How to generate an array of incremental values from a given array
the idea is to create a kind of diamond shape where the arrays start decreasing in size once they reach the middle of the array. In other words the longest array is going to be the one that is containing the middle value of the array or (array.length/2 + 1)
and in cases where the elements are short to complete the array on the second half just replace it with 'E' to indicate empty space just like on the second example.
example 1
var array = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p']
//the longest array in length is containing 'i' which is the value at
array.length/2 + 1
var output = [
['a'],
['b','c'],
['d','e','f'],
['g','h','i','j'],
['k','l','m'],
['n','o'],
['p']
]
example 2
var array = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t']
enter code here
//the longest array in length is containing 'k' which is the value at array.length/2 + 1
var output = [
['a'],
['b','c'],
['d','e','f'],
['g','h','i','j'],
['k','l','m','n','o'],
['p','q','r','s'],
['t','E','E'],
['E','E'],
['E]
]
Here is the code i have tried:
const values = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16];
const halfLen = values.length/2 + 1;
var topArr = [];
for(let i = 0; i < values.length; i ++) {
if(i <= halfLen) {
topArr.push(values[i])
}
}
console.log(topArr)
var filTopArr = [];
for(let i = 0; i <= topArr.length; i ++) {
let prevIndex = i - 1;
if(i === 0) {
filTopArr.push(topArr[i])
} else if(i === 1) {
filTopArr.push(topArr.slice(i, i + i + 1))
} else {
filTopArr.push(topArr.slice(i, i + i ))
}
}
console.log(filTopArr)
my idea here was to separate the array into two different arrays which are going to be the top part that is incrementing in size and the second/bottom part that is going to be decreasing in size.
The above code had this output
[1, [2, 3], [3, 4], [4, 5, 6], [5, 6, 7, 8], [6, 7, 8, 9], [7, 8, 9], [8, 9], [9], []]
Some observations:
The number of strings in the output (including the padding "E" strings) is always a perfect square (1, 4, 9, 16, 25, ...etc)
In order to know how many "E" strings need to be added, we thus need to know which is the least perfect square that is not less than the input size.
The longest (middle) subarray in the output has a size that is the square root of that perfect square.
The number of subarrays is the double of that number minus 1.
This leads to the following implementation:
function diamond(array) {
// Get least perfect square that is not less than the array length
const sqrt = Math.ceil(Math.sqrt(array.length));
const size = sqrt ** 2;
// Pad the array with "E" strings so to reach that perfect square size
const all = [...array, ..."E".repeat(size - array.length)];
const length = 2 * sqrt;
return Array.from({length}, (_, width) => {
return all.splice(0, Math.min(width, length - width));
}).slice(1); // Skip the first subarray that was produced (empty array)
}
// Demo using the two provided examples:
var array = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p'];
console.log(diamond(array));
var array = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t'];
console.log(diamond(array));Here's a recursive version. Note that display is just for presentation purposes.
The only real work is in diamond:
const diamond = (xs, [len = xs.length, up = true, n = 1] = []) => n == 0 ? [] : [
Object .assign (Array (n) .fill ('E'), xs .slice (0, n)),
...diamond (xs .slice (n), up && n * n < len ? [len, true, n + 1] : [len, false, n - 1])
]
const display = (xss) => console .log (`${xss .map (
(xs, i) => `${' '.repeat (Math .abs ((xss .length - 1) / 2 - i) + 1)}${xs .join (' ')
}`) .join ('\n')}`)
const demos = [
['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p'],
['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u'],
[1, 2, 3, 4, 5, 6, 7, 8]
]
demos .forEach (array => display (diamond (array))).as-console-wrapper {max-height: 100% !important; top: 0}We track the length of the current string (n, defaulting to 1), the length of the original array (len) , and a boolean flag to tell whether our length is moving up or down (up). We increase n on initial iterations, adding the next n characters from our input as the next subarray. When n hits zero we return an empty array. When n ** n is greater than or equal to len, we switch up to false and start subtracting one from n from then on. The only other necessity is to fill our remaining array with 'E's. We do this with an Object .assign call.
If you want formatted output more like an array literal form, you could use this version of display:
const display = (xss) => console .log (`[\n${xss .map (
(xs, i) => `${' '.repeat (Math .abs ((xss .length - 1) / 2 - i) + 1)}['${xs .join (`','`)
}']`) .join ('\n')}\n]`)
to get output like this:
[
['a']
['b','c']
['d','e','f']
['g','h','i','j']
['k','l','m','n','o']
['p','q','r','s']
['t','u','E']
['E','E']
['E']
]
Note though that this recursion is a little overbearing, with three separate defaulted recursive variables. I would be as likely to go with trincot's solution as this. But it's good to have alternatives.