Question 199909 in Javascript Question posted in
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JavaScript: How to match array values with each occurrence and not with all the matches

I have two arrays:

const array1 = ["A", "S", "S", "G"]; // multiple occurrences of 'S'.
const array2 = ["S", "F", "J", "A"]; // single occurrence of 'S'.

I want to match the array2 with array1 for each of the instances. So I made a function:

const matchedArray = (array1, array2) => {
    let finalArray = [];
    array1.forEach((char, idx) => array2.includes(char) ? finalArray[idx] = `✅${char}` : finalArray[idx] = char);
    return finalArray;
}

But as you can understand that .includes() matches with all the instances of that character. So a single instance of "S" on the array2 matches with the two Ss of the array1. So the function is returning:

["✅A", "✅S", "✅S", "G"]

How can I achieve the finalArray be like:

["✅A", "✅S", "S", "G"]
Tags:

4

Answers


  1. 0

    What I might do is create a dictionary of counts of the values in array2.

    { "A": 1, "S": 1, "F": 1, "J": 1 }

    Then use a loop to process each element of array1. If the count is greater than 0, replace and decrement. Otherwise, do nothing.

    const array1 = [ "A", "S", "S", "G" ];
const array2 = [ "S", "F", "J", "A" ];
const counts = { };

array2.forEach(
   _ => { counts[ _ ] = ( counts[ _ ] || 0 ) + 1; }
);

const finalArray = array1.map(
   ( _, i ) => counts[ _ ] && counts[ _ ]-- ? `✅$_` : _
);

console.log( finalArray );

    Very efficient, even for long lists (O(N+M)).

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  2. 0

    Just count and decrement the count.

    const
    array1 = ["A", "S", "S", "G"],
    array2 = ["S", "F", "J", "A"],
    hash = array2.reduce((h, s) => {
        h[s] = (h[s] || 0) + 1;
        return h;
    }, {}),
    result = array1.map(s => hash[s] && hash[s]-- ? '✅' + s : s);

console.log(result);
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  3. 0

    You could check if your array already contains the resulting product, and then not add that.

    const array1 = ["A", "S", "S", "G"]; // multiple occurrences of 'S'.
const array2 = ["S", "F", "J", "A"]; // single occurrence of 'S'.

const matchedArray = (array1, array2) => {
    let finalArray = [];
    array1.forEach((char, idx) => array2.includes(char) && !finalArray.includes(`✅${char}`) ? finalArray[idx] = `✅${char}` : finalArray[idx] = char);
    return finalArray;
}

console.log(matchedArray(array1, array2));
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  4. 0

    You can just use a Set to store all the char’s that have already been used. And use the .has to to check, finally add the used chars using .add.

    eg.

    const array1 = ["A", "S", "S", "G"]; // multiple occurrences of 'S'.
const array2 = ["S", "F", "J", "A"]; // single occurrence of 'S'.

const matchedArray = (array1, array2) => {
  let finalArray = [];
  const used = new Set();
  array1.forEach((char, idx) => {
    array2.includes(char) && !used.has(char) ? finalArray[idx] = `✅${char}` : finalArray[idx] = char;
    used.add(char);
  });
  return finalArray;
}

console.log(matchedArray(array1,array2));
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