Determine the shared elements within a single array by utilizing Javascript

Question

Determine the shared elements within a single array by utilizing Javascript

I have the following array of mergeUniqueItems:

var mergeUniqueItems = ["-JsDEcxz_ZSGFLKwd1QM", 
"-JsJ2NXGDYKI6QRsuXVK", 
"-JsJ2RK-kOG2eGcG04xF", 
"-JsJ2RK-kOG2eGcG04xF", 
"-JsJ2YLPiP6751zh8geS"]

After using this code snippet, I encountered an issue:

    var duplicateArray = [];
    for (var i = 0; i < mergeUniqueItems.length; i ++){
        for (var j = 1; j < mergeUniqueItems.length; j ++){
            if (mergeUniqueItems[i] == mergeUniqueItems[j]){
                duplicateArray.push(mergeUniqueItems[i]);
            }
        }
    }

    console.log(duplicateArray);

The resulting duplicateArray contains the following items:

["-JsJ2NXGDYKI6QRsuXVK", 
"-JsJ2RK-kOG2eGcG04xF", 
"-JsJ2RK-kOG2eGcG04xF", 
"-JsJ2RK-kOG2eGcG04xF", 
"-JsJ2RK-kOG2eGcG04xF", 
"-JsJ2YLPiP6751zh8geS"]

However, my expectation is to have only one array entry for each duplicated item like below:

 ["-JsJ2RK-kOG2eGcG04xF"]

If there are multiple duplicate values, the array should display them as follows:

["-JsJ2RK-kOG2eGcG04xF", "another_duplicate_1", "another_duplicate_2", ...]

I am unable to determine what's causing this issue with my code. Your assistance would be greatly appreciated.

Thank you

javascript arrays

Answer 1

Answer №1

Make a small change by initializing j as i+1 instead of just 1, and ensure that you are not adding duplicate values.

for (var i = 0; i < mergeUniqueItems.length; i ++){
    for (var j = i + 1; j < mergeUniqueItems.length; j ++){
        if (mergeUniqueItems[i] == mergeUniqueItems[j]){
            if (duplicateArray.indexOf(mergeUniqueItems[i]) < 0)
                duplicateArray.push(mergeUniqueItems[i]);
            break;
        }

Answer 2

Make a small change by initializing j as i+1 instead of just 1, and ensure that you are not adding duplicate values.

for (var i = 0; i < mergeUniqueItems.length; i ++){
    for (var j = i + 1; j < mergeUniqueItems.length; j ++){
        if (mergeUniqueItems[i] == mergeUniqueItems[j]){
            if (duplicateArray.indexOf(mergeUniqueItems[i]) < 0)
                duplicateArray.push(mergeUniqueItems[i]);
            break;
        }

Answer 3

Answer №2

Here's an alternative approach to Pointy's solution.

const uniqueValues = mergeUniqueItems.filter((item, index, originalArray) => {
    return originalArray.indexOf(item, index + 1) === -1;
});

console.log(uniqueValues);

This method is more concise and avoids explicit loops.

Keep in mind that this method keeps the last duplicate item. To retain the first one instead, use .indexOf(item, index - 1).

If you only want a single instance of each duplicate, you can do it like this:

const duplicates = mergeUniqueItems.filter((item, index, originalArray) => {
    const nextIndex = originalArray.indexOf(item, index + 1);
    return nextIndex !== -1 && originalArray.lastIndexOf(item) == nextIndex;
});

Alternatively, try this method:

const duplicates = mergeUniqueItems.filter((item, index, originalArray) => {
    const nextIndex = originalArray.indexOf(item, index + 1);
    return nextIndex !== -1 && originalArray.indexOf(item, nextIndex + 1) === -1;
});

Another option is to chain two .filter() methods:

const duplicates = mergeUniqueItems.filter((item, index, originalArray) => {
    return originalArray.indexOf(item, index + 1) !== -1;
}).filter((item, index, originalArray) => {
    return originalArray.indexOf(item, index + 1) === -1;
});

Answer 4

Here's an alternative approach to Pointy's solution.

const uniqueValues = mergeUniqueItems.filter((item, index, originalArray) => {
    return originalArray.indexOf(item, index + 1) === -1;
});

console.log(uniqueValues);

This method is more concise and avoids explicit loops.

Keep in mind that this method keeps the last duplicate item. To retain the first one instead, use .indexOf(item, index - 1).

If you only want a single instance of each duplicate, you can do it like this:

const duplicates = mergeUniqueItems.filter((item, index, originalArray) => {
    const nextIndex = originalArray.indexOf(item, index + 1);
    return nextIndex !== -1 && originalArray.lastIndexOf(item) == nextIndex;
});

Alternatively, try this method:

const duplicates = mergeUniqueItems.filter((item, index, originalArray) => {
    const nextIndex = originalArray.indexOf(item, index + 1);
    return nextIndex !== -1 && originalArray.indexOf(item, nextIndex + 1) === -1;
});

Another option is to chain two .filter() methods:

const duplicates = mergeUniqueItems.filter((item, index, originalArray) => {
    return originalArray.indexOf(item, index + 1) !== -1;
}).filter((item, index, originalArray) => {
    return originalArray.indexOf(item, index + 1) === -1;
});

Determine the shared elements within a single array by utilizing Javascript

Answer №1

Answer №2

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