Eliminate items that are not shared between two arrays using JavaScript

Question

Eliminate items that are not shared between two arrays using JavaScript

I have written a function in Javascript to eliminate non-common elements between two arrays. However, I am facing an issue where my code decreases the frequency of some items while increasing others. Here is the code snippet:

function findCommonElements(arr1, arr2) {
  let common = [];
  arr1.forEach(item => {
    arr2.forEach(element => {
      if (item == element) {
        common.push(item);
      }
    });
  });
  return common;
}
console.log(findCommonElements([1, 5, 6, 7, 5, 6, 5, 56], [1, 5, 6, 7, 8, 5, 6, 7, 8, 78]));

The output from the code above is:

[ 1, 5, 5, 6, 6, 7, 7, 5, 5, 6, 6, 5, 5 ]

Instead of the expected result:

[1, 1, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7]

After sorting

Can someone help me identify what is wrong with this code?

javascript arrays

Answer 1

Answer №1

Your current approach is to push the item every time there is a unique index match between the two arrays. For instance, when 7 matches at index 3 in both arrays, it only gets pushed once. However, when the next match occurs with index 3 (first array) and index 7 (second array), only two instances of the value 7 are pushed because there are no other index matches besides 3-3 and 3-7.

A more efficient method would be to create a Set from each array, combine them, filter out elements not present in both sets, and then sort the resulting array:

function canFormPairs(a, b) {
  const setA = new Set(a);
  const setB = new Set(b);
  return [...a, ...b]
    .filter(item => setA.has(item) && setB.has(item))
    .sort((a, b) => a - b);
}
console.log(canFormPairs([1, 5, 6, 7, 5, 6, 5, 56], [1, 5, 6, 7, 8, 5, 6, 7, 8, 78]));

Answer 2

Your current approach is to push the item every time there is a unique index match between the two arrays. For instance, when 7 matches at index 3 in both arrays, it only gets pushed once. However, when the next match occurs with index 3 (first array) and index 7 (second array), only two instances of the value 7 are pushed because there are no other index matches besides 3-3 and 3-7.

A more efficient method would be to create a Set from each array, combine them, filter out elements not present in both sets, and then sort the resulting array:

function canFormPairs(a, b) {
  const setA = new Set(a);
  const setB = new Set(b);
  return [...a, ...b]
    .filter(item => setA.has(item) && setB.has(item))
    .sort((a, b) => a - b);
}
console.log(canFormPairs([1, 5, 6, 7, 5, 6, 5, 56], [1, 5, 6, 7, 8, 5, 6, 7, 8, 78]));

Answer 3

Answer №2

Give this a shot

function pairUpSocks(clean, dirty) {
  let matched = [];
  let oneSide = clean.filter(el => dirty.includes(el));
  let otherSide = dirty.filter(el => clean.includes(el));
  matched = oneSide.concat(otherSide);
  matched.sort(function(a, b) {
    return a - b;
  });
  return matched;
}
console.log(pairUpSocks([1, 5, 6, 7, 5, 6, 5, 56], [1, 5, 6, 7, 8, 5, 6, 7, 8, 78]));

Answer 4

Give this a shot

function pairUpSocks(clean, dirty) {
  let matched = [];
  let oneSide = clean.filter(el => dirty.includes(el));
  let otherSide = dirty.filter(el => clean.includes(el));
  matched = oneSide.concat(otherSide);
  matched.sort(function(a, b) {
    return a - b;
  });
  return matched;
}
console.log(pairUpSocks([1, 5, 6, 7, 5, 6, 5, 56], [1, 5, 6, 7, 8, 5, 6, 7, 8, 78]));

Answer 5

Answer №3

Explaining the reason behind why the code provided by OP is producing inaccurate results.

The algorithm outlined in your explanation goes as follows:

Using 2 arrays, a and b.
Iterating through array a.
Iterating through array b.
If the element at index i in both arrays are equal (a[i] === b[i]), push the element from array a to the result array.

Issues identified:

The resulting array will have a total of m*n elements. For example, if 1 appears only once in both arrays, it will be included only once in the result. However, when dealing with multiple occurrences (e.g., 5, which occurs 2*3 times), the result will reflect this calculation and not just the sum of unique elements from both arrays (m+n instead).
Both input arrays may contain duplicates and are not necessarily in sorted order, leading to an unsorted final output.

