O(n) solution to Multiply Others problem in F#

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

The rear array starts with [ 0, 0, 0, 1 ] and starting from position 2 (ie, second to last), we can say rear[i] = rear[i+1] * input[i+1] and end up with [ 12, 4, 4, 1 ].

And finally we can work out the output array by multiplying the corresponding elements in the front and rear arrays, and that runs in O(n) of time and space.

Here’s the implementation in F#.

 

Try it Yourself

Links

10 Comments

  1. 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)

  2. Yan Cui   •  

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

  3. 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

  4. gentauro   •  

    Much better :)

  5. Yan Cui   •  

    mm.. getting the wrong result – [24; 24; 24; 24]

  6. gentauro   •  

    Yeah, my bad, forgot the ()


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

  7. gentauro   •  

    Yeah, my bad, forgot ()

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

  8. Prash   •  

    Hi 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?

  9. Prash   •  

    Here’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..]

  10. Pingback: Equilibrium Index problem in F# | theburningmonk.com

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