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Tolerim
a month ago

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

As a developer, I'm currently in the process of building a sorting algorithm for a list of items where each item can define which item(s) come after it. If there is no definition, the initial sorting sequence should remain. The provided data set contains multiple items with unique IDs and their corresponding "after" values. Using the "id" property to determine numerical order, the algorithm should recursively sort the array until it is completely resolved. If an item has an "after" value, it should be placed immediately after that item. The desired result should resemble the following:

[
    {
        "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,
    },
]
Can someone provide a sorting function that can achieve this desired result? Thank you!

Answers(2)
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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 - Programiz
avatar
Tolerim
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.
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