Javascript Question posted in
A very good W3school tutorial can be found here.

Javascript – Problem with finding out if the first element of an array is different from the rest of the identical elements in that array (due to a typo)

I am attempting to solve this challenge:

This function receives an array of fruits (orchard), where all fruits are spelled correctly except for one. The function is tasked with identifying the index position of the misspelled fruit. For instance, in the given example, the function should return 0.

['graep', 'grape', 'grape', 'grape']

This is what I have so far:

function findWrongWayFruit(orchard) {
  if (orchard.length < 3) {
    return 0;
  }

  const firstFruit = orchard[0];
  for (let i = 1; i < orchard.length; i++) {
    if (firstFruit !== orchard[i]) {
      return i;
    }
  }
  return 0;
}


console.log(findWrongWayFruit(['graep', 'grape', 'grape', 'grape']));

However, when I try to run this code, the only test case that failed was this: Should return the correct index when the wrong-way fruit is at start’ (AssertionError: expected 1 to equal +0)

Tags:

4

Answers


  1. 0

    Your logic is incorrect as you assumed firstFruit to be correct and the consequent fruits are wrong if they spell differently. Instead you can use 2 counts, one for correct spelling and the other for wrong spelling. The count==1 will be the one incorrectly spelled.

    Login or Signup to reply.
  2. 0

    When the inner if condition is true, but i is 1, there are still two options: the index to return could be either 0 or 1. To know which one, you should make one more comparison that involves another value in the array (like at index 2). If that one is different from firstFruit, then you know firstFruit is the deviating value, and you should return 0.

    A quick solution is to change your inner return statement to this:

          return i == 1 && firstFruit !== orchard[2] ? 0 : i;

    function findWrongWayFruit(orchard) {
  if (orchard.length < 3) {
    return 0;
  }

  const firstFruit = orchard[0];
  for (let i = 1; i < orchard.length; i++) {
    if (firstFruit !== orchard[i]) {
      return i == 1 && firstFruit !== orchard[2] ? 0 : i;
    }
  }
  return 0;
}


console.log(findWrongWayFruit(['graep', 'grape', 'grape', 'grape']));
    Login or Signup to reply.
  3. 0 
    function findWrongWayFruit(orchard) {

  const tempFruit = orchard[0];
  for (let i = 1; i < orchard.length; i++) {
    if (tempFruit !== orchard[i] && tempFruit !== orchard[i+1]) {
      return i-1;
    }
     else{
        tempFruit = orchard[i]
    }
  }
  return 0;
}


console.log(findWrongWayFruit(['graep', 'grape', 'grape', 'grape']));
    Login or Signup to reply.
  4. 0

    You could take an array for the indices, depending on the value which are stored in the references object and assign a false index if a reference is already found.

    Finally return if two values are in the indices array and if a false index is set.

    function findWrongWayFruit(orchard) {
    if (orchard.length < 3) return;
    const 
        indices = [],
        references = {};

    let falseIndex;

    for (let i = 0; i < orchard.length; i++) {
        const value = orchard[i];
        if (value in references) falseIndex = references[value];
        else references[value] = indices.push(i) - 1;
        if (falseIndex !== undefined && indices.length === 2) return indices[1 - falseIndex];
    }
    return;
}

console.log(findWrongWayFruit(['graep', 'grape', 'grape', 'grape']));
console.log(findWrongWayFruit(['grape', 'graep', 'grape', 'grape']));
console.log(findWrongWayFruit(['grape', 'grape', 'graep', 'grape']));
console.log(findWrongWayFruit(['grape', 'grape', 'grape', 'graep']));
    Login or Signup to reply.
Please signup or login to give your own answer.

Back To Top
Search