Quick Select
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"""
A Python implementation of the quick select algorithm, which is efficient for
calculating the value that would appear in the index of a list if it would be
sorted, even if it is not already sorted
https://en.wikipedia.org/wiki/Quickselect
"""
import random
def _partition(data: list, pivot) -> tuple:
"""
Three way partition the data into smaller, equal and greater lists,
in relationship to the pivot
:param data: The data to be sorted (a list)
:param pivot: The value to partition the data on
:return: Three list: smaller, equal and greater
"""
less, equal, greater = [], [], []
for element in data:
if element < pivot:
less.append(element)
elif element > pivot:
greater.append(element)
else:
equal.append(element)
return less, equal, greater
def quick_select(items: list, index: int):
"""
>>> quick_select([2, 4, 5, 7, 899, 54, 32], 5)
54
>>> quick_select([2, 4, 5, 7, 899, 54, 32], 1)
4
>>> quick_select([5, 4, 3, 2], 2)
4
>>> quick_select([3, 5, 7, 10, 2, 12], 3)
7
"""
# index = len(items) // 2 when trying to find the median
# (value of index when items is sorted)
# invalid input
if index >= len(items) or index < 0:
return None
pivot = items[random.randint(0, len(items) - 1)]
count = 0
smaller, equal, larger = _partition(items, pivot)
count = len(equal)
m = len(smaller)
# index is the pivot
if m <= index < m + count:
return pivot
# must be in smaller
elif m > index:
return quick_select(smaller, index)
# must be in larger
else:
return quick_select(larger, index - (m + count))
About this Algorithm
Problem Statement
Given an array, find the kth largest / smallest element in linear time complexity.
Approach
- Select a pivot element at random
- Apply partitioning as used in quick sort
- After partitioning, the pivot will be placed in its sorted location ie. All elements smaller than the pivot will be on its left and greater on its right
- If index of sorted pivot is k, then the pivot is our kth element and we return the number
- Else, check if 'k' is greater or smaller and choose a new pivot in that range.
- Repeat till we get the kth element at kth position
Time Complexity
-
O(n^2)Worst-Case Performance -
O(n)Best-case Performance -
O(n)Average Performance
Founder's Name
- This algorithm was developed by Tony Hoare and is also called
Hoare's Selection Algorithm.
Example
arr[] = {8,2,11,7,9,1,3}
Indexes: 0 1 2 3 4 5 6
Let's say k = 4. ie. We have to find 4th smallest element.
1. Choosing random pivot as 7
2. Swap 7 with the last element and apply the partitioning algorithm
3. After applying partition, all elements smaller than 7 will be placed to the left and greater in its right.
Thus we can say that 7 is in its sorted position arr[] = {2,3,1,7,8,9,11}
4. As position of '7' is 4th (ie. k). Thus we will simply return 7
Helpful Video Links
Video explaining how to find the Kth smallest/largest element in varying complexities
