Sorting a sequence by swapping adjacent elements using minimum swaps
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Sorting a Sequence with Minimum Adjacent Swaps

Explore algorithms to sort an array or sequence using only adjacent element swaps, focusing on minimizing the total number of swaps required.
Sorting a sequence is a fundamental problem in computer science. While many sorting algorithms exist, some scenarios impose specific constraints, such as only allowing swaps between adjacent elements. This article delves into how to sort a sequence under this constraint, with a particular focus on achieving the minimum number of swaps. This problem often arises in contexts like permutation puzzles, data reordering, or optimizing certain hardware operations where non-adjacent swaps are costly or impossible.
Understanding the Problem: Minimum Adjacent Swaps
When you can only swap adjacent elements, the minimum number of swaps required to sort a sequence is equivalent to the number of inversions in the sequence. An inversion is a pair of elements (A[i], A[j])
such that i < j
but A[i] > A[j]
. Each adjacent swap can reduce the number of inversions by exactly one. Therefore, counting inversions directly gives us the minimum number of adjacent swaps needed.
flowchart TD A[Start with Unsorted Array] --> B{Identify Inversions} B --> C{Count Inversions} C --> D{Minimum Swaps = Total Inversions} D --> E[End]
Conceptual flow for determining minimum adjacent swaps.
Algorithm 1: Brute-Force Inversion Counting
The most straightforward way to count inversions is to iterate through all possible pairs of elements. For each element A[i]
, we check all subsequent elements A[j]
(where j > i
) to see if A[i] > A[j]
. If this condition holds, we increment our inversion count. While simple to understand, this approach has a time complexity of O(n^2), which can be inefficient for large sequences.
def count_inversions_brute_force(arr):
inversions = 0
n = len(arr)
for i in range(n):
for j in range(i + 1, n):
if arr[i] > arr[j]:
inversions += 1
return inversions
# Example usage:
sequence = [2, 4, 1, 3, 5]
print(f"Minimum adjacent swaps: {count_inversions_brute_force(sequence)}") # Output: 3
Python implementation for brute-force inversion counting.
Algorithm 2: Counting Inversions using Merge Sort
A more efficient approach to count inversions leverages the Merge Sort algorithm. During the merge step of Merge Sort, when we combine two sorted subarrays, we can efficiently count inversions. If an element from the right subarray is picked before an element from the left subarray, it means all remaining elements in the left subarray form an inversion with the picked element from the right subarray. This method reduces the time complexity to O(n log n).
def merge_sort_and_count(arr):
if len(arr) <= 1:
return arr, 0
mid = len(arr) // 2
left, inv_left = merge_sort_and_count(arr[:mid])
right, inv_right = merge_sort_and_count(arr[mid:])
merged_arr = []
inv_merge = 0
i, j = 0, 0
while i < len(left) and j < len(right):
if left[i] <= right[j]:
merged_arr.append(left[i])
i += 1
else:
merged_arr.append(right[j])
j += 1
# All remaining elements in 'left' are greater than right[j]
inv_merge += (len(left) - i)
merged_arr.extend(left[i:])
merged_arr.extend(right[j:])
return merged_arr, inv_left + inv_right + inv_merge
# Example usage:
sequence = [2, 4, 1, 3, 5]
_, total_inversions = merge_sort_and_count(sequence)
print(f"Minimum adjacent swaps (Merge Sort): {total_inversions}") # Output: 3
Python implementation for counting inversions using a modified Merge Sort.
sequenceDiagram participant A as Array participant MS as Merge Sort participant IC as Inversion Counter A->>MS: `sort_and_count(arr)` MS->>MS: Divide array (recursive) MS->>MS: `sort_and_count(left_half)` MS->>MS: `sort_and_count(right_half)` MS->>IC: Merge and count inversions IC-->>MS: Return `inv_merge` MS-->>MS: Combine `inv_left + inv_right + inv_merge` MS-->>A: Return sorted array and total inversions
Sequence diagram for Merge Sort based inversion counting.