feat: add Reversort sorting algorithm (#11800)

- Add reversort.py with Reversort algorithm implementation
- Algorithm from Google Code Jam 2021 Qualification Round
- Time complexity: O(n²)
- Space complexity: O(n) due to slicing, can be O(1) with in-place reversal
- Includes reversort() for sorting and reversort_cost() for computing cost
- Comprehensive doctests included

Author: xingzihai

sorts/reversort.py

@@ -0,0 +1,159 @@
+"""
+Reversort is a sorting algorithm described in Google Code Jam 2021 Qualification Round.
+
+Algorithm:
+1. For i from 1 to N-1:
+   a. Find the position j of the minimum element in the subarray from position i to N
+   b. Reverse the subarray from position i to j
+
+Time Complexity: O(n²) - For each position, we find the minimum and reverse
+Space Complexity: O(n) - Due to list slicing in Python
+                  (can be O(1) with in-place reversal)
+
+For doctests run following command:
+python3 -m doctest -v reversort.py
+
+For manual testing run:
+python reversort.py
+"""
+
+from typing import Any
+
+
+def reversort(collection: list[Any]) -> list[Any]:
+    """
+    Sort a list using the Reversort algorithm.
+
+    Reversort works by repeatedly finding the minimum element in the unsorted
+    portion and reversing the subarray from the current position to where the
+    minimum element is located.
+
+    :param collection: A mutable ordered collection with comparable items
+    :return: The sorted collection in ascending order
+
+    Examples:
+    >>> reversort([4, 2, 1, 3])
+    [1, 2, 3, 4]
+    >>> reversort([0, 5, 3, 2, 2])
+    [0, 2, 2, 3, 5]
+    >>> reversort([])
+    []
+    >>> reversort([-2, -5, -45])
+    [-45, -5, -2]
+    >>> reversort([1])
+    [1]
+    >>> reversort([5, 4, 3, 2, 1])
+    [1, 2, 3, 4, 5]
+    >>> reversort([2, 1, 4, 3])
+    [1, 2, 3, 4]
+    >>> reversort([-23, 0, 6, -4, 34])
+    [-23, -4, 0, 6, 34]
+    >>> reversort([1, 2, 3, 4])
+    [1, 2, 3, 4]
+    >>> reversort([3, 3, 3, 3])
+    [3, 3, 3, 3]
+    >>> reversort([56])
+    [56]
+    >>> reversort([0, 5, 2, 3, 2]) == sorted([0, 5, 2, 3, 2])
+    True
+    >>> reversort([]) == sorted([])
+    True
+    >>> reversort([-2, -45, -5]) == sorted([-2, -45, -5])
+    True
+    >>> reversort([-23, 0, 6, -4, 34]) == sorted([-23, 0, 6, -4, 34])
+    True
+    >>> reversort(['d', 'a', 'b', 'e']) == sorted(['d', 'a', 'b', 'e'])
+    True
+    >>> reversort(['z', 'a', 'y', 'b', 'x', 'c'])
+    ['a', 'b', 'c', 'x', 'y', 'z']
+    >>> reversort([1.1, 3.3, 5.5, 7.7, 2.2, 4.4, 6.6])
+    [1.1, 2.2, 3.3, 4.4, 5.5, 6.6, 7.7]
+    >>> reversort([1, 3.3, 5, 7.7, 2, 4.4, 6])
+    [1, 2, 3.3, 4.4, 5, 6, 7.7]
+    >>> import random
+    >>> collection_arg = random.sample(range(-50, 50), 100)
+    >>> reversort(collection_arg) == sorted(collection_arg)
+    True
+    >>> import string
+    >>> collection_arg = random.choices(string.ascii_letters + string.digits, k=100)
+    >>> reversort(collection_arg) == sorted(collection_arg)
+    True
+    """
+    arr = collection[:]  # Create a copy to avoid modifying the original
+    n = len(arr)
+
+    for i in range(n - 1):
+        # Find the position of the minimum element in arr[i:]
+        min_index = i
+        for j in range(i + 1, n):
+            if arr[j] < arr[min_index]:
+                min_index = j
+
+        # Reverse the subarray from position i to min_index
+        if min_index != i:
+            arr[i:min_index + 1] = arr[i:min_index + 1][::-1]
+
+    return arr
+
+
+def reversort_cost(collection: list[Any]) -> int:
+    """
+    Calculate the cost of sorting using Reversort.
+
+    The cost is defined as the sum of the lengths of all reversed segments.
+    This is based on the Google Code Jam 2021 problem.
+
+    :param collection: A mutable ordered collection with comparable items
+    :return: The total cost of sorting
+
+    Examples:
+    >>> reversort_cost([4, 2, 1, 3])
+    6
+    >>> reversort_cost([1, 2])
+    1
+    >>> reversort_cost([7, 6, 5, 4, 3, 2, 1])
+    12
+    >>> reversort_cost([1, 2, 3, 4])
+    3
+    >>> reversort_cost([1])
+    0
+    >>> reversort_cost([])
+    0
+    """
+    arr = collection[:]  # Create a copy to avoid modifying the original
+    n = len(arr)
+    total_cost = 0
+
+    for i in range(n - 1):
+        # Find the position of the minimum element in arr[i:]
+        min_index = i
+        for j in range(i + 1, n):
+            if arr[j] < arr[min_index]:
+                min_index = j
+
+        # Reverse the subarray from position i to min_index
+        arr[i:min_index + 1] = arr[i:min_index + 1][::-1]
+
+        # Cost is the length of the reversed segment
+        total_cost += min_index - i + 1
+
+    return total_cost
+
+
+if __name__ == "__main__":
+    import doctest
+    from random import sample
+    from timeit import timeit
+
+    doctest.testmod()
+
+    user_input = input("Enter numbers separated by a comma:\n").strip()
+    unsorted = [int(item) for item in user_input.split(",")]
+    print(f"Sorted list: {reversort(unsorted)}")
+    print(f"Sort cost: {reversort_cost(unsorted)}")
+
+    # Benchmark
+    num_runs = 1000
+    test_arr = sample(range(-50, 50), 100)
+    timer = timeit("reversort(test_arr[:])", globals=globals(), number=num_runs)
+    print(f"\nProcessing time: {timer:.5f}s for {num_runs:,} runs on 100 elements")