Quick Select

Quick select often comes up when the question asks for the first/last K elements or the Kth largest/smallest element


T(n)=T(n2)+O(n)T(n) = T(\frac{n}{2}) + O(n)
  • Best/Average: O(n)O(n) can guaranteed by shuffling the array

  • Worst: O(n2)O(n^2)if sorted


Kth Element

Arrange largest elements in front by inverting the pivot comparison. This is a common problem type that tests if you know exactly how the partitioning algorithm works.

Partitioning Algorithms

Pick a pivot, partition elements smaller than the pivot to the left, and elements larger to the right of the pivot

idx-1 is the pivot index after performing the partitioning the array. Think of idx as the position for the next element that is smaller than the pivot value

  • Easier to implement

def select(start, end):
    # Randomly choose a pivot value to guarantee O(n)
    random = randint(start, end)
    # Swap the pivot value to the end
    points[end], points[random] = points[random], points[end]
    # Calculate the value of the pivot
    pivot_value = calculate(points[end])
    idx = start
    for i in range(start, end + 1):
        # If current value less than pivot, then swap to idx position
        if calculate(points[i]) <= pivot_value:
            points[i], points[idx] = points[idx], points[i]
            idx += 1
    return idx - 1

s, e, pos = 0, len(points) - 1, len(points)
while pos != k:
    pos = select(s, e)
    if pos < k:
        s = pos + 1
        e = pos - 1
return points[:k]

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