Remove any repeated elements from the array and ensure that the element which occurs the most is placed at the beginning of the new array

Question

Remove any repeated elements from the array and ensure that the element which occurs the most is placed at the beginning of the new array

Given an array "source" with values [2, 9, 9, 1, 6], we use the filter method to remove duplicate elements. The variable 'ans' stores the result.

Now, how can we rearrange the elements in such a way that the highest repeated value (9) comes first in the new array?

If you have any suggestions or solutions, your help would be greatly appreciated!

The desired output should be [9, 2, 1, 6]

javascript arrays filter

Answer 1

Answer №1

This solution is designed to handle scenarios where an array needs to be sorted based on the frequency of its elements.

const source = [2,1,9,9,6];
const indexes = [];

var ans = source.filter((item,index,arr)=>{
    if(arr.indexOf(item) === index){
        indexes.push({item:item,count:1})
        
        return true;
    }
    else if(arr.indexOf(item) !== index){
        
        indexes[indexes.findIndex(object=>object.item === item)].count++
        return false;
    }
    return false;

})

ans =(indexes.sort((a,b)=>{return b.count - a.count}).map(obj =>obj.item))

console.log(ans)

Answer 2

This solution is designed to handle scenarios where an array needs to be sorted based on the frequency of its elements.

const source = [2,1,9,9,6];
const indexes = [];

var ans = source.filter((item,index,arr)=>{
    if(arr.indexOf(item) === index){
        indexes.push({item:item,count:1})
        
        return true;
    }
    else if(arr.indexOf(item) !== index){
        
        indexes[indexes.findIndex(object=>object.item === item)].count++
        return false;
    }
    return false;

})

ans =(indexes.sort((a,b)=>{return b.count - a.count}).map(obj =>obj.item))

console.log(ans)

Answer 3

Answer №2

To increase efficiency, you can utilize a hash map to count elements and then transform it into an array if additional space is not a concern.

Here is an example:

let arr =  [2, 9, 9, 1, 6];
// Count the elements
const map = arr.reduce((acc, e) => acc.set(e, (acc.get(e) || 0) + 1), new Map());

// Sort by values and convert back to array
const res = [...map.entries()].sort((a, b) => b[0] - a[0]).map((a) => {
    return a[0]
});


console.log(res)

Answer 4

To increase efficiency, you can utilize a hash map to count elements and then transform it into an array if additional space is not a concern.

Here is an example:

let arr =  [2, 9, 9, 1, 6];
// Count the elements
const map = arr.reduce((acc, e) => acc.set(e, (acc.get(e) || 0) + 1), new Map());

// Sort by values and convert back to array
const res = [...map.entries()].sort((a, b) => b[0] - a[0]).map((a) => {
    return a[0]
});


console.log(res)

Answer 5

Answer №3

Give This a Go

function organizeAndFilter(array) {
    var count = {};
    for (var index = 0; index < array.length; index++) {
        var element = array[index];
        if (element in count) {
            count[element]++;
        } else {
            count[element] = 1;
        }
    }
    var finalResult = [];
    for (element in count) {
        finalResult.push(element);
    }
    function sortByOccurrence(x, y) {
        return count[y] - count[x];
    }
    return finalResult.sort(sortByOccurrence);
}

console.log(organizeAndFilter([7, 3, 3, 9, 2]));

It's Actually Functioning, Verify It

Answer 6

Give This a Go

function organizeAndFilter(array) {
    var count = {};
    for (var index = 0; index < array.length; index++) {
        var element = array[index];
        if (element in count) {
            count[element]++;
        } else {
            count[element] = 1;
        }
    }
    var finalResult = [];
    for (element in count) {
        finalResult.push(element);
    }
    function sortByOccurrence(x, y) {
        return count[y] - count[x];
    }
    return finalResult.sort(sortByOccurrence);
}

console.log(organizeAndFilter([7, 3, 3, 9, 2]));

It's Actually Functioning, Verify It

Answer 7

Answer №4

To eliminate duplicates using your current code, it is important to create a frequency map to store information about the duplicate elements. In this scenario, I have created a map using an object structure like this...

const freq = { 9: [9, 2], 1: [1, 1] ... };

In this setup, each iterated element serves as a key in the object, while the corresponding value includes the element itself and the number of duplicates. You can access these arrays by using Object.values, then sort them based on the duplicate value with sort, and finally iterate over the sorted nested array using map to generate the final output.

