Count exact occurrences of a certain number in array and return either True or False

Question

I have two arrays Arr1 = [1,1,1,2,2,2,3,3] Arr2 =[1,1,2,1]

Comparing both arrays should return True as there are same occurrences of no. 1.

However If Arr2 = [1,1,2] it should return false as the no. Of occurrences of 1 or 2 don't match with the no. Of occurrences of 1 and 2 in Arr1

Even Arr2 = [1,1,2,3,1] should return True.

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));
}

You mean the number of occurrences must match? Please make the criteria for returning true clear. — Unmitigated, Jun 03 '23 at 04:14
@MisterJojo actually I push the second array on click of button.. So there can be another array containing 2's or 3's.. But the first array Is already present with Me.. So just comparing the occurrences if they match or not. — Prathamesh Koyande, Jun 03 '23 at 05:09

milklizard · Accepted Answer · 2023-06-04T21:48:35.677

1

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, matchAll = false) {
  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 (matchAll && (!counts1st[count] || counts1st[count] !== counts2nd[count])) return false;
    if (!matchAll && (counts1st[count] && counts1st[count] === counts2nd[count])) return true;
  }
  return matchAll ? true : false;
}

edited Jun 04 '23 at 21:48

answered Jun 03 '23 at 04:38

milklizard

228
1
7

This is how I wanted! Thankss.. Although for blank array it return True.. – Prathamesh Koyande Jun 03 '23 at 05:31
1

Updated: now it will only return true if they're both blank. – milklizard Jun 03 '23 at 05:36
Great.. Thanks a ton!! – Prathamesh Koyande Jun 03 '23 at 06:17
Hi.. Just another doubt if the second array contains 1,1,2,1,1,1..it should still return True as 5 1s are there but it's returning false – Prathamesh Koyande Jun 03 '23 at 09:06
It returns false because the number of **2**s don't match. Are you only concerned with **1**s? – milklizard Jun 03 '23 at 12:17
No I am concerned with occurrences.. If 1 has 5 occurrences in any order it should return True same goes with the 2.. If there are 4 2s then True else false.. Like [1,1,2,1,1,1] should return True and also [1,1,2,1,2,1, 1,2,2] should return True rest cases false – Prathamesh Koyande Jun 03 '23 at 13:57
Do you want it to return true if **any** occurrence count matches, not necessarily all of them? – milklizard Jun 03 '23 at 22:26
Yeah right.. It should check for occurrences simultaneously – Prathamesh Koyande Jun 04 '23 at 02:04
I've added a parameter that makes checking all occurrences optional. If I understand correctly, the default `false` behavior is the one you want. – milklizard Jun 04 '23 at 15:01
For which case would it return false?Actually it's returning True even if the second array has 1 occurrence of 1 – Prathamesh Koyande Jun 04 '23 at 17:49
My mistake, fixed an operator on the `!matchAll` line: `||` changed to `&&` – milklizard Jun 04 '23 at 21:50
Works really well thanks for the code! – Prathamesh Koyande Jun 05 '23 at 02:55

score 0 · Answer 2 · edited Jun 03 '23 at 07:56

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 does not have all elements
    return -1;
  }
}

Hope this helps

dandavis · Answer 3 · 2023-06-03T05:20:06.503

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)
}

edited Jun 03 '23 at 05:20

answered Jun 03 '23 at 05:00

dandavis

16,370
5
40
36

Could you please provide a snippet of sorts.. – Prathamesh Koyande Jun 03 '23 at 05:17
sure, all you have to do is wrap the function boilerplate around it. – dandavis Jun 03 '23 at 05:21

Count exact occurrences of a certain number in array and return either True or False

3 Answers3