Traverse the array and exchange elements with adjacent neighbors

Question

Traverse the array and exchange elements with adjacent neighbors

I'm facing a challenge with my BIP39 passphrase which consists of 24 words for recovering my cryptocurrency wallet. Unfortunately, the passphrase does not validate due to a checksum error in the last word. All the words in the passphrase are part of the available BIP39 words list, so my initial assumption is that I may have mistakenly switched the order of two adjacent words in the phrase/array.

My approach involves iterating through each word in an array, swapping it with the neighboring word, running a function check(phrase), reverting the swap operation, and moving on to the next word.

Here's an example:

initialPhrase = 'wordOne wordTwo wordThree'
firstAttempt = 'wordTwo wordOne wordThree'
secondAttempt = 'wordOne wordThree wordTwo'

How can I implement this logic using JavaScript assuming all words are stored in an array?

javascript arrays algorithm

Answer 1

Answer №1

Below is a snippet of code that can be used:

var sentence = originalText.split(' ');

for(var index=0; index < sentence.length - 1; ++index){

sentence[index] = sentence.splice(index+1, 1, sentence[index])[0];

var finalSentence = sentence.join(' '); // your desired text

sentence = originalText.split(' ');

}

Answer 2

Below is a snippet of code that can be used:

var sentence = originalText.split(' ');

for(var index=0; index < sentence.length - 1; ++index){

sentence[index] = sentence.splice(index+1, 1, sentence[index])[0];

var finalSentence = sentence.join(' '); // your desired text

sentence = originalText.split(' ');

}

Answer 3

Answer №2

This example includes an undo feature:

Here is a JavaScript code snippet that demonstrates a custom validator and an undo functionality:

function validatePhrase(phrase) {
  //custom validation logic
  var phraseStr = phrase.join(' ');
  console.log(phraseStr);
  if (phraseStr == 'wordOne wordTwo wordFour wordThree') {
    console.log('valid!');
    return true;
  }
  return false;
}

function getValidPhraseWithUndo(phrase) {
  for (i=1; i<phrase.length; i++) {
    //swap elements
    phrase[i] = phrase.splice(i-1, 1, phrase[i])[0];

    if (validatePhrase(phrase)) {
      return phrase;
    }

    //undo the swap
    phrase[i] = phrase.splice(i-1, 1, phrase[i])[0];
  }
  return null;
}

var validatedPhrase = getValidPhraseWithUndo(['wordOne', 'wordTwo', 'wordThree', 'wordFour']);

Answer 4

This example includes an undo feature:

Here is a JavaScript code snippet that demonstrates a custom validator and an undo functionality:

function validatePhrase(phrase) {
  //custom validation logic
  var phraseStr = phrase.join(' ');
  console.log(phraseStr);
  if (phraseStr == 'wordOne wordTwo wordFour wordThree') {
    console.log('valid!');
    return true;
  }
  return false;
}

function getValidPhraseWithUndo(phrase) {
  for (i=1; i<phrase.length; i++) {
    //swap elements
    phrase[i] = phrase.splice(i-1, 1, phrase[i])[0];

    if (validatePhrase(phrase)) {
      return phrase;
    }

    //undo the swap
    phrase[i] = phrase.splice(i-1, 1, phrase[i])[0];
  }
  return null;
}

var validatedPhrase = getValidPhraseWithUndo(['wordOne', 'wordTwo', 'wordThree', 'wordFour']);

Traverse the array and exchange elements with adjacent neighbors

Answer №1

Answer №2

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