feat: solution for kth order statistic

Python/kth_order_statistic.py

@@ -0,0 +1,66 @@
+"""
+Find the kth smallest element in linear time using divide and conquer.
+Recall we can do this trivially in O(nlogn) time. Sort the list and
+access kth element in constant time.
+
+This is a divide and conquer algorithm that can find a solution in O(n) time.
+
+For more information of this algorithm:
+https://web.stanford.edu/class/archive/cs/cs161/cs161.1138/lectures/08/Small08.pdf
+"""
+
+from __future__ import annotations
+
+from random import choice
+
+
+def random_pivot(lst):
+    """
+    Choose a random pivot for the list.
+    We can use a more sophisticated algorithm here, such as the median-of-medians
+    algorithm.
+    """
+    return choice(lst)
+
+
+def kth_number(lst: list[int], k: int) -> int:
+    """
+    Return the kth smallest number in lst.
+    >>> kth_number([2, 1, 3, 4, 5], 3)
+    3
+    >>> kth_number([2, 1, 3, 4, 5], 1)
+    1
+    >>> kth_number([2, 1, 3, 4, 5], 5)
+    5
+    >>> kth_number([3, 2, 5, 6, 7, 8], 2)
+    3
+    >>> kth_number([25, 21, 98, 100, 76, 22, 43, 60, 89, 87], 4)
+    43
+    """
+    # pick a pivot and separate into list based on pivot.
+    pivot = random_pivot(lst)
+
+    # partition based on pivot
+    # linear time
+    small = [e for e in lst if e < pivot]
+    big = [e for e in lst if e > pivot]
+
+    # if we get lucky, pivot might be the element we want.
+    # we can easily see this:
+    # small (elements smaller than k)
+    # + pivot (kth element)
+    # + big (elements larger than k)
+    if len(small) == k - 1:
+        return pivot
+    # pivot is in elements bigger than k
+    elif len(small) < k - 1:
+        return kth_number(big, k - len(small) - 1)
+    # pivot is in elements smaller than k
+    else:
+        return kth_number(small, k)
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()