Add merge insertion sort (#2211)

* Add merge insertion sort

* Fix python naming conventions

* Add wikipedia link

* Add type hint

* Fix python to python3

Co-authored-by: Christian Clauss <[email protected]>

* Refactor doubled process in if-condition into one outside of if-condition

Co-authored-by: Christian Clauss <[email protected]>

* Refactor make python3 prior to python

Co-authored-by: Christian Clauss <[email protected]>

* Fix name of is_surplus into has_last_odd_item

* Add comment

* Fix long comment to shorten

Co-authored-by: Christian Clauss <[email protected]>

Author: ryuta69

sorts/merge_insertion_sort.py

@@ -0,0 +1,179 @@
+"""
+This is a pure Python implementation of the merge-insertion sort algorithm
+Source: https://en.wikipedia.org/wiki/Merge-insertion_sort
+
+For doctests run following command:
+python3 -m doctest -v merge_insertion_sort.py
+or
+python -m doctest -v merge_insertion_sort.py
+
+For manual testing run:
+python3 merge_insertion_sort.py
+"""
+
+from typing import List
+
+
+def merge_insertion_sort(collection: List[int]) -> List[int]:
+    """Pure implementation of merge-insertion sort algorithm in Python
+
+    :param collection: some mutable ordered collection with heterogeneous
+    comparable items inside
+    :return: the same collection ordered by ascending
+
+    Examples:
+    >>> merge_insertion_sort([0, 5, 3, 2, 2])
+    [0, 2, 2, 3, 5]
+
+    >>> merge_insertion_sort([99])
+    [99]
+
+    >>> merge_insertion_sort([-2, -5, -45])
+    [-45, -5, -2]
+    """
+
+    def binary_search_insertion(sorted_list, item):
+        left = 0
+        right = len(sorted_list) - 1
+        while left <= right:
+            middle = (left + right) // 2
+            if left == right:
+                if sorted_list[middle] < item:
+                    left = middle + 1
+                break
+            elif sorted_list[middle] < item:
+                left = middle + 1
+            else:
+                right = middle - 1
+        sorted_list.insert(left, item)
+        return sorted_list
+
+    def sortlist_2d(list_2d):
+        def merge(left, right):
+            result = []
+            while left and right:
+                if left[0][0] < right[0][0]:
+                    result.append(left.pop(0))
+                else:
+                    result.append(right.pop(0))
+            return result + left + right
+
+        length = len(list_2d)
+        if length <= 1:
+            return list_2d
+        middle = length // 2
+        return merge(sortlist_2d(list_2d[:middle]), sortlist_2d(list_2d[middle:]))
+
+    if len(collection) <= 1:
+        return collection
+
+    """
+    Group the items into two pairs, and leave one element if there is a last odd item.
+
+    Example: [999, 100, 75, 40, 10000]
+                -> [999, 100], [75, 40]. Leave 10000.
+    """
+    two_paired_list = []
+    has_last_odd_item = False
+    for i in range(0, len(collection), 2):
+        if i == len(collection) - 1:
+            has_last_odd_item = True
+        else:
+            """
+            Sort two-pairs in each groups.
+
+            Example: [999, 100], [75, 40]
+                        -> [100, 999], [40, 75]
+            """
+            if collection[i] < collection[i + 1]:
+                two_paired_list.append([collection[i], collection[i + 1]])
+            else:
+                two_paired_list.append([collection[i + 1], collection[i]])
+
+    """
+    Sort two_paired_list.
+
+    Example: [100, 999], [40, 75]
+                -> [40, 75], [100, 999]
+    """
+    sorted_list_2d = sortlist_2d(two_paired_list)
+
+    """
+    40 < 100 is sure because it has already been sorted.
+    Generate the sorted_list of them so that you can avoid unnecessary comparison.
+
+    Example:
+           group0 group1
+           40     100
+           75     999
+        ->
+           group0 group1
+           [40,   100]
+           75     999
+    """
+    result = [i[0] for i in sorted_list_2d]
+
+    """
+    100 < 999 is sure because it has already been sorted.
+    Put 999 in last of the sorted_list so that you can avoid unnecessary comparison.
+
+    Example:
+           group0 group1
+           [40,   100]
+           75     999
+        ->
+           group0 group1
+           [40,   100,   999]
+           75
+    """
+    result.append(sorted_list_2d[-1][1])
+
+    """
+    Insert the last odd item left if there is.
+
+    Example:
+           group0 group1
+           [40,   100,   999]
+           75
+        ->
+           group0 group1
+           [40,   100,   999,   10000]
+           75
+    """
+    if has_last_odd_item:
+        pivot = collection[-1]
+        result = binary_search_insertion(result, pivot)
+
+    """
+    Insert the remaining items.
+    In this case, 40 < 75 is sure because it has already been sorted.
+    Therefore, you only need to insert 75 into [100, 999, 10000],
+    so that you can avoid unnecessary comparison.
+
+    Example:
+           group0 group1
+           [40,   100,   999,   10000]
+            ^ You don't need to compare with this as 40 < 75 is already sure.
+           75
+        ->
+           [40,   75,    100,   999,   10000]
+    """
+    is_last_odd_item_inserted_before_this_index = False
+    for i in range(len(sorted_list_2d) - 1):
+        if result[i] == collection[-i]:
+            is_last_odd_item_inserted_before_this_index = True
+        pivot = sorted_list_2d[i][1]
+        # If last_odd_item is inserted before the item's index,
+        # you should forward index one more.
+        if is_last_odd_item_inserted_before_this_index:
+            result = result[: i + 2] + binary_search_insertion(result[i + 2 :], pivot)
+        else:
+            result = result[: i + 1] + binary_search_insertion(result[i + 1 :], pivot)
+
+    return result
+
+
+if __name__ == "__main__":
+    user_input = input("Enter numbers separated by a comma:\n").strip()
+    unsorted = [int(item) for item in user_input.split(",")]
+    print(merge_insertion_sort(unsorted))