I am trying to substitute letters of the alphabet with numbers that look similar in style in typescript. I.e The letter I to 1. I have a working function named replaceLettersWithNumber which completes this with all letters, however, I am struggling to grasp and achieve a working solution where it extends from just, Hello -> H3llo -> Hell0, to finally, H3ll0.

The Function named howManyNumberLookingCharacters is called first, and the returned arrays are parameters in the bottom function. I am new to StackOverflow so please let me know if there is any more information I can provide. I feel like this is fairly straightforward, however my brain is not co-operating! Thanks.

const replacementLetters = ['O', 'I', 'E', 'A', 'T'];
const replacementNumbers = ['0', '1', '3', '4', '7'];


export function howManyNumberLookingCharacters(stringToBeRead: string): {alphabeticalCharacterPosition: number[], alphabeticalCharacter: string[]} {
let alphabeticalCharacterPosition: number[] = [];
let alphabeticalCharacter: string[] = [];

for (let x = 0; x < stringToBeRead.length; x  ) {
    for (let i = 0; i < replacementLetters.length; i  ) {
        if (stringToBeRead.toLocaleUpperCase().charAt(x) == replacementLetters[i]) {
            alphabeticalCharacterPosition.push(x);
            alphabeticalCharacter.push(replacementLetters[i])
        }
    }
}
return {alphabeticalCharacterPosition, alphabeticalCharacter};
}


export function replaceLettersWithNumber(stringToBeRead: string, alphabeticalCharacterPosition: number[], alphabeticalCharacter: string[]): string[] {

let stringArray: string[] = [];
for (let x = 0; x < alphabeticalCharacter.length; x  ) {
    var indexInArray = replacementLetters.indexOf(alphabeticalCharacter[x].toString());
    stringArray[x] = stringToBeRead.slice(0, alphabeticalCharacterPosition[x])   replacementNumbers[indexInArray]   stringToBeRead.slice(alphabeticalCharacterPosition[x]   1);
}
return stringArray;
}

CodePudding user response：

The easiest way to generate combinations is to use recursion.

At each level where a replacement is available, you call the function with both versions, like this:

const replacementNumbers = {
  'O': '0', 
  'I': '1', 
  'E': '3', 
  'A': '4', 
  'T': '7',
};

function replacementCombinations(input: string): string[] {
  const uppercase = input.toUpperCase()
  const combinations: string[] = []

  function r(str: string): void{
    if(str.length >= input.length){
      combinations.push(str)
      return
    }

    r(str   input[str.length])

    if(uppercase[str.length] in replacementNumbers)
      r(str   replacementNumbers[uppercase[str.length]])
  }
  r('')
  return combinations
}

Note that this code assumes that you're doing single character replacements.