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Home/ Questions/Q 8866919
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Editorial Team
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Editorial Team
Asked: June 14, 20262026-06-14T16:53:56+00:00 2026-06-14T16:53:56+00:00

I’ve been trying different ways for a while and I’ve reached that point where

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I’ve been trying different ways for a while and I’ve reached that point where whatever I do just gets wrong.

So this is what I’ve tried to do. First I create a random number and add it to an array:

for(.......){
  random[i] = Math.floor(Math.random() * (word.length));

Then another random number is added to the array:

randomkey[i] = Math.floor(Math.random() * (word.length));

Then I create this function:

var accept=true;
// word is a text (length:22)
function acceptRandom(random,word){
  stop:
    for(var i=0;i<randomkey.length+1; i++){
      if(word[i] != word[random])
        accept = true;
      else {
        accept = false;
        break stop;
      }
    }
    if(accept == false){
      newrandom = Math.floor(Math.random() * (word.length));
      // Random number exists so we call the function again with a new value
      acceptRandom(newrandom,word);
    } else
      return random;
}

Now, the problem is that it won’t return the new value when the random number already exists.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-14T16:53:57+00:00Added an answer on June 14, 2026 at 4:53 pm

    Since you’re iterating through the whole list, there will always be a point where word[i] == word[random] (because you’ve comparing the word to itself.) A quick fix would be:

    for(var i=0;i<randomkey.length+1; i++){
        if(word[i] == word[random] && i !== random) {
            accept = false;
            break;
        }
    }
    

    You’ll also need to return your recursive call:

    if(accept == false){
          newrandom = Math.floor(Math.random() * (word.length));
          // Random number exists so we call the function again with a new value
          return acceptRandom(newrandom,word);
    }
    

    Honestly, I think you’ve run into the XY problem. What exactly are you trying to do here? There’s probably a better way.

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