Here is my array of objects
const array = [
{id: 1, data: "foo"},
{id: 1, data: "bar"},
{id: 2, data: "baz"}
]
I want to remove all duplicate objects by its id and return only the array of objects that have an unique id.
Expected result:
[
{id: 2, data: "baz"}
]
This is what I have now: O(n^2)
function getUnique(array) {
const newArray = []
for (let obj of array) {
if (array.filter(x => x.id === obj.id).length === 1) {
newArray.push(obj)
}
}
return newArray
}
Whats the more efficient way to achieve this?
Is it possible to get the time-complexity to O(n) or O(n log n)?
const array = [{
id: 1,
data: "foo"
},
{
id: 1,
data: "bar"
},
{
id: 2,
data: "baz"
}
]
let map = {};
array.forEach(x => {
map[x.id] = (map[x.id] || 0) + 1
});
console.log(array.filter(x => map[x.id] === 1))I would suggest to count the number of occurrences in your array and store them in a Map. Next you filter all element, which count is 1
function getUnique(arr) {
const count = new Map();
arr.forEach((element) => {
count.set(element.id, (count.get(element.id) || 0) + 1);
});
return array.filter((element) => {
return count.get(element.id) === 1;
});
}
This has a runtime of 2(n) since you have to iterate over them twice