Using `quick-look` to find certain element

Question

First of all, I want to say that this is a school assignment and I am only seeking for some guidance.

My task was to write an algorithm that finds the k:th smallest element in a seq using quickselect. This should be easy enough but when running some tests I hit a wall. For some reason if I use input (List(1, 1, 1, 1), 1) it goes into infinite loop.

Here is my implementation:

  val rand = new scala.util.Random()

  def find(seq: Seq[Int], k: Int): Int = {
    require(0 <= k && k < seq.length)        
    val a: Array[Int] = seq.toArray[Int]      // Can't modify the argument sequence

    val pivot = rand.nextInt(a.length)
    val (low, high) = a.partition(_ < a(pivot))
    if (low.length == k) a(pivot)
    else if (low.length < k) find(high, k - low.length)
    else find(low, k) 
  }

For some reason (or because I am tired) I cannot spot my mistake. If someone could hint me where I go wrong I would be pleased.

Walk through it in a debugger. But a hint `a.partition(_ < a(pivot))` when all the elements are the same will always produce an empty `low` and original a in `high` — The Archetypal Paul, Sep 29 '16 at 11:42

score 1 · Answer 1 · answered Sep 29 '16 at 12:00

Basically you are depending on this line - val (low, high) = a.partition(_ < a(pivot)) to split the array into 2 arrays. The first one containing the continuous sequence of elements smaller than pivot-element and the second contains the rest.

Then you say that if the first array has length k that means you have already seen k elements smaller that your pivot-element. Which means pivot-element is actually k+1th smallest and you are actually returning k+1th smallest element instead of kth. This is your first mistake.

Also... A greater problem occurs when you have all elements which are same because your first array will always have 0 elements.

Not only that... your code will give you wrong answer for inputs where you have repeating elements among k smallest ones like - (1, 3, 4, 1, 2).

The solution lies in obervation that in the sequence (1, 1, 1, 1) the 4th smallest element is the 4th 1. Meaning you have to use <= instead of <.

Also... Since the partition function will not split the array until your boolean condition is false, you can not use partition for achieving this array split. you will have to write the split yourself.

Using `quick-look` to find certain element

1 Answers1