Algorithm
30 Days of JavaScript
Group By

2631. Group By

Tags

  • JSON

Link

https://leetcode.com/problems/group-by/description/?envType=study-plan-v2&envId=30-days-of-javascript (opens in a new tab)

Question

Write code that enhances all arrays such that you can call the array.groupBy(fn) method on any array and it will return a grouped version of the array.

A grouped array is an object where each key is the output of fn(arr[i]) and each value is an array containing all items in the original array which generate that key.

The provided callback fn will accept an item in the array and return a string key.

The order of each value list should be the order the items appear in the array. Any order of keys is acceptable.

Please solve it without lodash's _.groupBy function.

Example 1:
Input:
array = [
  {"id":"1"},
  {"id":"1"},
  {"id":"2"}
],
fn = function (item) {
  return item.id;
}
Output:
{
  "1": [{"id": "1"}, {"id": "1"}],
  "2": [{"id": "2"}]
}
Explanation:
Output is from array.groupBy(fn).
The selector function gets the "id" out of each item in the array.
There are two objects with an "id" of 1. Both of those objects are put in the first array.
There is one object with an "id" of 2. That object is put in the second array.
Example 2:
Input:
array = [
  [1, 2, 3],
  [1, 3, 5],
  [1, 5, 9]
]
fn = function (list) {
  return String(list[0]);
}
Output:
{
  "1": [[1, 2, 3], [1, 3, 5], [1, 5, 9]]
}
Explanation:
The array can be of any type. In this case, the selector function defines the key as being the first element in the array.
All the arrays have 1 as their first element so they are grouped together.
{
  "1": [[1, 2, 3], [1, 3, 5], [1, 5, 9]]
}
Example 3:
Input:
array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
fn = function (n) {
  return String(n > 5);
}
Output:
{
  "true": [6, 7, 8, 9, 10],
  "false": [1, 2, 3, 4, 5]
}
Explanation:
The selector function splits the array by whether each number is greater than 5.
Constraints:
  • 0 <= array.length <= 105
  • fn returns a string

Answer

JavaScript

/**
 * @param {Function} fn
 * @return {Object}
 */
Array.prototype.groupBy = function (fn) {
  // Reduce the array into a single object
  return this.reduce((grouped, item) => {
    // Apply the provided callback function to get the key
    const key = fn(item);
 
    // If the key doesn't exist in the grouped object, create a new array for it
    if (!grouped[key]) {
      grouped[key] = [];
    }
 
    // Push the current item to the array associated with the key
    grouped[key].push(item);
 
    // Return the updated grouped object for the next iteration
    return grouped;
  }, {});
};
 
/**
 * [1,2,3].groupBy(String) // {"1":[1],"2":[2],"3":[3]}
 */
Function CompositionInterval Cancellation