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Suppose we have a set A where all elements from 1 to n are present. And P(A) represents all permutations of elements present in A. We have to find number of elements in P(A) which satisfies the given conditions

- For all i in range [1, n], A[i] is not same as i
- There exists a set of k indices {i1, i2, ... ik} such that A[ij] = ij+1 for all j < k and A[ik] = i1 (cyclic).

So, if the input is like n = 3 k = 2, then the output will be 0 because -

Consider the Array's are 1 indexed. As N = 3 and K = 2, we can find 2 sets of A that satisfy the first property a[i] ≠ i, they are [3,1,2] and [2,3,1]. Now as K = 2, we can have 6 such elements.

[1,2], [1,3],[2,3], [2,1], [3,1], [3,2]. Now if we consider the first element of

P(A) -> [3,1,2]

- [1,2], A[1] ≠ 2
- [1,3], A[1] = 3 but A[3] ≠ 1
- [2,3], A[2] ≠ 3
- [2,1], A[2] = 1 but A[1] ≠ 2
- [3,1], A[3] = 1 but A[1] ≠ 3
- [3,2], A[3] ≠ 2

P(A) -> [2,3,1]

- [1,2], A[1] = 2 but A[2] ≠ 1
- [1,3], A[1] ≠ 3
- [2,3], A[2] = 3 but A[3] ≠ 3
- [2,1], A[2] ≠ 1
- [3,1], A[3] = but A[1] ≠ 3
- [3,2], A[3] ≠ 2

As none of the elements of a satisfy the properties above, hence 0.

To solve this, we will follow these steps −

- ps := all permutations of arrays with elements in range [1, n]
- c := 0
- for each p in ps, do
- for each index i and value a in p, do
- if a is same as i, then
- come out from the loop

- if a is same as i, then
- otherwise,
- for j in range 0 to n - 1, do
- current := p[j]
- cycle_length := 1
- while current is not same as j, do
- current := p[current]
- cycle_length := cycle_length + 1

- if cycle_length is same as k, then
- c := c + 1
- come out from the loop

- for j in range 0 to n - 1, do

- for each index i and value a in p, do
- return c

Let us see the following implementation to get better understanding −

import itertools def solve(n, k): ps = itertools.permutations(range(n), n) c = 0 for p in ps: for i, a in enumerate(p): if a == i: break else: for j in range(n): current = p[j] cycle_length = 1 while current != j: current = p[current] cycle_length += 1 if cycle_length == k: c += 1 break return c n = 3 k = 2 print(solve(n, k))

3, 2

0

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