2257 - MinusOne (explanation with other's code)

[baeharam]

I couldn't solve so I studied @Chen0x00 's answer.

1. To check current string is 0-9, let's define digit type.

type Digit = '0' | '1' | '2' | '3' | '4' | '5' | '6' | '7' | '8' | '9';

2. To get number from string, we need array because can get length from array. This is the key of this problem. So, we need converter to convert digit type to array.

type ConvertDigitToArray<N extends Digit, DigitArray extends any[] = []>
  = `${DigitArray['length']}` extends N ? DigitArray : ConvertDigitToArray<N, [...DigitArray, 0]>;

0 can be any value because it just for length property.

3. To express 10, 100, 1000.. digits, we need util type to multiply 10 type element.

type Multiply10<T extends any[]> = [...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T, ...T];

4. To convert number to array, let's combine above types into single util type. In this type, we use recursive typing with template literal string. At each digit, we use Multiply10 to get proper digit and conver it to digitarray to express proper number.

type ToArray<S extends string | number, T extends any[] = []>
  = `${S}` extends `${infer F}${infer L}` ? F extends Digit ? ToArray<L, [...Multiply10<T>, ...ConvertDigitToArray<F>]> : never
  : T;

5. Finally, let's destruct number which is converted to array and compute "minus 1".

type Minus<S extends number, T extends number>
  = ToArray<S> extends [...ToArray<T>, ...infer L] ? L['length'] : 0;

type MinusOne<S extends number> = Minus<S, 1>;