Overview

This problem asks us to find the minimum possible “number of inversions (measure of disorder)” of the resulting sequence after performing the operation “swap two adjacent elements” exactly \(K\) times on a given permutation \(P\) of \(1\) to \(N\).

Observations

1. Relationship Between Adjacent Swaps and Inversions

The number of inversions is the number of pairs \((i, j)\) where \(i < j\) and \(P_i > P_j\) (i.e., pairs that are out of order). When we swap two adjacent elements \(P_i\) and \(P_{i+1}\), the relative order and size relationship of all pairs other than these two elements remain unchanged. Only the relationship between \(P_i\) and \(P_{i+1}\) changes.

• When \(P_i > P_{i+1}\): Swapping them results in \(P_i < P_{i+1}\), so the number of inversions decreases by exactly \(1\).
• When \(P_i < P_{i+1}\): Swapping them results in \(P_i > P_{i+1}\), so the number of inversions increases by exactly \(1\).

Therefore, regardless of which adjacent elements are swapped, the change in the number of inversions is always \(+1\) or \(-1\).

2. Comparing the Number of Operations \(K\) and the Initial Inversions \(I\)

Let \(I\) be the number of inversions in the initial state. How much we can reduce the number of inversions in \(K\) operations depends on the relationship between \(I\) and \(K\), which can be divided into the following two cases:

Case 1: \(I \ge K\)

As long as the number of inversions is at least \(1\), there always exists an adjacent pair such that \(P_i > P_{i+1}\). By swapping these, we can definitely decrease the number of inversions by \(1\). Therefore, it is possible to continuously decrease the number of inversions by \(1\) in each operation, reducing the number of inversions to \(I - K\) in exactly \(K\) operations. This is the minimum possible value.

Case 2: \(I < K\)

First, we can reduce the number of inversions to \(0\) (the state sorted in ascending order) in \(I\) operations. At this point, the number of remaining operations is \(K - I\).

If we perform \(1\) operation from the sorted state (with \(0\) inversions), the number of inversions will always become \(1\). If we perform another operation from there, we can reverse the previous swap to bring the inversions back to \(0\), or swap a different pair to make it \(2\). In this way, the parity (even/odd) of the number of inversions always flips with each operation.

Therefore, the minimum achievable value is determined by the parity of the remaining operations \(K - I\). - When \(K - I\) is even: By repeatedly swapping the same two elements back and forth, we can make the number of inversions \(0\). - When \(K - I\) is odd: Since the final operation must inevitably increase the number of inversions by \(1\), the minimum value will be \(1\).

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Algorithm

The steps to solve this problem are as follows:

1. Find the initial number of inversions \(I\)
   • Comparing all pairs naively takes \(O(N^2)\) time complexity, which will result in a Time Limit Exceeded (TLE) for \(N \le 2 \times 10^5\).
   • Therefore, we use a Fenwick Tree (Binary Indexed Tree, BIT) to calculate the number of inversions quickly in \(O(N \log N)\).
2. Determine the minimum value using conditional branching
   • If \(I \ge K\), the answer is \(I - K\).
   • If \(I < K\), the answer is \(0\) if \((K - I)\) is even, and \(1\) if it is odd.

Calculating Inversions Using a Fenwick Tree (BIT)

We iterate through the sequence from left to right, recording the numbers encountered so far in the Fenwick Tree. When processing the element \(P_i\) at index \(i\), the number of elements seen so far that are larger than \(P_i\) can be calculated as (current number of elements i) - (number of elements in the Fenwick Tree less than or equal to P_i). By summing this up for all elements, we can obtain the total number of inversions.

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Complexity

• Time Complexity: \(O(N \log N)\)
  • We perform element additions and queries on the Fenwick Tree \(N\) times. Since each operation can be executed in \(O(\log N)\), the overall time complexity is \(O(N \log N)\), which easily runs within the time limit.
• Space Complexity: \(O(N)\)
  • A Fenwick Tree of size \(N + 1\) and an array to store the input sequence are required, so the space complexity is \(O(N)\).

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Implementation Points

• The data type and size of \(K\): \(K\) can be as large as \(10^{18}\), which is a very large value. In Python, it is automatically handled as an arbitrary-precision integer, so there is no need to worry about overflow. However, when implementing in other languages like C++, you must use a 64-bit integer type such as long long.

• 1-based indexing: Since the Fenwick Tree inherently uses 1-based indexing, you can directly use the elements \(P_i\) (\(1 \le P_i \le N\)) of the sequence as indices.

  Source Code

import sys

def solve():
    input_data = sys.stdin.read().split()
    if not input_data:
        return
    N = int(input_data[0])
    K = int(input_data[1])
    P = [int(x) for x in input_data[2:]]
    
    tree = [0] * (N + 1)
    
    def add(i, x):
        while i <= N:
            tree[i] += x
            i += i & -i
            
    def query(i):
        s = 0
        while i > 0:
            s += tree[i]
            i -= i & -i
        return s

    inv = 0
    for i, x in enumerate(P):
        inv += i - query(x)
        add(x, 1)
        
    if inv >= K:
        print(inv - K)
    else:
        if (K - inv) % 2 == 0:
            print(0)
        else:
            print(1)

if __name__ == '__main__':
    solve()

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