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.

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


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