What is the best approach to gather both the intersection and the complements of two arrays in a single operation?

Question

What is the best approach to gather both the intersection and the complements of two arrays in a single operation?

Comparing Arrays -

const source = [1, 1, 1, 3, 4];
const target = [1, 2, 4, 5, 6];

Initializing Arrays -

const matches1 = [];
const matches2 = [];
const unmatches1 = [];

Array Matching Process -

for (const line of source) {
  const line2Match = target.find((line2) => {
    const isMatch = line === line2;
    return isMatch;
  });

  if (line2Match) {
    matches1.push(
      line
    );
    matches2.push(line2Match);
  } else {
    unmatches1.push(line);
  }
}

Current Output -

[ 1, 1, 1, 4 ] found in matches1

[ 3 ] added to unmatches1

[ 1, 1, 1, 4 ] at matches2

Desired Output -

[ 1, 4 ] should be in matches1

[ 1, 1, 3 ] should be in unmatches1

[ 1, 4 ] should be in matches2

Additional Functionality Request -

If a match occurs between source and target, the matched value should be removed from the target array. How can this be achieved?

javascript arrays intersection array-difference complement

Answer 1

Answer №1

The original poster raised a question about handling matching values between two arrays.

"What I would like to add is when source has a match with target, the value will be deleted from the target array, how can I achieve that?"

One approach is to manually remove the common item from the target array, essentially creating the relative complement of source in target, also known as the target difference.

Another method involves making a shallow copy of the target array and using it as an initial value for a reduce operation to avoid modifying the original target reference.

function collectIntersectionAndComplements(collector, sourceItem) {
  const { targetDiff } = collector;

  const targetIndex = targetDiff
    .findIndex(targetItem => targetItem === sourceItem);

  if (targetIndex === -1) {

    // collect the relative complement of target
    // in source ... vulgo ... the source difference.
    (collector.sourceDiff ??= [])
      .push(sourceItem)
  } else {

    // collect the intersection of both source and target ...
    (collector.intersection ??= [])
      .push(
        // ... by rejecting the common item from the original
        // target, thus actively mutating the latter, which leads
        // to ending up additionally with the relative complement
        // of source in target ... vulgo ... the target difference.
        targetDiff.splice(targetIndex, 1)[0]
      );
  }
  return collector;
}

const source = [1, 1, 1, 3, 4];
const target = [1, 2, 4, 5, 6];
const {

  intersection,
  sourceDiff,
  targetDiff,

} = source.reduce(collectIntersectionAndComplements, { targetDiff: [...target] });

console.log({ source, target, intersection, sourceDiff, targetDiff });

.as-console-wrapper { min-height: 100%!important; top: 0; }

@heyheyhey2 ... By the way, in the context of the Simple Theory of Sets, duplicate values are not permitted within arrays/lists/sets. Sets should only contain unique values.

Answer 2

The original poster raised a question about handling matching values between two arrays.

"What I would like to add is when source has a match with target, the value will be deleted from the target array, how can I achieve that?"

One approach is to manually remove the common item from the target array, essentially creating the relative complement of source in target, also known as the target difference.

Another method involves making a shallow copy of the target array and using it as an initial value for a reduce operation to avoid modifying the original target reference.

function collectIntersectionAndComplements(collector, sourceItem) {
  const { targetDiff } = collector;

  const targetIndex = targetDiff
    .findIndex(targetItem => targetItem === sourceItem);

  if (targetIndex === -1) {

    // collect the relative complement of target
    // in source ... vulgo ... the source difference.
    (collector.sourceDiff ??= [])
      .push(sourceItem)
  } else {

    // collect the intersection of both source and target ...
    (collector.intersection ??= [])
      .push(
        // ... by rejecting the common item from the original
        // target, thus actively mutating the latter, which leads
        // to ending up additionally with the relative complement
        // of source in target ... vulgo ... the target difference.
        targetDiff.splice(targetIndex, 1)[0]
      );
  }
  return collector;
}

const source = [1, 1, 1, 3, 4];
const target = [1, 2, 4, 5, 6];
const {

  intersection,
  sourceDiff,
  targetDiff,

} = source.reduce(collectIntersectionAndComplements, { targetDiff: [...target] });

console.log({ source, target, intersection, sourceDiff, targetDiff });

.as-console-wrapper { min-height: 100%!important; top: 0; }

@heyheyhey2 ... By the way, in the context of the Simple Theory of Sets, duplicate values are not permitted within arrays/lists/sets. Sets should only contain unique values.

Answer 3

Answer №2

Only considering matches for matches1: (similar process for matches2)

const source = [1, 1, 1, 3, 4];
const target = [1, 2, 4, 5, 6];

let matches1 = source.reduce((res, curr)  => {
    if (target.includes(curr) && !res.includes(curr)) {
        res.push(curr);
    }
    return res;
}, []
);

console.log(matches1)
//[1,4]

For unmatches1:

let unmatches1 = source.reduce((res, curr)  => {
    if ((target.includes(curr) && res[0].includes(curr)) || !target.includes(curr)) {    
        res[1].push(curr);
    }
    if (target.includes(curr)) {
        res[0].push(curr)
    }
    return res;
}, [[],[]]
)[1]

console.log(unmatches1)
//[1,1,3]

To get the unmatches2 result of [2,5,6] - simply switch out the source and target in the input or within the code, or enhance the code to produce both outputs simultaneously. It's straightforward.

