Question 181244 in Javascript Question posted in
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Javascript – Count exact instances of element in array and return True or False

I have two arrays

first= [1,1,1,1,1,2,2,2,2]

second =[1,1,1,1,1]

The javascript function should return True only when all elements in second array is present in first array and if second =[1,1,1,1] or [1,1,1],[1,1],[1],[] it should return false.

Thanks in advance! Cheers

I tried this but doesn’t work for other instances.

function allElementsPresent(first, second) {
  return second.every((element) => first.includes(element));
}
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3

Answers


  1. 0

    First conditional expression should be checking length

        if(first.length != second.length) {
       //They don't have the same number of element thus we don't need to check further
        return -1
      }

    Then, try re-arranging the array in order.
    using an ordering algorithm

    Then with two loop check if each element of the same index matches each array like below

        for(int i =0;i<length;i++)
{
  if(first[i] != second[i]) {

//if they don't at this instance both array doesnot have all elements
return -1;
  }
}

    Hope this helps

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  2. 0

    I believe I understand what you want to accomplish. You want to see if the number of occurrences in the second array matches the first. If that’s the case, I’ve used this answer as a basis

    function allElementsPresent(first, second) {
  if (first.length > 0 && second.length === 0) return false;
  var counts1st = {};
  var counts2nd = {};

  for (var num of first) {
    counts1st[num] = counts1st[num] ? counts1st[num] + 1 : 1;
  }
  for (var num of second) {
    counts2nd[num] = counts2nd[num] ? counts2nd[num] + 1 : 1;
  }

  for (var count in counts2nd) {
    if (!counts1st[count] || counts2nd[count] !== counts1st[count]) return false;
  }
  return true;
}
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  3. 0

    simple JS-magic, (if I understand the problem):

    String(first).includes(second) && // second in first
/[^1,]/.test(second) // not all 1s or empty

    Sometimes dynamic types really simplify things by treating data generically; use that strength where you can.

    If the second array is not allowed to be uniform elements, but could be all non-1 digits, then it’s more complicated:

    String(first).includes(second) && // second in first
RegExp(`[^${second[0]},]`).test(second) // not all uniform or empty

    as a function:

    function allElementsPresent(first, second) {  
  return String(first).includes(second) && /[^1,]/.test(second)
}
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