Check if a specific number appears exactly once in an array and output either True or False

Question

Check if a specific number appears exactly once in an array and output either True or False

I am facing a challenge with comparing two arrays

Array1 = [1,1,1,2,2,2,3,3] 
Array2 =[1,1,2,1]

When comparing both arrays, the desired result is True if the number of occurrences of Integer 1 are the same.


Array2 = [1,1,2]  //Expecting False

For the above case, the expected result is false because the occurrences of 1 and 2 in Array2 do not match with the occurrences in Array1.

Array2 = [1,1,2,3,1] //Expecting True.

I have attempted a solution but it is not effective for all instances.

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

Any help or suggestions are greatly appreciated. Thank you in advance!

javascript arrays loops

Answer 1

Answer №1

I think I grasp the concept you're aiming to achieve. Your goal is to compare the occurrence of elements in the second array with those in the first array. To address this, I have referenced this particular answer.

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;
}

Answer 2

I think I grasp the concept you're aiming to achieve. Your goal is to compare the occurrence of elements in the second array with those in the first array. To address this, I have referenced this particular answer.

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;
}

Answer 3

Answer №2

Here is a simple solution using JavaScript:

if (String(first).indexOf(second) !== -1 && /^[^1,]+$/.test(second)) {
    // Perform some actions
}

Dynamic types in programming can be quite helpful in simplifying tasks, especially when handling data in a generic manner. It's always good to leverage this flexibility to your advantage.

If the requirement is for the second array to have non-uniform elements (excluding only 1s), the solution becomes a bit more intricate:

if (String(first).indexOf(second) !== -1 && new RegExp('[^' + second[0] + ',]').test(second)) {
    // Perform some actions
}

This logic can also be encapsulated in a function for reusability:

function checkElementsPresence(first, second) {
    return String(first).indexOf(second) !== -1 && /^[^1,]+$/.test(second);
}

Answer 4

Here is a simple solution using JavaScript:

if (String(first).indexOf(second) !== -1 && /^[^1,]+$/.test(second)) {
    // Perform some actions
}

Dynamic types in programming can be quite helpful in simplifying tasks, especially when handling data in a generic manner. It's always good to leverage this flexibility to your advantage.

If the requirement is for the second array to have non-uniform elements (excluding only 1s), the solution becomes a bit more intricate:

if (String(first).indexOf(second) !== -1 && new RegExp('[^' + second[0] + ',]').test(second)) {
    // Perform some actions
}

This logic can also be encapsulated in a function for reusability:

function checkElementsPresence(first, second) {
    return String(first).indexOf(second) !== -1 && /^[^1,]+$/.test(second);
}

Answer 5

Answer №3

Make sure to first check the length of the array

if (first.length != second.length) {
  // If the number of elements is different, there is no need to proceed further
  return -1
}

Next, consider rearranging the array in a specific order using an algorithm

Then, compare each element of the arrays at the same index using two loops as shown below

for(int i =0;i<length;i++)
{
  if (first[i] != second[i]) {
    // If any elements don't match at this stage, the arrays are not equal
    return -1;
  }
}

I hope this explanation is clear

Answer 6

Make sure to first check the length of the array

if (first.length != second.length) {
  // If the number of elements is different, there is no need to proceed further
  return -1
}

Next, consider rearranging the array in a specific order using an algorithm

Then, compare each element of the arrays at the same index using two loops as shown below

for(int i =0;i<length;i++)
{
  if (first[i] != second[i]) {
    // If any elements don't match at this stage, the arrays are not equal
    return -1;
  }
}

I hope this explanation is clear

Check if a specific number appears exactly once in an array and output either True or False

Answer №1

Answer №2

Answer №3

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