Answer 4

Only considering matches for matches1: (similar process for matches2)

const source = [1, 1, 1, 3, 4];
const target = [1, 2, 4, 5, 6];

let matches1 = source.reduce((res, curr)  => {
    if (target.includes(curr) && !res.includes(curr)) {
        res.push(curr);
    }
    return res;
}, []
);

console.log(matches1)
//[1,4]

For unmatches1:

let unmatches1 = source.reduce((res, curr)  => {
    if ((target.includes(curr) && res[0].includes(curr)) || !target.includes(curr)) {    
        res[1].push(curr);
    }
    if (target.includes(curr)) {
        res[0].push(curr)
    }
    return res;
}, [[],[]]
)[1]

console.log(unmatches1)
//[1,1,3]

To get the unmatches2 result of [2,5,6] - simply switch out the source and target in the input or within the code, or enhance the code to produce both outputs simultaneously. It's straightforward.

Answer 5

Answer №3

It is important to note that in order for a single pass to occur, both arrays need to be sorted in ascending order.

function processArrays(){
    const arr1 = [0,1, 1, 1, 3, 4, 6,8,10];
    const arr2 = [-3,-4,1, 2, 4, 5, 6,25,26];

    let noMatchArray1=[];
    let matchingElements=[];
    let noMatchArray2=[];
    let i=0,j=0;
    while(true){
        if(i>=arr1.length){
            for (;j<arr2.length;j++) noMatchArray2.push(arr2[j]);
            break;
        }
        else if (j>=arr2.length){
            for (;i<arr1.length;i++) noMatchArray1.push(arr1[j]);
            break;
        }
        else if(arr1[i]==arr2[j]){
            matchingElements.push(arr1[i++]);
            j++;
        }
        else if (arr1[i]<arr2[j]){
            let val=arr1[i++];
            if(noMatchArray1.length==0)noMatchArray1.push(val);
            else if(val != noMatchArray1[noMatchArray1.length-1])
                noMatchArray1.push(val);
        }
        else if (arr2[j]<arr1[i]){
            let val=arr2[j++];
            if(noMatchArray2.length==0)noMatchArray2.push(val);
            else if(val != noMatchArray2[noMatchArray2.length-1])
                noMatchArray2.push(val);
        }


    }

    console.log("matching elements: "+matchingElements);
    console.log("noMatchArray1: "+noMatchArray1);
    console.log("noMatchArray2: "+noMatchArray2);

}

}

Output:

matching elements: 1,4,6 noMatchArray1: 0,1,3,8,10 noMatchArray2: -3,-4,2,5,25,26

Answer 6

It is important to note that in order for a single pass to occur, both arrays need to be sorted in ascending order.

function processArrays(){
    const arr1 = [0,1, 1, 1, 3, 4, 6,8,10];
    const arr2 = [-3,-4,1, 2, 4, 5, 6,25,26];

    let noMatchArray1=[];
    let matchingElements=[];
    let noMatchArray2=[];
    let i=0,j=0;
    while(true){
        if(i>=arr1.length){
            for (;j<arr2.length;j++) noMatchArray2.push(arr2[j]);
            break;
        }
        else if (j>=arr2.length){
            for (;i<arr1.length;i++) noMatchArray1.push(arr1[j]);
            break;
        }
        else if(arr1[i]==arr2[j]){
            matchingElements.push(arr1[i++]);
            j++;
        }
        else if (arr1[i]<arr2[j]){
            let val=arr1[i++];
            if(noMatchArray1.length==0)noMatchArray1.push(val);
            else if(val != noMatchArray1[noMatchArray1.length-1])
                noMatchArray1.push(val);
        }
        else if (arr2[j]<arr1[i]){
            let val=arr2[j++];
            if(noMatchArray2.length==0)noMatchArray2.push(val);
            else if(val != noMatchArray2[noMatchArray2.length-1])
                noMatchArray2.push(val);
        }


    }

    console.log("matching elements: "+matchingElements);
    console.log("noMatchArray1: "+noMatchArray1);
    console.log("noMatchArray2: "+noMatchArray2);

}

}

Output:

matching elements: 1,4,6 noMatchArray1: 0,1,3,8,10 noMatchArray2: -3,-4,2,5,25,26

What is the best approach to gather both the intersection and the complements of two arrays in a single operation?

Answer №1

Answer №2

Answer №3

Similar questions

Scraping dynamic content with Python for websites using JavaScript-generated elements

Angular UI modal on close event

Including variables in package.jsonORInjecting variables into package

Utilizing jQuery with variable assignment: A beginner's guide

What methods and applications are available for utilizing the AbortController feature within Next.js?

Merging object keys and values from JSON arrays based on their keys, using JavaScript

Tips for accessing the 'styled' function in Material UI ReactHow to utilize the 'styled' function

How can I extract an integer and match it to a corresponding name within JSON data using parsing?

JQuery Anchor Link Scrolling Problem on Specific Devices

Is there a way to disable auto rotation on a website when accessed from a mobile phone?

Detecting Specific Web Browsers on My Website: What's the Best Approach?

The array containing numbers or undefined values cannot be assigned to an array containing only numbers

What is the best way to make the children of a parent div focusable without including the grandchildren divs in the focus?

Exploring MongoDB through User Interface Requests

Struggling to retrieve the most recent emitted value from another component in Angular?

"Utilizing a function from an external file within the Main App function - a step-by-step guide

Obtain the value of a checkbox using jQuery

Unable to switch. Is there an issue with the initialization of Bootstrap 5 JavaScript?

How about wrapping a large amount of text three times and including a "read more" link?

Adjust Mui Autocomplete value selection in real-time