Efficient algorithm to order objects nested in an array

Question

It seems easy enough but when I think of how I would write it, it turns into some big mess and I want it to be as efficient as possible (within a reasonable amount at least). I also have two questions pertaining to this problem.

1) So I have an array of objects that look like this

[{id: apple, position: 0},{id: orange, position: 1},{id: banana, position: 2}]

But the position is not guaranteed to be in order, like this

[{id: apple, position: 2},{id: orange, position: 0},{id: banana, position: 1}]

And I'd like the algorithm to loop through the array and grab the id's in ascending order based on the positions, naturally I'd want to do something with the id I receive in the loop, so I could have some sort of loop and within the loop print out 'I ate' + id so it would come out as

I ate orange
I ate banana
I ate apple

2) Think of this as a different scenario but using the same data structure as before

[{id: apple, position: 0},{id: orange, position: 1},{id: banana, position: 2}, {id: grape, position: 4}, {id: mango, position: 3}]

And I take out the object with id orange so it looks like this

[{id: apple, position: 0},{id: banana, position: 2}, {id: grape, position: 4}, {id: mango, position: 3}]

But now the positions don't make sense because there's 0 -> 2 -> 3 -> 4 but no position 1, so I'd like to put them back in proper ascending order (It must hold the same sequence, so it still has to go apple -> banana -> mango -> grape) so it turns into

[{id: apple, position: 0},{id: banana, position: 1}, {id: grape, position: 3}, {id: mango, position: 2}]

How would I do this best as well?

Note: There could be a scenario where I have to take out more than one object, but I guess it doesn't matter since I can just sequentially do it or something.

Thanks!

Thanks for the question. Please take a moment to share with us what you've already tried. We're here to help you debug existing code, but your question could be be met with some negativity, as you're asking for a programming solution without showing an effort of your own. — Tyler Roper
– Tyler Roper, Commented Jan 4, 2019 at 4:37
Why not just utilize something akin to Array.prototype.sort()? — Ben
– Ben, Commented Jan 4, 2019 at 4:38

ic3b3rg · Accepted Answer · 2019-01-04 04:46:39Z

3

I broke this out into

Sorting
Position recalculation
Working with sorted fruit

const fruit = [{id: "apple", position: 0},{id: "banana", position: 2}, {id: "grape", position: 4}, {id: "mango", position: 3}];

// 1. Sort the fruit by position
fruit.sort((a, b) => a.position - b.position);

console.log(fruit)

// 2. Recalculate position
fruit.forEach((o, idx) => {
  o.position = idx;
});

console.log(fruit)

// 3. Work with the sorted fruit
console.log(fruit.map(({ id }) => `I ate ${id}`))

answered Jan 4, 2019 at 4:46

ic3b3rg

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1 Comment

Hello Mellow Over a year ago

This is perfect, I had loops inside loops and it was just a mess. Totally forgot I had these functions available, thank you!

Code Maniac · Accepted Answer · 2019-01-04 04:40:57Z

0

You can do it with sort and map

let arr = [{id: 'apple', position: 2},{id: 'orange', position: 0},{id: 'banana', position: 1}];
let op = arr.sort((a,b)=> a.position-b.position).map(e=> `I ate ${e.id}`);

console.log(op)

answered Jan 4, 2019 at 4:40

Code Maniac

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2 Comments

Ben Over a year ago

Why use map() here and create a whole new array if OP simply wants to print the phrase "I ate {id}"? There would be less overhead with a simple iteration of the sorted array via forEach().

Tyler Roper Over a year ago

@BenBeck "Premature optimization is the root of all evil" :)

p u · Accepted Answer · 2019-01-04 05:10:50Z

amm..too late to post! but hope it may help others!..i have merged all in same function :)

var arr = [{
    id: 'apple',
    position: 0
}, {
    id: 'banana',
    position: 2
}, {
    id: 'grape',
    position: 4
}, {
    id: 'grapeeee',
    position: 5
}]
var old = 0,
    next = 1;
arr.sort(function(a, b) {

    if (old != a.position) {
        a.position = old;
    }
    if (next != b.position) {
        b.position = next;
    }

    var keyA = a.position,
        keyB = b.position;
    console.log(keyA, "--", keyB)
    old++;
    next++;

    if (keyA < keyB) return -1;
    if (keyA > keyB) return 1;
    return 0;

});
console.log("arr--", arr);
for (var i = 0; i < arr.length; i++) {
    console.log("I ate ", arr[i].id)
}

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