Finding the single index of an element in a multidimensional array (number of elements that exist before that element)

Question

Assume we have the following multidimensional array:

const myArray = [[[1, 2], 3], [4, 5], [6, 7], [7, 8], [9, 10]];

and I want to focus on element myArray[2][1], which is 7.

How can I find the single index of this element, despite the multidimensional array, in the fastest possible algorithm?

i.e. How can I find how many indivisible elements exist before this element?

So in the example above, the previous indivisible elements would be [0] = 1, [1] = 2, [2] = 3, [3] = 4, [4] = 5, [5] = 6 making the overall index of myArray[2][1], 6 .

Edit: Even though this example shows the largest dimension being 3-dimensions, is it possible to have an algorithm working for up to n-dimensions.

Ideally a function of the following form:

Array.prototype.getOverallIndex(i, j, k, ...)
{
...
return overallIndex;
}

Create a function that takes an array and indexes as arguments and returns the element at the specified indexes: function getElementAt(arr, i, j) { return arr[i][j]; } console.log(getElementAt(myArray, 2, 1)); // Output: 7 — liontass
– liontass, Commented Jan 22, 2023 at 12:28
How would you treat empty arrays? For example, this: [ [ [1, [ ] , [ [ [ [ ] ] ] ] , 2], 3], [4, 5], [6, 7], [7, 8], [9, 10]] — FiddlingAway
– FiddlingAway, Commented Jan 22, 2023 at 12:28
please how do you get six as result? please add a detailed calculation for it. — Nina Scholz
– Nina Scholz, Commented Jan 22, 2023 at 12:33
If the output is 6 (the overall index of the input), then what is the input? [ 2, 1 ] (the index path) or 7 (the value)? This would make a difference for repeated values, e.g. for [ [ 1, 2 ], [ 3, 2 ] ], I’d expect the overall index of the path [ 1, 1 ] (representing array[1][1]) to be 3, but the overall index of the value 2 to be 1 (because it’s found at array[0][1]). — Sebastian Simon
– Sebastian Simon, Commented Jan 22, 2023 at 12:33
If the empty elements would be skipped, you might get away with using Array.prototype.flat(), but the trick would be in guessing the depth to pass to the .flat() method. EDIT It would also depend on your setup to adhere to these rules: i.imgur.com/ybBsdq1.png — FiddlingAway
– FiddlingAway, Commented Jan 22, 2023 at 12:46

Roko C. Buljan · Accepted Answer · 2023-01-22 19:10:47Z

3

use Array.prototype.flat(Infinity) to flatten (any depth) your array
use Array.prototype.indexOf() to get the index of the desired item

Example:

const myArray = [[[1, 2], 3], [4, 5], [6, 7], [7, 8], [9, 10]];
const flattened = myArray.flat(Infinity);
console.log(flattened.indexOf(7)) // 6

Then, if you want all the indexes of N, you could use:

const indexesOf = (arr, num) => arr.flat(Infinity).reduce((a, n, i) => (num === n && a.push(i), a), []);

// Use like:
const myArray = [[[1, 2], 3], [4, 5], [6, 7], [7, 8], [9, 10]];
console.log(indexesOf(myArray, 7)) // [6, 7]

If in need to get the max depth of your array you could use:

const getDepth = (a) => Array.isArray(a) ? 1 + Math.max(0, ...a.map(getDepth)) : 0;

const myArray = [[[1, 2], 3], [4, 5], [6, 7], [7, 8], [9, 10]];

// levels deep:
const depth = getDepth(myArray);
console.log(depth); // 3

edited Jan 22, 2023 at 19:10

answered Jan 22, 2023 at 13:00

Roko C. Buljan

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3 Comments

FiddlingAway Over a year ago

Ah, but what do we do if we were looking for the second 7? Or nth 7?

Roko C. Buljan Over a year ago

@FiddlingAway added another example that returns an Array of all the indexes where 7 is present in the flattened array. This way you can get the length of appearances and you can then query the nth.

Roko C. Buljan Over a year ago

@SUTerliakov you're absolutely right. Corrected as per docs examples.

Nina Scholz · Accepted Answer · 2023-01-22 13:19:11Z

1

You could count.

const
    getCount = v => Array.isArray(v)
        ? v.reduce((t, w) => t + getCount(w), 0)
        : 1,
    getCountOfItemsBefore = (array, target) => {
        let i = 0,
            count = 0;

        while (i < target[0]) count += getCount(array[i++]);

        if (target.length > 1 && Array.isArray(array[target[0]])) {
            count += getCountOfItemsBefore(array[target[0]], target.slice(1));
        }
        return count;
    },
    data = [[[1, 2], 3], [4, 5], [6, 7], [7, 8], [9, 10]],
    target = [2, 1],
    result = getCountOfItemsBefore(data, target);

console.log(result);

answered Jan 22, 2023 at 13:19

Nina Scholz

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