Overview

This problem asks us to find the minimum number of operations needed to sort an array using only adjacent element swaps. This reduces to finding the inversion count of the array.

Analysis

Key Insight

The operation of swapping two adjacent elements is the same operation performed in “bubble sort.” It is well known that the number of swaps required to complete bubble sort equals the inversion count of the array.

What is inversion count? It is the number of pairs \((i, j)\) in an array where \(i < j\) and \(A_i > A_j\).

For example, for the array \([3, 1, 2]\): - \((1, 2)\): \(A_1 = 3 > A_2 = 1\) → inversion pair - \((1, 3)\): \(A_1 = 3 > A_3 = 2\) → inversion pair - \((2, 3)\): \(A_2 = 1 < A_3 = 2\) → not an inversion pair

Therefore, the inversion count is \(2\). Indeed, we can sort it with 2 adjacent swaps: \([3,1,2] \to [1,3,2] \to [1,2,3]\).

Problem with the Naive Approach

If we check all pairs \((i, j)\) to see whether \(i < j\) and \(A_i > A_j\), the total number of pairs is \(\frac{N(N-1)}{2}\), resulting in \(O(N^2)\) time complexity. With the constraint \(N \leq 2 \times 10^5\), this would result in TLE (Time Limit Exceeded).

Solution

By using merge sort, we can count inversions in \(O(N \log N)\) time while sorting the array.

Algorithm

We count inversions during the “merge” step of merge sort.

In merge sort, we divide the array into left and right halves, sort each half, and then merge them. During the merge:

1. When taking an element from the left half: no inversion occurs
2. When taking an element from the right half: it forms inversion pairs with all remaining elements in the left half that are larger than it

Example: Merging left = [2, 4], right = [1, 3]
- Take right[0]=1 → smaller than remaining left elements [2,4] → inversions +2
- Take left[0]=2 → inversions +0
- Take right[1]=3 → smaller than remaining left elements [4] → inversions +1
- Take left[1]=4 → inversions +0
Total: 3

Complexity

• Time Complexity: \(O(N \log N)\)

  • Same as the complexity of merge sort
  • The array is divided \(\log N\) times, with \(O(N)\) total processing at each level

• Space Complexity: \(O(N)\)

  • Due to the temporary array used during merging

Implementation Notes

1. Base case for recursion: When the array length is 1 or less, return with inversion count 0

2. Counting during merge: When selecting an element from the right array, add the number of “elements not yet merged from the left array” to the inversion count (len(left) - i)

3. Stable sort condition: By using left[i] <= right[j] with the equality sign, we prioritize the left side when values are equal, preventing unnecessary counts (in this problem, since we have a permutation, no duplicate values exist, but this is important for general implementations)

   Source Code

def merge_sort_count(arr):
    if len(arr) <= 1:
        return arr, 0
    
    mid = len(arr) // 2
    left, left_inv = merge_sort_count(arr[:mid])
    right, right_inv = merge_sort_count(arr[mid:])
    
    merged = []
    inv_count = left_inv + right_inv
    i = j = 0
    
    while i < len(left) and j < len(right):
        if left[i] <= right[j]:
            merged.append(left[i])
            i += 1
        else:
            merged.append(right[j])
            inv_count += len(left) - i
            j += 1
    
    merged.extend(left[i:])
    merged.extend(right[j:])
    
    return merged, inv_count

def solve():
    N = int(input())
    A = list(map(int, input().split()))
    
    _, inversions = merge_sort_count(A)
    print(inversions)

solve()

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