"""

Given the array queries of positive integers between 1 and m, you have to process all queries[i] (from i=0 to i=queries.length-1) according to the following rules:

In the beginning, you have the permutation P=[1,2,3,...,m].
For the current i, find the position of queries[i] in the permutation P (indexing from 0) and then move this at the beginning of the permutation P. Notice that the position of queries[i] in P is the result for queries[i].
Return an array containing the result for the given queries.



Example 1:

Input: queries = [3,1,2,1], m = 5
Output: [2,1,2,1]
Explanation: The queries are processed as follow:
For i=0: queries[i]=3, P=[1,2,3,4,5], position of 3 in P is 2, then we move 3 to the beginning of P resulting in P=[3,1,2,4,5].
For i=1: queries[i]=1, P=[3,1,2,4,5], position of 1 in P is 1, then we move 1 to the beginning of P resulting in P=[1,3,2,4,5].
For i=2: queries[i]=2, P=[1,3,2,4,5], position of 2 in P is 2, then we move 2 to the beginning of P resulting in P=[2,1,3,4,5].
For i=3: queries[i]=1, P=[2,1,3,4,5], position of 1 in P is 1, then we move 1 to the beginning of P resulting in P=[1,2,3,4,5].
Therefore, the array containing the result is [2,1,2,1].
Example 2:

Input: queries = [4,1,2,2], m = 4
Output: [3,1,2,0]
Example 3:

Input: queries = [7,5,5,8,3], m = 8
Output: [6,5,0,7,5]


Constraints:

1 <= m <= 10^3
1 <= queries.length <= m
1 <= queries[i] <= m

"""

from typing import List


class Solution:

    def findQueryPosition(self, p: List[int], query: int) -> int:
        for i in range(len(p)):
            if p[i] == query:
                return i

    def processQueries(self, queries: List[int], m: int) -> List[int]:
        P = [i for i in range(1, m + 1)]
        out = []
        for i in range(len(queries)):
            position = self.findQueryPosition(P, queries[i]);
            out.append(position)
            print(position)
            P.insert(0, P.pop(position))
        return out

"""

Given the array queries of positive integers between 1 and m, you have to process all queries[i] (from i=0 to i=queries.length-1) according to the following rules:

In the beginning, you have the permutation P=[1,2,3,...,m].
For the current i, find the position of queries[i] in the permutation P (indexing from 0) and then move this at the beginning of the permutation P. Notice that the position of queries[i] in P is the result for queries[i].
Return an array containing the result for the given queries.



Example 1:

Input: queries = [3,1,2,1], m = 5
Output: [2,1,2,1]
Explanation: The queries are processed as follow:
For i=0: queries[i]=3, P=[1,2,3,4,5], position of 3 in P is 2, then we move 3 to the beginning of P resulting in P=[3,1,2,4,5].
For i=1: queries[i]=1, P=[3,1,2,4,5], position of 1 in P is 1, then we move 1 to the beginning of P resulting in P=[1,3,2,4,5].
For i=2: queries[i]=2, P=[1,3,2,4,5], position of 2 in P is 2, then we move 2 to the beginning of P resulting in P=[2,1,3,4,5].
For i=3: queries[i]=1, P=[2,1,3,4,5], position of 1 in P is 1, then we move 1 to the beginning of P resulting in P=[1,2,3,4,5].
Therefore, the array containing the result is [2,1,2,1].
Example 2:

Input: queries = [4,1,2,2], m = 4
Output: [3,1,2,0]
Example 3:

Input: queries = [7,5,5,8,3], m = 8
Output: [6,5,0,7,5]


Constraints:

1 <= m <= 10^3
1 <= queries.length <= m
1 <= queries[i] <= m

"""

from typing import List


class Solution:

    def findQueryPosition(self, p: List[int], query: int) -> int:
        for i in range(len(p)):
            if p[i] == query:
                return i

    def processQueries(self, queries: List[int], m: int) -> List[int]:
        P = [i for i in range(1, m + 1)]
        out = []
        for i in range(len(queries)):
            position = self.findQueryPosition(P, queries[i]);
            out.append(position)
            print(position)
            P.insert(0, P.pop(position))
        return out

Given the array queries of positive integers between 1 and m, you have to process all queries[i] (from i=0 to i=queries.length-1) according to the following rules:  In the beginning, you have the permutation P=[1,2,3,...,m]. For the current i, find the position of queries[i] in the permutation P (indexing from 0) and then move this at the beginning of the permutation P. Notice that the position of queries[i] in P is the result for queries[i]. Return an array containing the result for the given queries.    Example 1:  Input: queries = [3,1,2,1], m = 5 Output: [2,1,2,1] Explanation: The queries are processed as follow: For i=0: queries[i]=3, P=[1,2,3,4,5], position of 3 in P is 2, then we move 3 to the beginning of P resulting in P=[3,1,2,4,5]. For i=1: queries[i]=1, P=[3,1,2,4,5], position of 1 in P is 1, then we move 1 to the beginning of P resulting in P=[1,3,2,4,5]. For i=2: queries[i]=2, P=[1,3,2,4,5], position of 2 in P is 2, then we move 2 to the beginning of P resulting in P=[2,1,3,4,5]. For i=3: queries[i]=1, P=[2,1,3,4,5], position of 1 in P is 1, then we move 1 to the beginning of P resulting in P=[1,2,3,4,5]. Therefore, the array containing the result is [2,1,2,1]. Example 2:  Input: queries = [4,1,2,2], m = 4 Output: [3,1,2,0] Example 3:  Input: queries = [7,5,5,8,3], m = 8 Output: [6,5,0,7,5]   Constraints:  1 <= m <= 10^3 1 <= queries.length <= m 1 <= queries[i] <= m