Reduce until run in Javascript

After ES6, we tend to use map, filter a lot. And sometimes if we really need to get hands dirty, we can use reduce where we take an initial state and reduce it to next state going through all elements.

Looping through the entire array can be a bit expensive, especially when you want to stop when certain condition is met. This conditional reduce can be implemented using a Lazy library, where each element traverse is deferred till needed, therefore only loops to the point when a match is found. But how do we do this in a classical way without knowing what Lazy is?

Problem 1d

If you can go up or down one floor at a time, stated in an array [1,1,-1,-1,-1,1,1,1], when will you reach the basement -1 starting from floor 0.

// advent of code 2015, day 1[1,1,-1,-1,-1,1,1,1].run(
(acc, v) => acc + v, 0,
acc => acc == -1

Wish we could have a function run like that, similar to reduce only with additional until function.

const run = (arr, fn, init, until = () => false) => {
let i = 0, acc = init, done = false, n = arr.length
while (i < n && !done) {
acc = fn(acc, arr[i], i, arr)
done = until(acc, arr[i], i, arr)
return [acc, i]
} = function (fn, init, until) {
return run(this, fn, init, until)

By default until function return false, thus it behaviors exactly like reduce does. But when it returns true, it'll finish the loop early and return the last state and index, as in a [acc, i] pair. Pretty handy, huh?

Problem N-d

The power of reduce and run doesn't stop at one dimensional array for sure. Consider to find out what are the two numbers in [1721,979,366,299,675,1456] that add up to 2020.

Imagine that we can have a similar function runN that iterates through a multi dimensional array, in our case the same array twice so that we can test two numbers in each iteration.

// advent of code 2020, day 1let N = 2
new Array(N).fill(0)
.map(v => [1721,979,366,299,675,1456])
(acc, vs) => vs, null,
acc => (acc.sum() == 2020)

Here Array.sum is a utility function which adds numbers together from all elements, basically another reduce call reduce((acc, v) => acc + v, 0). It’s easier to read this way.

If you set N = 2, we'll find out two numbers. And what if we need to find out which three numbers that sum up to be 2020. Yeah you guessed right, just change N = 3. Wow, how powerful is that in terms of semantic writing?

Ok now is the implementation of runN which is very similar in structure as the previous version, whereas the complex part is more from how to do multiple loops in a non nested way.

const runN = (arr, fn, init, until = () => false) => {
let re = 0, acc = init, done = false, d = arr.length
const is = new Array(d).fill(0)

while (re < 1 && !done) {

value =, j) => {
const dn = arr[j].length
if (dn < 1) return undefined
return arr[j][k]

acc = fn(acc, value, is, arr)
done = until(acc, value, is, arr)
if (done) continue

let j = d - 1
re = 1
while (j >=0 && re > 0) {
const dn = arr[j].length
if (dn < 1) { j--; continue }
if (is[j] == arr[j].length) {
is[j] = 0
} else {
return [acc, is]
Array.prototype.runN = function (fn, init, until) {
return runN(this, fn, init, until)

Performance wise, this multi-loop approach isn’t the best one for this classical problem, since it’s in the order of O(n^N), n being the length of the array. Therefore here we’re only focusing on the semantics to favor the logic sense, which is inspired more or less by Haskell.


We extend reduce with an additional until function to adjust when the loop can be finished. And then we apply run and runN to both 1-d and N-d problems and observe that the process can be drastically simplified.

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