extends `${infer L}${From}${infer R}`
  ? S
  : S;

type ReplaceAll<
  S extends string,
  From extends string,
  To extends string,
> = S extends `${infer L}${From}${infer R}`
  ?  `${L}${To}${R}`
  : S;

type ReplaceAll<
  S extends string,
  From extends string,
  To extends string,
> = S extends `${infer L}${From}${infer R}`
  ?  `${L}${From extends "" ? "" : To}${R}`
  : S;

type ReplaceAll<
  S extends string,
  From extends string,
  To extends string,
> = S extends `${infer L}${From}${infer R}`
  ? `${L}${From extends "" ? "" : To}${ReplaceAll<`${R}`, From, To>}`
  : S;

Solution by awkmia #32236 

type ReplaceAll<S extends string, From extends string, To extends string> =
  S extends `${infer F}${From extends '' ? never : From}${infer L}`
  ? `${F}${To}${ReplaceAll<L, From, To>}`
  : S

We infer the first appear of From type, replace it with To and concatenate the string from the beginning to from with recursively ReplaceAll of Tail

Solution by joyanedel #32155 

type ReplaceAll<S extends string, From extends string, To extends string, Acc extends string = ""> = From extends ""
	? S
	: S extends `${Acc}${infer F}${From}${infer L}`
	? L extends ""
		? `${Acc}${F}${To}`
		: ReplaceAll<`${Acc}${F}${To}${L}`, From, To, `${Acc}${F}${To}`>
	: S;

Solution by gasmg #32060 

// your answers
type ReplaceAll<S extends string, From extends string, To extends string> = From extends '' ? S : S extends `${infer F}${From}${infer R}` ? `${F}${To}${ReplaceAll<R, From, To>}` : S;

Solution by pea-sys #32020 

type ReplaceAll<S extends string, From extends string, To extends string> = 
  From extends '' ? S : 
    S extends `${infer First}${From}${infer Rest}` ? 
      `${First}${To}${ReplaceAll<`${Rest}`, From, To>}` : 
        S

Solution by jinyoung234 #31702 

type ReplaceAll<S extends string, From extends string, To extends string> = S extends `${infer F}${From extends '' ? never : From}${infer R}` 
  ? `${ReplaceAll<F, From, To>}${To}${ReplaceAll<R, From, To>}`
  : S

Solution by kai-phan #31629 

// your answers

type ReplaceAll<
  S extends string,
  From extends string,
  To extends string
> = S extends `${infer L}${From}${infer R}` ? `${L}${To}${ReplaceAll<R, From, To>}` : S;

Solution by AhmedRagabKamal #31509