5381_queries_on_permutation_key.py

'''
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]

1 <= m <= 10^3
1 <= queries.length <= m
1 <= queries[i] <= m
'''
class Solution:
    def processQueries(self, queries, m: int):
        arr = [i + 1 for i in range(m)]
        res = []
        for q in queries:
            for i in range(m):
                if arr[i] == q:
                    res.append(i)
                    break
            i = res[-1]
            while i > 0:
                arr[i] = arr[i-1]
                i -= 1
            arr[0] = q
            print(arr)
        return res

s = Solution()
# queries = [7,5,5,8,3]
# m = 8
queries = [3,1,2,1]
m = 5
res = s.processQueries(queries, m)
print(res)