Another tasty challenge that I ran into during my preparation for technical interviews is this seemingly simple problem from Facebook.

input [2,3,1,4]

output [12,8,24,6]

Multiply all fields except it’s own position.

Restrictions:

1. no use of division

2. complexity in O(n)

The main challenge is in making the algorithm run in O(n) time.

One solution I saw (and thought it was quite clever) is to maintain two temporary arrays, calculated by multiplying the elements of the input array from front-to-back, and back-to-front.

ie. input is [ 2, 3, 1, 4 ]

The * front* array starts with [ 1, 0, 0, 0 ], and starting from position 1, we can say

*and end up with [ 1, 2, 6, 6 ] (which is the product of all the input elements before that position.*

**front[i] = front[i-1] * input[i-1]**The * rear* array starts with [ 0, 0, 0, 1 ] and starting from position 2 (ie, second to last), we can say

*and end up with [ 12, 4, 4, 1 ].*

**rear[i] = rear[i+1] * input[i+1]**And finally we can work out the output array by multiplying the corresponding elements in the * front* and

*arrays, and that runs in O(n) of time and space.*

**rear**Here’s the implementation in F#.

## Try it Yourself

## Links

- Interview question on CareerCup
- DotNetFiddle snippet
- All my Project Euler solutions in F#
- All my Advent of Code solutions in F#

gentauro:)

let multOthers : int list -> int list =

fun xs ->

let rec hlp n = function

| [ ] | [_] -> [ ] | x :: xs -> let i = x*n in i :: (hlp i xs)

(1 :: xs |> hlp 1,

1 :: xs |> List.rev |> hlp 1 |> List.rev)

||> List.zip

|> List.map(fun (x,y) -> x*y)

[ 2; 3; 1; 4; ] |> multOthers

(it probably doesn’t look nice cos of no monospace font)

Yan Cuiyeah, you’re right, the font makes it hard to read, can you try editing it and put them inside a code block? see here

gentaurolet multOthers : int list -> int list =

fun xs ->

let rec hlp n = function

| [ ] | [_] -> [ ] | x :: xs -> let i = x*n in i :: (hlp i xs)

(1 :: xs |> hlp 1,

1 :: xs |> List.rev |> hlp 1 |> List.rev)

||> List.zip

|> List.map(fun (x,y) -> x*y)

`[ 2; 3; 1; 4; ] |> multOthers`

gentauroMuch better :)

Yan Cuimm.. getting the wrong result – [24; 24; 24; 24]

gentauroYeah, my bad, forgot the ()

...

(1 :: xs |> hlp 1,

1 :: (xs |> List.rev) |> hlp 1 |> List.rev)

...

gentauroYeah, my bad, forgot ()

...

(1 :: xs |> hlp 1,

1 :: (xs |> List.rev) |> hlp 1 |> List.rev)

...

PrashHi Yan. Your site brings my browser to a crashing halt with all of the ads. By disabling Flash, I was able to make the page stable enough to leave this comment, but it’s still using loads of CPU. I believe it’s pretty much unusable on mobile for this reason. Would you consider removing some ads and/or widgets?

PrashHere’s one using effectively the same algorithm but with more standard library functions and no mutation:

let multiplyOthers xs =

let front = Array.scan (*) 1 xs

let back = Array.scanBack (*) xs 1

Array.map2 (*) front.[..front.Length - 2] back.[1..]

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