Technologies:

javascript

Tolerim

a month ago

### How can a JavaScript sorting algorithm be developed that orders listed items immediately following itself?

```
[
{
"id": 1871,
"after": null,
},
{
"id": 1872,
"after": null,
},
{
"id": 1873,
"after": 1872,
},
{
"id": 1875,
"after": 1873,
},
{
"id": 1876,
"after": 1875,
},
{
"id": 1877,
"after": 1876,
},
{
"id": 1878,
"after": 1877,
},
{
"id": 1874,
"after": 1872,
},
{
"id": 1880,
"after": 1874,
},
{
"id": 1879,
"after": null,
},
]
```

Answers(2)

Vogi

a month ago

Verified Answer

Certainly! Here is a sorting function that can achieve the desired result:

```
function sortItems(items) {
const map = new Map(items.map(item => [item.id, item]));
const result = [];
for (const item of items) {
if (!result.includes(item)) {
const after = item.after;
if (after === null || result.includes(map.get(after))) {
result.push(item);
} else {
const index = result.indexOf(map.get(after));
result.splice(index + 1, 0, item);
}
}
}
return result;
}
```

This function takes an array of items as input and returns the sorted array. It uses a map to store each item by its ID and then iterates through the items to sort them. If an item has no "after" value or its "after" value is already in the sorted array, it is added to the end of the array. Otherwise, it is inserted after the item with the corresponding "after" value.
I hope this helps! Let me know if you have any other questions.

#### Sources

javascript - Sorting an array of objects by property values - Stack ...How to Sort an ArrayList of Objects by Property in Java?Sorting Algorithms - GeeksforGeeksStable and Unstable Sorting Algorithms - GeeksforGeeksSorting Algorithm - ProgramizTolerim

a month ago

Here's a possible implementation for the sort function that solves the problem:

```
function sortItems(items) {
const map = items.reduce((acc, item) => ({...acc, [item.id]: item}), {});
function sortHelper(a, b) {
if (!b) {
return a;
}
const next = sortHelper(map[b.after], b);
return [a, ...next];
}
return items.reduce((acc, item) => {
if (!item.after) {
return [...acc, item];
}
const index = acc.findIndex(x => x.id === item.after);
if (index === -1) {
return [...acc, item];
}
const result = sortHelper(acc[index], item);
return [...acc.slice(0, index), ...result, ...acc.slice(index + 1)];
}, []);
}
```

The

**sortItems**function takes an array of items, and first creates a**map**object for quick lookup of items based on their**id**property. It then uses a recursive helper function**sortHelper**that takes an item and keeps searching for its "after" item until it reaches the end of the chain (i.e. when an item has no "after" property). The function returns an array of items that represent the chain in the correct order. Finally, the**sortItems**function goes over the items again, and for each item that has an "after" property, it finds the index of the "after" item in the accumulated array so far, and then inserts the current item and its chain according the correct sequence. Note that this implementation assumes that there are no circular dependencies in the "after" values. If there are, the function will go into infinite recursion. Also note that this function has time complexity O(n^2), which could be a problem for large data sets. If performance is a concern, a better approach would be to use a graph data structure and perform a topological sort algorithm on it.