Answer 6

Explaining the reason behind why the code provided by OP is producing inaccurate results.

The algorithm outlined in your explanation goes as follows:

Using 2 arrays, a and b.
Iterating through array a.
Iterating through array b.
If the element at index i in both arrays are equal (a[i] === b[i]), push the element from array a to the result array.

Issues identified:

The resulting array will have a total of m*n elements. For example, if 1 appears only once in both arrays, it will be included only once in the result. However, when dealing with multiple occurrences (e.g., 5, which occurs 2*3 times), the result will reflect this calculation and not just the sum of unique elements from both arrays (m+n instead).
Both input arrays may contain duplicates and are not necessarily in sorted order, leading to an unsorted final output.

Answer 7

Answer №4

After making some adjustments to the function, it is now operational:

function checkSockPairs(cleanSocks, dirtySocks) {
    let comparedSocks = [];
    let uniqueCleanSocks = [];

    cleanSocks.forEach((sock, index) => {
        cleanSocks.slice(index+1).forEach(otherSock => {
            if (sock === otherSock && !cleanSocks.slice(0, index).includes(sock)) {
               dirtySocks.push(otherSock);
            }
        });
    });

    uniqueCleanSocks = Array.from(new Set(cleanSocks));

    uniqueCleanSocks.forEach(cleanSock => {
        dirtySocks.forEach(dirtySock => {
            if (cleanSock === dirtySock) {
                if (!comparedSocks.includes(cleanSock)) {
                    comparedSocks.push(cleanSock);
                }
                comparedSocks.push(cleanSock);
            }
        });
   });
   
   return comparedSocks.sort();
}

console.log(checkSockPairs([1, 5, 6, 7, 5, 6, 5, 56], [1, 5, 6, 7, 8, 5, 6, 7, 8, 78]));

Answer 8

After making some adjustments to the function, it is now operational:

function checkSockPairs(cleanSocks, dirtySocks) {
    let comparedSocks = [];
    let uniqueCleanSocks = [];

    cleanSocks.forEach((sock, index) => {
        cleanSocks.slice(index+1).forEach(otherSock => {
            if (sock === otherSock && !cleanSocks.slice(0, index).includes(sock)) {
               dirtySocks.push(otherSock);
            }
        });
    });

    uniqueCleanSocks = Array.from(new Set(cleanSocks));

    uniqueCleanSocks.forEach(cleanSock => {
        dirtySocks.forEach(dirtySock => {
            if (cleanSock === dirtySock) {
                if (!comparedSocks.includes(cleanSock)) {
                    comparedSocks.push(cleanSock);
                }
                comparedSocks.push(cleanSock);
            }
        });
   });
   
   return comparedSocks.sort();
}

console.log(checkSockPairs([1, 5, 6, 7, 5, 6, 5, 56], [1, 5, 6, 7, 8, 5, 6, 7, 8, 78]));

Answer 9

Answer №5

Check out this straightforward approach that utilizes the flatMap method.

Array.prototype.comparedPairs = function(arr2) {
  var arr1 = this;
  
  let compared = [arr1, arr2].flatMap(function(x) {
    return x.filter(a => arr1.indexOf(a) !== -1 && arr2.indexOf(a) !== -1);
  }).sort((a, b) => a - b);

  return compared;
}

var result = [1, 5, 6, 7, 5, 6, 5, 56].comparedPairs([1, 5, 6, 7, 8, 5, 6, 7, 8, 78]);
console.log(result);

Answer 10

Check out this straightforward approach that utilizes the flatMap method.

Array.prototype.comparedPairs = function(arr2) {
  var arr1 = this;
  
  let compared = [arr1, arr2].flatMap(function(x) {
    return x.filter(a => arr1.indexOf(a) !== -1 && arr2.indexOf(a) !== -1);
  }).sort((a, b) => a - b);

  return compared;
}

var result = [1, 5, 6, 7, 5, 6, 5, 56].comparedPairs([1, 5, 6, 7, 8, 5, 6, 7, 8, 78]);
console.log(result);

Eliminate items that are not shared between two arrays using JavaScript

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

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