Equilibrium Index problem in F#

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A good friend pointed me to a nice coding challenge she encountered – finding the Equilibrium Index of an array in O(n) time.

After a quick discussion we ended up with 2 solutions, and naturally I had to code them up in F#

 

Solution 1

This is the same approach to the one you saw in the O(n) solution I posted the other day for the Multiply Others problem.

Here, we’ll create two temporary arrays – one that’s the running sum from the front, and the other the running sum from the rear, both are O(n) operations.

Once we have both, we can return all the indices where the front and rear arrays are equal.

(the questions usually only ask for one, but it’s nice to see all of them :-P)

This is a O(n) solution in both time and space.

 

Solution 2

A more space efficient, O(1), solution is to:

  • do one pass to sum the array (as the sum to the right)
  • then starting from the front and iteratively subtract elements from that sum until you find an equilibrium index

for example, given the input array [ -1; 3; -4; 5; 1; -6; 2; 1 ], the sum is 1, so if we start from the front of the array:

  • -1 : sum to the left is 0 (no elements), sum to the right is 1 – -1 = 2, no match
  • 3 : sum to the left is -1, sum to the right is 2 – 3 = -1, match, so index 1 is an equilibrium index
  • -4 : sum to the left is -1 + 3 = 2, sum to the right is -1 – -4 = 3, no match
  • 5 : sum to the left is 2 + -4 = -2, sum to the right is 3 – 5 = -2, match, index 3 is also an equilibrium index
  • 1 : sum to the left is -2 + 5 = 3, sum to the right is -2 – 1 = -3, no match
  • -6 : sum to the left is 3 + 1 = 4, sum to the right is -3 – -6 = 3, no match
  • 2 : sum to the left is 4 + -6 = -2, sum to the right is 3 – 2 = 1, no match
  • 1 : sum to the left is -2 + 2 = 0, sum to the right is 1 – 1 = 0, match, index 7 is also an equilibrium index

So, applying this in F# I ended up with this:

notice that I’m generating the output array via comprehensions, this can look a bit odd to people new to F# so I tend to shy away from it and usually go for seq { … } |> Seq.toArray instead.

 

Try it Yourself

 

Links

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