However, please note that due to the sorting process, the resulting order will be [9, 1, 2, 6].

const source = [2, 9, 9, 1, 6];

// Apply `reduce` method on the array to create key/value pairs where
// the value is an array consisting of the element value and its duplicate count
const nested = source.reduce((acc, c) => {

  // If the object property identified by the element
  // value doesn't exist, assign an array to initialize
  // the duplicate count as zero 
  acc[c] ??= [c, 0];

  // Increase the duplicate count
  ++acc[c][1];

  // Return the updated object for the next iteration
  return acc;

}, {});

// Obtain the `Object.values` from the object (array of nested arrays)
// Sort them based on their duplicate count,
// Finally `map` over the sorted arrays to extract
// the first element from each
const out = Object.values(nested).sort((a, b) => {
  return b[1] - a[1];
}).map(arr => arr[0]);

console.log(out);

For more information, please refer to:

Logical nullish assignment

Answer 8

To eliminate duplicates using your current code, it is important to create a frequency map to store information about the duplicate elements. In this scenario, I have created a map using an object structure like this...

const freq = { 9: [9, 2], 1: [1, 1] ... };

In this setup, each iterated element serves as a key in the object, while the corresponding value includes the element itself and the number of duplicates. You can access these arrays by using Object.values, then sort them based on the duplicate value with sort, and finally iterate over the sorted nested array using map to generate the final output.

However, please note that due to the sorting process, the resulting order will be [9, 1, 2, 6].

const source = [2, 9, 9, 1, 6];

// Apply `reduce` method on the array to create key/value pairs where
// the value is an array consisting of the element value and its duplicate count
const nested = source.reduce((acc, c) => {

  // If the object property identified by the element
  // value doesn't exist, assign an array to initialize
  // the duplicate count as zero 
  acc[c] ??= [c, 0];

  // Increase the duplicate count
  ++acc[c][1];

  // Return the updated object for the next iteration
  return acc;

}, {});

// Obtain the `Object.values` from the object (array of nested arrays)
// Sort them based on their duplicate count,
// Finally `map` over the sorted arrays to extract
// the first element from each
const out = Object.values(nested).sort((a, b) => {
  return b[1] - a[1];
}).map(arr => arr[0]);

console.log(out);

For more information, please refer to:

Logical nullish assignment

Answer 9

Answer №5

function filterAndSort(list) {
    let counts = {};
    // calculate the occurrences
    list.filter((item, index, arr) => {
        if (arr.indexOf(item) != index) 
            return counts[item] = (counts[item] || 0) + 1;
        counts[item] = 0;
    })
    // sort the numbers based on the count of occurrences
    return Object.keys(counts).map(a => parseInt(a)).sort((a, b) => counts[b] - counts[a]);
}

Output: [ 7, 4, 3, 2 ]

This function should work well for this task.

Answer 10

function filterAndSort(list) {
    let counts = {};
    // calculate the occurrences
    list.filter((item, index, arr) => {
        if (arr.indexOf(item) != index) 
            return counts[item] = (counts[item] || 0) + 1;
        counts[item] = 0;
    })
    // sort the numbers based on the count of occurrences
    return Object.keys(counts).map(a => parseInt(a)).sort((a, b) => counts[b] - counts[a]);
}

Output: [ 7, 4, 3, 2 ]

This function should work well for this task.

Answer 11

Answer №6

Look no further, this is the ultimate solution for your query

def remove_duplicates(lst):
    seen = set()
    result = []
    for item in lst:
        if item not in seen:
            result.append(item)
            seen.add(item)
    return result

source = [2, 9, 9, 1, 6]
print(remove_duplicates(source))

Answer 12

Look no further, this is the ultimate solution for your query

def remove_duplicates(lst):
    seen = set()
    result = []
    for item in lst:
        if item not in seen:
            result.append(item)
            seen.add(item)
    return result

source = [2, 9, 9, 1, 6]
print(remove_duplicates(source))

Answer 13

Answer №7

function myFunc() {
  const arr = [5, 2, 7, 7, 9];
  
  const uniqueArr = arr.filter((item, index, array)=> array.indexOf(item) === index);

  uniqueArr.sort((a, b) => b-a);

  console.log(uniqueArr);
}

Result: [ 9, 7, 5, 2 ]

Answer 14

function myFunc() {
  const arr = [5, 2, 7, 7, 9];
  
  const uniqueArr = arr.filter((item, index, array)=> array.indexOf(item) === index);

  uniqueArr.sort((a, b) => b-a);

  console.log(uniqueArr);
}

Result: [ 9, 7, 5, 2 ]

Remove any repeated elements from the array and ensure that the element which occurs the most is placed at the beginning of the new array

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

Answer №7

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