1409. Queries on a Permutation With Key

#### QUESTION:

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
``````

#### SOLUTION:

``````class Solution {
func processQueries(_ queries: [Int], _ m: Int) -> [Int] {
var arr: [Int] = []
for i in 1...m {
arr.append(i)
}
var result:[Int] = []
for query in queries {
var index = arr.firstIndex(of: query)
let tmp = arr[index!]
arr.remove(at: index!)
arr.insert(tmp, at: 0)
result.append(index!)
}
return result
}
}
``````