# Program to find number of magic sets from a permutation of first n natural numbers in Python

PythonServer Side ProgrammingProgramming

Suppose we have an array A with first n natural numbers, and one permutation P{p1, p2, ... pn} of array A. We have to check how many magic sets are there. A permutation is said to be magic set, if this satisfies these few rules −

• If we have k, then the elements in positions a, a, ... a[k] are less than their adjacent elements [P[a[i] - 1] > P[a[i]] < P[a[i] + 1]]
• If we have l, then the elements in positions b, b, ... b[l] are greater than their adjacent elements [P[b[i] - 1] < P[b[i]] > P[b[i] + 1]]

So, if the input is like n = 4 k = 1 l = 1 k_vals =  l_vals = , then the output will be 5 because: N = 4, a = 2 and b = 3. So the five permutations are [2,1,4,3], [3,2,4,1], [4,2,3,1], [3,1,4,2], [4,1,3,2].

To solve this, we will follow these steps −

• p := 10^9+7
• F := An array of size n+2 and fill with 0
• for each a in k_vals, do
• if F[a - 1] is 1 or F[a + 1] is 1, then
• if F[a - 1] is 1 or F[a + 1] is 1, then
• p := null
• F[a] := 1
• for each b in l_vals, do
• if F[b] is 1 or F[b - 1] is -1 or F[b + 1] is -1, then
• p := null
• F[b] := -1
• if p is same as null, then
• return 0
• otherwise,
• A := An array of size n+1 and fill with 0
• B := An array of size n+1 and fill with 0
• FF := An array of size n+1 and fill with null
• for i in range 1 to n, do
• FF[i] := F[i] - F[i - 1]
• A := 1
• for i in range 2 to n, do
• for j in range 1 to i, do
• if FF[i] > 0, then
• B[j] :=(B[j - 1] + A[j - 1]) mod p
• otherwise when FF[i] < 0, then
• B[j] :=(B[j - 1] + A[i - 1] - A[j - 1]) mod p
• otherwise,
• B[j] :=(B[j - 1] + A[i - 1]) mod p
• swap A and B
• return A[n]

## Example

Let us see the following implementation to get better understanding −

def solve(n, k, l, k_vals, l_vals):
p = 10**9+7
F =  * (n + 2)
for a in k_vals:
if F[a - 1] == 1 or F[a + 1] == 1:
p = None
F[a] = 1
for b in l_vals:
if F[b] == 1 or F[b - 1] == -1 or F[b + 1] == -1:
p = None
F[b] = -1
if p == None:
return 0
else:
A =  * (n + 1)
B =  * (n + 1)
FF = [None] * (n + 1)
for i in range(1, n + 1):
FF[i] = F[i] - F[i - 1]
A = 1
for i in range(2, n + 1):
for j in range(1, i + 1):
if FF[i] > 0:
B[j] = (B[j - 1] + A[j - 1]) % p
elif FF[i] < 0:
B[j] = (B[j - 1] + A[i - 1] - A[j - 1]) % p
else:
B[j] = (B[j - 1] + A[i - 1]) % p
A, B = B, A
return A[n]

n = 4
k = 1
l = 1
k_vals = 
l_vals = 
print(solve(n, k, l, k_vals, l_vals))

## Input

4, 1, 1, , 


## Input

5